Number & Algebra

QUESTION 1 6 marks

Easy

The table shows the first four terms of three sequences: $u_n$, $v_n$, and $w_n$.

n	1	2	3	4
$u_n$	15	25	35	45
$v_n$	15	30	60	120
$w_n$	15	25	37	55

a. State which sequence is (i) arithmetic; (ii) geometric.

[2]

b. Find the sum of the first 20 terms of the arithmetic sequence.

[2]

c. Find the exact value of the 10th term of the geometric sequence.

[2]

Show complete worked solution

(a)

$u_n$ is arithmetic (constant difference of $10$). $v_n$ is geometric (constant ratio of $2$).

(b)

$$S_{20} = \frac{n}{2}(2u_1+(n-1)d) = 2200$$

(c)

$$v_{10} = u_1 r^{n-1} = 15 \times 2^{9} = 7680$$

QUESTION 2 6 marks

Medium

Consider the sum $$S = \sum_{k=3}^{l} (5k+1), \text{ where } l \text{ is a positive integer greater than } 3.$$

a. Write down the first three terms of the series.

[2]

b. Write down the number of terms in the series, in terms of $l$.

[1]

c. Given that $S = 1270$, find the value of $l$.

[3]

Show complete worked solution

(a)

$$16, 21, 26, \ldots$$

(b)

$$\text{number of terms} = l - 3 + 1$$

(c)

Using $S_n = \frac{n}{2}(u_1+u_n)$ with $u_1=16$, and solving for $l$ gives $$l = 22$$

QUESTION 3 6 marks

Easy

A 3D printer builds a set of 49 scale-model towers of decreasing size. The height of the largest tower is 80 cm. The height of each successive smaller tower is 95% of the preceding tower's height.

a. Find the height of the smallest tower in this set.

[3]

b. Find the total height if all 49 towers were placed one on top of another.

[3]

Show complete worked solution

(a)

$$h_{49} = 80 \times (0.95)^{48} = 6.821\text{ cm}$$

(b)

$$S_{49} = \frac{80(1-0.95^{49})}{1-0.95} = 1470\text{ cm}$$

QUESTION 4 4 marks

Medium

The first term of an arithmetic sequence is 8 and the common difference is 4. The first and fourth terms of this arithmetic sequence are the first two terms of a geometric sequence.

a. Find $u_{60}$, the 60th term of the arithmetic sequence.

[2]

b. Calculate the sixth term of the geometric sequence.

[2]

Show complete worked solution

(a)

$$u_{60} = u_1 + 59d = 8 + 59(4) = 244$$

(b)

The geometric sequence starts $8, 20, \ldots$, so $r = \frac{20}{8} = 2.5$. $$v_6 = v_1 r^5 = 8 \times (2.5)^5 = 781.2$$

QUESTION 5 6 marks

Medium

Consider the following sequence of figures made of connected line segments, where each new figure adds a fixed number of segments to the previous one. Figure 1 contains 7 line segments, and each subsequent figure adds 5 more segments than the previous.

a. Given that Figure $n$ contains 122 line segments, show that $n = 24$.

[3]

b. Find the total number of line segments in the first 20 figures.

[3]

Show complete worked solution

(a)

Segments in Figure $n$: $u_n = 7 + (n-1)(5)$. Setting $u_n = 122$: $$7+(n-1)(5)=122 \Rightarrow n = 24$$

(b)

$$S_{20} = \frac{20}{2}(2(7)+(20-1)(5)) = 1090$$

QUESTION 6 6 marks

Medium

Maya walks to school each morning. During the first minute, she travels 150 metres. In each subsequent minute, she travels 6 metres less than the distance travelled during the previous minute. The distance from home to school is 800 metres. Maya leaves home at 8:05 am and must be at school by 8:15 am. Will Maya arrive on time? Justify your answer.

Show complete worked solution

This is an arithmetic series: the distance travelled in minute $n$ is $u_n = 150 + (n-1)(-6)$, so the first term is $u_1 = 150$ and the common difference is $d = -6$. Maya has 10 minutes available (8:05 am to 8:15 am), so find the total distance travelled in 10 minutes using the sum of an arithmetic series: $$S_n = \frac{n}{2}\big(2u_1+(n-1)d\big)$$ Substituting $u_1 = 150$, $d = -6$, $n = 10$: $$S_{10} = \frac{10}{2}\big(2(150)+(10-1)(-6)\big)$$ $$S_{10} = 5\big(300-54\big) = 5(246)$$ $$S_{10} = 1230 \text{ m}$$ Comparing with the distance to school: $$1230 \text{ m} > 800 \text{ m}$$ Since the total distance Maya covers in the 10 minutes available exceeds the distance to school, she reaches school before using all 10 minutes. $$\boxed{\text{Yes, Maya will arrive on time (1230 m} > \text{800 m).}}$$

QUESTION 7 6 marks

Medium

The sum of the first three terms of a geometric sequence is 414.5, and the sum of the infinite sequence is 571.4. Find the common ratio.

Show complete worked solution

Let the first term be $u_1$ and the common ratio be $r$. Using the formula for the sum of the first three terms and the formula for the sum to infinity: $$u_1(1+r+r^2) = 414.5 \qquad \text{...(1)}$$ $$\frac{u_1}{1-r} = 571.4 \qquad \text{...(2)}$$ From (2): $u_1 = 571.4(1-r)$. Note that $1+r+r^2 = \dfrac{1-r^3}{1-r}$, so substituting into (1): $$571.4(1-r) \times \frac{1-r^3}{1-r} = 414.5$$ $$571.4(1-r^3) = 414.5$$ Dividing both sides by 571.4: $$1-r^3 = \frac{414.5}{571.4} = 0.725586\ldots$$ $$r^3 = 1-0.725586\ldots = 0.274414\ldots$$ $$r = \sqrt[3]{0.274414\ldots}$$ $$\boxed{r = 0.65}$$

QUESTION 8 6 marks

Easy

A geometric sequence has $u_1 = 20$, $u_2 = 40$, $u_3 = 80$.

a. Find the common ratio, $r$.

[2]

b. Find $u_{6}$.

[2]

c. Find $S_{7}$.

[2]

Show complete worked solution

(a)

$$r = \frac{u_2}{u_1} = \frac{40}{20} = 2$$

(b)

$$u_{6} = u_1 r^{n-1} = 20 \times \left(2\right)^{5} = 640$$

(c)

$$S_{7} = \frac{u_1(r^n-1)}{r-1} = 2540$$

QUESTION 9 6 marks

Easy

In an arithmetic sequence, $u_{3} = -8$, $u_{4} = -13$.

a. Find the common difference.

[2]

b. Find the first term.

[2]

c. Find the sum of the first 100 terms.

[2]

Show complete worked solution

(a)

$$d = \frac{u_{4} - u_{3}}{4-3} = \frac{-13 - (-8)}{1} = -5$$

(b)

$$u_1 = u_{3} - (3-1)d = -8 - (2)(-5) = 2$$

(c)

$$S_{100} = \frac{n}{2}(2u_1+(n-1)d) = -24550$$

QUESTION 10 4 marks

Medium

The first term of an arithmetic sequence is 10 and the common difference is 4. The first and fourth terms of this arithmetic sequence are the first two terms of a geometric sequence.

a. Find $u_{60}$, the 60th term of the arithmetic sequence.

[2]

b. Calculate the sixth term of the geometric sequence.

[2]

Show complete worked solution

(a)

$$u_{60} = u_1 + 59d = 10 + 59(4) = 246$$

(b)

The geometric sequence starts $10, 22, \ldots$, so $r = \frac{22}{10} = 2.2$. $$v_6 = v_1 r^5 = 10 \times (2.2)^5 = 515.4$$

QUESTION 11 6 marks

Medium

The sum of the first three terms of a geometric sequence is 277.5, and the sum of the infinite sequence is 480. Find the common ratio.

Show complete worked solution

Let the first term be $u_1$ and the common ratio be $r$. Using the formula for the sum of the first three terms and the formula for the sum to infinity: $$u_1(1+r+r^2) = 277.5 \qquad \text{...(1)}$$ $$\frac{u_1}{1-r} = 480 \qquad \text{...(2)}$$ From (2): $u_1 = 480(1-r)$. Using $1+r+r^2 = \dfrac{1-r^3}{1-r}$ and substituting into (1): $$480(1-r) \times \frac{1-r^3}{1-r} = 277.5$$ $$480(1-r^3) = 277.5$$ Dividing both sides by 480: $$1-r^3 = \frac{277.5}{480} = 0.578125$$ $$r^3 = 1-0.578125 = 0.421875$$ $$r = \sqrt[3]{0.421875}$$ $$\boxed{r = 0.75}$$

QUESTION 12 5 marks

Hard

The first two terms of an infinite geometric sequence, in order, are $$3\log_5 x, \quad 2\log_5 x, \quad \text{where } x > 0.$$

a. Find the common ratio, $r$.

[2]

b. Show that the sum of the infinite sequence is $9\log_{5} x$.

[3]

Show complete worked solution

(a)

$$r = \frac{2\log_5 x}{3\log_5 x} = \frac{2}{3}$$

(b)

$$S_\infty = \frac{u_1}{1-r} = \frac{3\log_5 x}{1-\frac{2}{3}} = \frac{3\log_5 x}{\frac{1}{3}} = 9\log_5 x$$

QUESTION 13 6 marks

Easy

A geometric sequence has $u_1 = 7$, $u_2 = 4.667$, $u_3 = 3.111$.

a. Find the common ratio, $r$.

[2]

b. Find $u_{6}$.

[2]

c. Find $S_\infty$.

[2]

Show complete worked solution

(a)

$$r = \frac{4.667}{7} = \frac{2}{3}$$

(b)

$$u_{6} = 7\times\left(\frac{2}{3}\right)^{5} = 0.9218$$

(c)

$$S_\infty = \frac{7}{1-(\frac{2}{3})} = 21$$

QUESTION 14 6 marks

Easy

Only one of the following four sequences is arithmetic and only one of them is geometric: $$d_n = 11, 23, 24, 27, \ldots,\quad c_n = 2, 1, 0.5, 0.25, \ldots,\quad b_n = 1, 4, 9, 16, \ldots,\quad a_n = 3, 5, 7, 9, \ldots$$

a. State which sequence is arithmetic and find the common difference.

[2]

b. State which sequence is geometric and find the common ratio.

[2]

c. For the geometric sequence, find the exact value of the 6th term. Give your answer as a fraction.

[2]

Show complete worked solution

(a)

Sequence $a_n$ is arithmetic, with common difference $d = 2$.

(b)

Sequence $c_n$ is geometric, with common ratio $r = \frac{1}{2}$.

(c)

$$c_{6} = 2 \times \left(\frac{1}{2}\right)^{5} = \frac{1}{16}$$

QUESTION 15 5 marks

Medium

An arithmetic sequence has first term 40 and common difference -1.

a. Given that the $k$th term of the sequence is zero, find the value of $k$.

[2]

b. Let $S_n$ denote the sum of the first $n$ terms. Find the largest value of $n$ such that $S_n > 100$.

[3]

Show complete worked solution

(a)

$$0 = u_1+(k-1)d \Rightarrow k = 1 - \frac{u_1}{d} = 41$$

(b)

$$S_n = \frac{n}{2}(2u_1+(n-1)d) = -0.5n^2 + 40.5n$$ Setting up the inequality $$-0.5n^2 + 40.5n > 100$$ and solving (using the quadratic formula, then checking nearby integers since $n$ must be a positive integer), the largest value of $n$ satisfying this is $$n = 78$$

QUESTION 16 4 marks

Medium

The first term of an arithmetic sequence is 10 and the common difference is 2. The first and fourth terms of this arithmetic sequence are the first two terms of a geometric sequence.

a. Find $u_{60}$, the 60th term of the arithmetic sequence.

[2]

b. Calculate the sixth term of the geometric sequence.

[2]

Show complete worked solution

(a)

$$u_{60} = u_1 + 59d = 10 + 59(2) = 128$$

(b)

The geometric sequence starts $10, 16, \ldots$, so $r = \frac{16}{10} = 1.6$. $$v_6 = v_1 r^5 = 10 \times (1.6)^5 = 104.9$$

QUESTION 17 6 marks

Easy

A geometric sequence has $u_1 = 12$, $u_2 = -6$, $u_3 = 3$.

a. Find the common ratio, $r$.

[2]

b. Find $u_{7}$.

[2]

c. Find the sum to infinity, $S_\infty$.

[2]

Show complete worked solution

(a)

$$r = \frac{u_2}{u_1} = \frac{-6}{12} = \frac{-1}{2}$$

(b)

$$u_{7} = u_1 r^{n-1} = 12 \times \left(\frac{-1}{2}\right)^{6} = 0.1875$$

(c)

$$S_\infty = \frac{u_1}{1-r} = \frac{12}{1-(\frac{-1}{2})} = 8$$

QUESTION 18 6 marks

Easy

In an arithmetic sequence, $u_{4} = -11$, $u_{5} = -4$.

a. Find the common difference.

[2]

b. Find the first term.

[2]

c. Find the sum of the first 25 terms.

[2]

Show complete worked solution

(a)

$$d = \frac{u_{5} - u_{4}}{5-4} = \frac{-4 - (-11)}{1} = 7$$

(b)

$$u_1 = u_{4} - (4-1)d = -11 - (3)(7) = -32$$

(c)

$$S_{25} = \frac{n}{2}(2u_1+(n-1)d) = 1300$$

QUESTION 19 6 marks

Easy

An arithmetic sequence has $u_1 = 24$, $u_2 = 17$, $u_3 = 10$.

a. Find the common difference, $d$.

[2]

b. Find $u_{9}$.

[2]

c. Find $S_{8}$.

[2]

Show complete worked solution

(a)

$$d = u_2 - u_1 = 17 - (24) = -7$$

(b)

$$u_{9} = u_1 + (n-1)d = 24 + (9-1)(-7) = -32$$

(c)

$$S_{8} = \frac{n}{2}(2u_1+(n-1)d) = \frac{8}{2}(2(24)+(8-1)(-7)) = -4$$

QUESTION 20 6 marks

Easy

In an arithmetic sequence, $u_{5} = -14$, $u_{11} = 4$.

a. Find the common difference.

[2]

b. Find the first term.

[2]

c. Find the sum of the first 15 terms.

[2]

Show complete worked solution

(a)

$$d = \frac{u_{11} - u_{5}}{11-5} = \frac{4 - (-14)}{6} = 3$$

(b)

$$u_1 = u_{5} - (5-1)d = -14 - (4)(3) = -26$$

(c)

$$S_{15} = \frac{n}{2}(2u_1+(n-1)d) = -75$$

QUESTION 21 6 marks

Easy

An arithmetic sequence has $u_1 = -5$, $u_2 = -14$, $u_3 = -23$.

a. Find the common difference, $d$.

[2]

b. Find $u_{11}$.

[2]

c. Find $S_{11}$.

[2]

Show complete worked solution

(a)

$$d = u_2-u_1 = -14-(-5) = -9$$

(b)

$$u_{11} = -5+(11-1)(-9) = -95$$

(c)

$$S_{11} = \frac{11}{2}(2(-5)+(11-1)(-9)) = -550$$

QUESTION 22 5 marks

Medium

An arithmetic sequence has first term 20 and common difference -1.5.

a. Given that the $k$th term of the sequence is zero, find the value of $k$.

[2]

b. Let $S_n$ denote the sum of the first $n$ terms. Find the largest value of $n$ such that $S_n > 100$.

[3]

Show complete worked solution

(a)

$$0 = u_1+(k-1)d \Rightarrow k = 1 - \frac{u_1}{d} = 14.33$$

(b)

$$S_n = \frac{n}{2}(2u_1+(n-1)d) = -0.75n^2 + 20.75n$$ Setting up the inequality $$-0.75n^2 + 20.75n > 100$$ and solving (using the quadratic formula, then checking nearby integers since $n$ must be a positive integer), the largest value of $n$ satisfying this is $$n = 21$$

QUESTION 23 8 marks

Hard

The sum, $S_n$, of the first $n$ terms of a geometric sequence, whose $n$th term is $u_n$, is given by $$S_n = \frac{5^n - a^n}{5^n}, \text{ where } a > 0.$$

a. Find an expression for $u_n$.

[3]

b. Find the first term and common ratio of the sequence.

[2]

c. (i) Determine the values of $a$ such that the sum to infinity exists. (ii) Find the sum to infinity when it exists.

[3]

Show complete worked solution

(a)

$$u_n = S_n - S_{n-1} = \left(1-\left(\frac{a}{5}\right)^n\right) - \left(1-\left(\frac{a}{5}\right)^{n-1}\right) = \left(\frac{a}{5}\right)^{n-1}\left(1-\frac{a}{5}\right)$$

(b)

First term: $u_1 = 1-\frac{a}{5}$. Common ratio: $r = \frac{a}{5}$.

(c)

(i) $S_\infty$ exists when $|r|<1$, i.e. $\left|\frac{a}{5}\right|<1$. Since $a>0$, this means $0

QUESTION 24 4 marks

Medium

The first term of an arithmetic sequence is 9 and the common difference is 5. The first and fourth terms of this arithmetic sequence are the first two terms of a geometric sequence.

a. Find $u_{60}$, the 60th term of the arithmetic sequence.

[2]

b. Calculate the sixth term of the geometric sequence.

[2]

Show complete worked solution

(a)

$$u_{60} = 9+59(5) = 304$$

(b)

$r = \frac{24}{9} = 2.667$. $$v_6 = 9\times(2.667)^5 = 1214$$

QUESTION 25 5 marks

Hard

The largest rectangle has dimensions 6 cm by 3 cm. A smaller rectangle is formed inside it with half the area, and this process is repeated indefinitely, with every other ring of the pattern shaded (similar to the pattern shown).

a. Find the total area of the first 3 shaded regions.

[3]

b. If the process is repeated indefinitely, find the total area of the shaded regions.

[2]

Show complete worked solution

(a)

Each shaded ring's area follows a geometric sequence with first term $\frac{18}{4} = 4.5$ cm$^2$ and ratio $\frac{1}{2}$. $$S_{3} = \frac{4.5(1-0.5^{3})}{0.5} = 7.875\text{ cm}^2$$

(b)

$$S_\infty = \frac{4.5}{0.5} = 9\text{ cm}^2$$

QUESTION 26 5 marks

Medium

A ball is dropped from a height of 1.5 m onto a concrete floor. After hitting the floor, the ball rebounds back up to 85% of its previous height, and this pattern continues indefinitely.

a. Show that the total distance travelled by the ball can be expressed as $$1.5 + 2(1.5)(0.85) + 2(1.5)(0.85)^2 + \cdots$$

[2]

b. Find the total distance travelled by the ball.

[3]

Show complete worked solution

(a)

After falling 1.5 m, the ball rebounds to 1.5×0.85 m and falls that same distance again (up and down), giving $2(1.5)(0.85)$ for that bounce, then $2(1.5)(0.85)^2$ for the next, and so on.

(b)

Total $$= 1.5 + \frac{2(1.5)(0.85)}{1-0.85} = 18.5\text{ m}$$

QUESTION 27 7 marks

Hard

The sides of a square are 10 cm long. A new square is formed by joining the midpoints of the sides of the original square, and two of the four resulting corner triangles are shaded. This process is repeated 5 more times (each time joining midpoints of the previous inner square), forming a spiral pattern of shaded triangles.

a. Show that the area shaded at each stage forms a geometric sequence with common ratio $\frac{1}{2}$, and find the first term.

[3]

b. Find the total area of the shaded region after all 6 stages (the original subdivision plus 5 repeats).

[2]

c. Find the total area of the shaded region if the process is repeated indefinitely.

[2]

Show complete worked solution

(a)

Joining the midpoints of a square with area $A$ creates an inner square of area $\frac{A}{2}$, leaving 4 equal corner triangles totalling area $\frac{A}{2}$ (so shading 2 of them shades $\frac{A}{4}$). Since the area being subdivided halves at every stage, the shaded amount at each stage also halves — a geometric sequence with ratio $\frac{1}{2}$. First term (from the original square, area $100$): $$u_1 = \frac{100}{4} = 25\text{ cm}^2$$

(b)

$$S_{6} = \frac{u_1(1-r^{6})}{1-r} = \frac{25(1-0.5^{6})}{0.5} = 49.22\text{ cm}^2$$

(c)

$$S_\infty = \frac{u_1}{1-r} = \frac{25}{0.5} = 50\text{ cm}^2$$

QUESTION 28 6 marks

Easy

An arithmetic sequence has $u_1 = -28$, $u_2 = -21$, $u_3 = -14$.

a. Find the common difference, $d$.

[2]

b. Find $u_{9}$.

[2]

c. Find $S_{8}$.

[2]

Show complete worked solution

(a)

$$d = u_2 - u_1 = -21 - (-28) = 7$$

(b)

$$u_{9} = u_1 + (n-1)d = -28 + (9-1)(7) = 28$$

(c)

$$S_{8} = \frac{n}{2}(2u_1+(n-1)d) = \frac{8}{2}(2(-28)+(8-1)(7)) = -28$$

QUESTION 29 6 marks

Easy

A geometric sequence has $u_1 = 20$, $u_2 = 5$, $u_3 = 1.25$.

a. Find the common ratio, $r$.

[2]

b. Find $u_{9}$.

[2]

c. Find the sum to infinity, $S_\infty$.

[2]

Show complete worked solution

(a)

$$r = \frac{u_2}{u_1} = \frac{5}{20} = \frac{1}{4}$$

(b)

$$u_{9} = u_1 r^{n-1} = 20 \times \left(\frac{1}{4}\right)^{8} = 0.0003052$$

(c)

$$S_\infty = \frac{u_1}{1-r} = \frac{20}{1-(\frac{1}{4})} = 26.67$$

QUESTION 30 6 marks

Easy

An arithmetic sequence has $u_1 = -40$, $u_2 = -32$, $u_3 = -24$.

a. Find the common difference, $d$.

[2]

b. Find $u_{12}$.

[2]

c. Find $S_{12}$.

[2]

Show complete worked solution

(a)

$$d = u_2 - u_1 = -32 - (-40) = 8$$

(b)

$$u_{12} = u_1 + (n-1)d = -40 + (12-1)(8) = 48$$

(c)

$$S_{12} = \frac{n}{2}(2u_1+(n-1)d) = \frac{12}{2}(2(-40)+(12-1)(8)) = 48$$

QUESTION 31 6 marks

Medium

The first term and the common ratio of a geometric series are denoted, respectively, by $u_1$ and $r$, where $u_1, r \in \mathbb{Q}$. Given that the fourth term is 40 and the sum to infinity is 481.7, find the value of $u_1$ and the value of $r$.

Show complete worked solution

Let the first term be $u_1$ and the common ratio be $r$, with $u_1, r \in \mathbb{Q}$. Using the formula for the $n$th term of a geometric sequence and the sum to infinity: $$u_1 r^3 = 40 \qquad \text{...(1)}$$ $$\frac{u_1}{1-r} = 481.7 \qquad \text{...(2)}$$ From (2): $u_1 = 481.7(1-r)$. Substituting into (1): $$481.7(1-r)\,r^3 = 40$$ Solving this numerically (e.g. using a GDC) for $r$ in $(0,1)$: $$r \approx 0.5848$$ Substituting back into (2): $$u_1 = 481.7(1-0.5848) \approx 481.7(0.4152) \approx 200$$ Check: with $u_1 = 200$, $u_1 r^3 = 200(0.5848)^3 \approx 200(0.20004) \approx 40.0$, which agrees with the given fourth term, and $\dfrac{u_1}{1-r} = \dfrac{200}{0.4152} \approx 481.7$, which agrees with the given sum to infinity. $$\boxed{u_1 = 200, \quad r \approx 0.5848}$$

QUESTION 32 6 marks

Easy

In an arithmetic sequence, $u_{9} = 10$, $u_{11} = 16$.

a. Find the common difference.

[2]

b. Find the first term.

[2]

c. Find the sum of the first 25 terms.

[2]

Show complete worked solution

(a)

$$d = \frac{u_{11} - u_{9}}{11-9} = \frac{16 - (10)}{2} = 3$$

(b)

$$u_1 = u_{9} - (9-1)d = 10 - (8)(3) = -14$$

(c)

$$S_{25} = \frac{n}{2}(2u_1+(n-1)d) = 550$$

QUESTION 33 6 marks

Easy

The table shows the first four terms of three sequences: $u_n$, $v_n$, and $w_n$.

n	1	2	3	4
$u_n$	24	36	48	60
$v_n$	24	48	96	192
$w_n$	24	36	50	66

a. State which sequence is (i) arithmetic; (ii) geometric.

[2]

b. Find the sum of the first 20 terms of the arithmetic sequence.

[2]

c. Find the exact value of the 12th term of the geometric sequence.

[2]

Show complete worked solution

(a)

$u_n$ is arithmetic (constant difference of $12$). $v_n$ is geometric (constant ratio of $2$).

(b)

$$S_{20} = \frac{n}{2}(2u_1+(n-1)d) = 2760$$

(c)

$$v_{12} = u_1 r^{n-1} = 24 \times 2^{11} = 49152$$

QUESTION 34 6 marks

Easy

An arithmetic sequence has $u_1 = -28$, $u_2 = -20$, $u_3 = -12$.

a. Find the common difference, $d$.

[2]

b. Find $u_{12}$.

[2]

c. Find $S_{12}$.

[2]

Show complete worked solution

(a)

$$d = u_2 - u_1 = -20 - (-28) = 8$$

(b)

$$u_{12} = u_1 + (n-1)d = -28 + (12-1)(8) = 60$$

(c)

$$S_{12} = \frac{n}{2}(2u_1+(n-1)d) = \frac{12}{2}(2(-28)+(12-1)(8)) = 192$$

QUESTION 35 6 marks

Easy

A geometric sequence has $u_1 = 2$, $u_2 = 4$, $u_3 = 8$.

a. Find the common ratio, $r$.

[2]

b. Find $u_{8}$.

[2]

c. Find $S_{7}$.

[2]

Show complete worked solution

(a)

$$r = \frac{u_2}{u_1} = \frac{4}{2} = 2$$

(b)

$$u_{8} = u_1 r^{n-1} = 2 \times \left(2\right)^{7} = 256$$

(c)

$$S_{7} = \frac{u_1(r^n-1)}{r-1} = 254$$

QUESTION 36 6 marks

Medium

Consider the sum $$S = \sum_{k=5}^{l} (3k-1), \text{ where } l \text{ is a positive integer greater than } 5.$$

a. Write down the first three terms of the series.

[2]

b. Write down the number of terms in the series, in terms of $l$.

[1]

c. Given that $S = 711$, find the value of $l$.

[3]

Show complete worked solution

(a)

$$14, 17, 20, \ldots$$

(b)

$$\text{number of terms} = l - 5 + 1$$

(c)

Using $S_n = \frac{n}{2}(u_1+u_n)$ with $u_1=14$, and solving for $l$ gives $$l = 22$$

QUESTION 37 6 marks

Medium

The first term and the common ratio of a geometric series are denoted, respectively, by $u_1$ and $r$, where $u_1, r \in \mathbb{Q}$. Given that the fourth term is 30 and the sum to infinity is 426.7, find the value of $u_1$ and the value of $r$.

Show complete worked solution

Let the first term be $u_1$ and the common ratio be $r$, with $u_1, r \in \mathbb{Q}$. Using the formula for the $n$th term of a geometric sequence and the sum to infinity: $$u_1 r^3 = 30 \qquad \text{...(1)}$$ $$\frac{u_1}{1-r} = 426.7 \qquad \text{...(2)}$$ From (2): $u_1 = 426.7(1-r)$. Substituting into (1): $$426.7(1-r)\,r^3 = 30$$ This equation cannot be solved algebraically in one step, so solve it numerically (e.g. using a GDC) for $r$ in $(0,1)$: $$r \approx 0.5313$$ Substituting back into (2): $$u_1 = 426.7(1-0.5313) \approx 426.7(0.4687) \approx 200$$ Check: with $u_1 = 200$, $u_1 r^3 = 200(0.5313)^3 \approx 200(0.14996) \approx 30.0$, which agrees with the given fourth term, and $\dfrac{u_1}{1-r} = \dfrac{200}{0.4687} \approx 426.7$, which agrees with the given sum to infinity. $$\boxed{u_1 = 200, \quad r \approx 0.5313}$$

QUESTION 38 6 marks

Easy

An arithmetic sequence has $u_1 = 40$, $u_2 = 47$, $u_3 = 54$.

a. Find the common difference, $d$.

[2]

b. Find $u_{8}$.

[2]

c. Find $S_{15}$.

[2]

Show complete worked solution

(a)

$$d = u_2 - u_1 = 47 - (40) = 7$$

(b)

$$u_{8} = u_1 + (n-1)d = 40 + (8-1)(7) = 89$$

(c)

$$S_{15} = \frac{n}{2}(2u_1+(n-1)d) = \frac{15}{2}(2(40)+(15-1)(7)) = 1335$$

QUESTION 39 6 marks

Easy

An arithmetic sequence has $u_1 = 40$, $u_2 = 44$, $u_3 = 48$.

a. Find the common difference, $d$.

[2]

b. Find $u_{8}$.

[2]

c. Find $S_{15}$.

[2]

Show complete worked solution

(a)

$$d = u_2 - u_1 = 44 - (40) = 4$$

(b)

$$u_{8} = u_1 + (n-1)d = 40 + (8-1)(4) = 68$$

(c)

$$S_{15} = \frac{n}{2}(2u_1+(n-1)d) = \frac{15}{2}(2(40)+(15-1)(4)) = 1020$$

QUESTION 40 6 marks

Medium

Maya walks to school each morning. During the first minute, she travels 160 metres. In each subsequent minute, she travels 6 metres less than the distance travelled during the previous minute. The distance from home to school is 1000 metres. Maya leaves home at 8:05 am and must be at school by 8:15 am. Will Maya arrive on time? Justify your answer.

Show complete worked solution

This is an arithmetic series: the distance travelled in minute $n$ is $u_n = 160 + (n-1)(-6)$, so the first term is $u_1 = 160$ and the common difference is $d = -6$. Maya has 10 minutes available (8:05 am to 8:15 am), so find the total distance travelled in 10 minutes using the sum of an arithmetic series: $$S_n = \frac{n}{2}\big(2u_1+(n-1)d\big)$$ Substituting $u_1 = 160$, $d = -6$, $n = 10$: $$S_{10} = \frac{10}{2}\big(2(160)+(10-1)(-6)\big)$$ $$S_{10} = 5\big(320-54\big) = 5(266)$$ $$S_{10} = 1330 \text{ m}$$ Comparing with the distance to school: $$1330 \text{ m} > 1000 \text{ m}$$ Since the total distance Maya covers in the 10 minutes available exceeds the distance to school, she reaches school before using all 10 minutes. $$\boxed{\text{Yes, Maya will arrive on time (1330 m} > \text{1000 m).}}$$

QUESTION 41 6 marks

Easy

In an arithmetic sequence, $u_{3} = -14$, $u_{9} = 4$.

a. Find the common difference.

[2]

b. Find the first term.

[2]

c. Find the sum of the first 15 terms.

[2]

Show complete worked solution

(a)

$$d = \frac{u_{9} - u_{3}}{9-3} = \frac{4 - (-14)}{6} = 3$$

(b)

$$u_1 = u_{3} - (3-1)d = -14 - (2)(3) = -20$$

(c)

$$S_{15} = \frac{n}{2}(2u_1+(n-1)d) = 15$$

QUESTION 42 6 marks

Easy

Only one of the following four sequences is arithmetic and only one of them is geometric: $$c_n = 4, 12, 36, 108, \ldots,\quad a_n = 3, 7, 11, 15, \ldots,\quad d_n = 2, 5, 10, 14, \ldots,\quad b_n = 1, 4, 9, 16, \ldots$$

a. State which sequence is arithmetic and find the common difference.

[2]

b. State which sequence is geometric and find the common ratio.

[2]

c. For the geometric sequence, find the exact value of the 8th term. Give your answer as a fraction.

[2]

Show complete worked solution

(a)

Sequence $a_n$ is arithmetic, with common difference $d = 4$.

(b)

Sequence $c_n$ is geometric, with common ratio $r = 3$.

(c)

$$c_{8} = 4 \times \left(3\right)^{7} = 8748$$

QUESTION 43 9 marks

Medium

Diego is planning to eat a tub of ice cream, one spoonful at a time. He believes the mass of ice cream remaining can be modelled by an arithmetic sequence. He puts the tub on a set of scales and records the following data.

Number of spoonfuls	1	2	3	4	5	6
Mass remaining (g)	307	285	266	241	219	197

a. Find an estimate of the common difference, by using an average of the differences.

[2]

b. Find a model for the $n$th term $u_n$ (mass remaining after $n$ spoonfuls).

[2]

c. Use your model to estimate the original mass of ice cream in the tub (before any spoonfuls), and give a reason to support whether this is an over- or under-estimate.

[3]

d. Use your model to estimate the number of spoonfuls required to eat all the ice cream in the tub.

[2]

Show complete worked solution

(a)

Average of consecutive differences $= -22.0$ g per spoonful.

(b)

$$u_n = 307 + (n-1)(-22.0)$$

(c)

Extrapolating one step back ($n=0$): $$u_0 = 307 - (-22.0) = 329\text{ g}$$ This is a reasonably accurate estimate, since the first actual difference (-22 g) matches the average difference used in the model almost exactly, so there's no strong evidence of bias in either direction.

(d)

Setting $u_n = 0$: $$0 = 307 + (n-1)(-22.0) \Rightarrow n \approx 14.95$$ so approximately 15 spoonfuls.

QUESTION 44 6 marks

Easy

A geometric sequence has $u_1 = 12$, $u_2 = 4$, $u_3 = 1.333$.

a. Find the common ratio, $r$.

[2]

b. Find $u_{9}$.

[2]

c. Find the sum to infinity, $S_\infty$.

[2]

Show complete worked solution

(a)

$$r = \frac{u_2}{u_1} = \frac{4}{12} = \frac{1}{3}$$

(b)

$$u_{9} = u_1 r^{n-1} = 12 \times \left(\frac{1}{3}\right)^{8} = 0.001829$$

(c)

$$S_\infty = \frac{u_1}{1-r} = \frac{12}{1-(\frac{1}{3})} = 18$$

QUESTION 45 5 marks

Medium

A ball is dropped from a height of 2 m onto a concrete floor. After hitting the floor, the ball rebounds back up to 75% of its previous height, and this pattern continues indefinitely.

a. Show that the total distance travelled by the ball can be expressed as $$2 + 2(2)(0.75) + 2(2)(0.75)^2 + \cdots$$

[2]

b. Find the total distance travelled by the ball.

[3]

Show complete worked solution

(a)

After falling 2 m, the ball rebounds to 2×0.75 m and falls that same distance again (up and down), giving $2(2)(0.75)$ for that bounce, then $2(2)(0.75)^2$ for the next, and so on.

(b)

Total $$= 2 + \frac{2(2)(0.75)}{1-0.75} = 14\text{ m}$$

QUESTION 46 6 marks

Easy

The table shows the first four terms of three sequences: $u_n$, $v_n$, and $w_n$.

n	1	2	3	4
$u_n$	12	22	32	42
$v_n$	12	24	48	96
$w_n$	12	22	34	52

a. State which sequence is (i) arithmetic; (ii) geometric.

[2]

b. Find the sum of the first 30 terms of the arithmetic sequence.

[2]

c. Find the exact value of the 12th term of the geometric sequence.

[2]

Show complete worked solution

(a)

$u_n$ is arithmetic (constant difference of $10$). $v_n$ is geometric (constant ratio of $2$).

(b)

$$S_{30} = \frac{n}{2}(2u_1+(n-1)d) = 4710$$

(c)

$$v_{12} = u_1 r^{n-1} = 12 \times 2^{11} = 24576$$

QUESTION 47 6 marks

Medium

Consider the following sequence of figures made of connected line segments, where each new figure adds a fixed number of segments to the previous one. Figure 1 contains 6 line segments, and each subsequent figure adds 5 more segments than the previous.

a. Given that Figure $n$ contains 81 line segments, show that $n = 16$.

[3]

b. Find the total number of line segments in the first 20 figures.

[3]

Show complete worked solution

(a)

Segments in Figure $n$: $u_n = 6 + (n-1)(5)$. Setting $u_n = 81$: $$6+(n-1)(5)=81 \Rightarrow n = 16$$

(b)

$$S_{20} = \frac{20}{2}(2(6)+(20-1)(5)) = 1070$$

QUESTION 48 5 marks

Medium

A rope of length 75 metres is cut into $n$ pieces of increasing lengths that form an arithmetic sequence with a common difference of $d$ metres. Given that the lengths of the shortest and longest pieces are 2 metres and 8 metres respectively, find the values of $n$ and $d$.

Show complete worked solution

Let $u_1 = 2$ m be the shortest piece and $u_n = 8$ m be the longest ($n$th) piece, with common difference $d$ and total length $S_n = 75$ m. Using the formula for the sum of an arithmetic series in terms of the first and last terms: $$S_n = \frac{n}{2}(u_1+u_n)$$ Substituting $S_n = 75$, $u_1 = 2$, $u_n = 8$: $$75 = \frac{n}{2}(2+8)$$ $$75 = \frac{n}{2}(10) = 5n$$ $$n = \frac{75}{5}$$ $$\boxed{n = 15}$$ Using the formula for the $n$th term of an arithmetic sequence: $$u_n = u_1+(n-1)d$$ Substituting $u_n = 8$, $u_1 = 2$, $n = 15$: $$8 = 2+(15-1)d$$ $$8 = 2+14d$$ $$14d = 6$$ $$d = \frac{6}{14} = \frac{3}{7}$$ $$\boxed{d = \frac{3}{7} \approx 0.4286\text{ m}}$$

QUESTION 49 6 marks

Easy

Only one of the following four sequences is arithmetic and only one of them is geometric: $$c_n = 4, 2, 1, 0.5, \ldots,\quad b_n = 1, 4, 9, 16, \ldots,\quad d_n = 20, 25, 26, 28, \ldots,\quad a_n = 2, 7, 12, 17, \ldots$$

a. State which sequence is arithmetic and find the common difference.

[2]

b. State which sequence is geometric and find the common ratio.

[2]

c. For the geometric sequence, find the exact value of the 8th term. Give your answer as a fraction.

[2]

Show complete worked solution

(a)

Sequence $a_n$ is arithmetic, with common difference $d = 5$.

(b)

Sequence $c_n$ is geometric, with common ratio $r = \frac{1}{2}$.

(c)

$$c_{8} = 4 \times \left(\frac{1}{2}\right)^{7} = \frac{1}{32}$$

QUESTION 50 5 marks

Medium

An arithmetic sequence has first term 45 and common difference -2.5.

a. Given that the $k$th term of the sequence is zero, find the value of $k$.

[2]

b. Let $S_n$ denote the sum of the first $n$ terms. Find the smallest value of $n$ such that $S_n < 0$.

[3]

Show complete worked solution

(a)

$$0 = u_1+(k-1)d \Rightarrow k = 1 - \frac{u_1}{d} = 19$$

(b)

$$S_n = \frac{n}{2}(2u_1+(n-1)d) = -1.25n^2 + 46.25n$$ Setting up the inequality $$-1.25n^2 + 46.25n < 0$$ and solving (using the quadratic formula, then checking nearby integers since $n$ must be a positive integer), the smallest value of $n$ satisfying this is $$n = 38$$

QUESTION 51 6 marks

Easy

A geometric sequence has $u_1 = 9$, $u_2 = 2.25$, $u_3 = 0.5625$.

a. Find the common ratio, $r$.

[2]

b. Find $u_{7}$.

[2]

c. Find $S_\infty$.

[2]

Show complete worked solution

(a)

$$r = \frac{2.25}{9} = \frac{1}{4}$$

(b)

$$u_{7} = 9\times\left(\frac{1}{4}\right)^{6} = 0.002197$$

(c)

$$S_\infty = \frac{9}{1-(\frac{1}{4})} = 12$$

QUESTION 52 6 marks

Easy

An arithmetic sequence has $u_1 = 55$, $u_2 = 53$, $u_3 = 51$.

a. Find the common difference, $d$.

[2]

b. Find $u_{9}$.

[2]

c. Find $S_{14}$.

[2]

Show complete worked solution

(a)

$$d = u_2-u_1 = 53-(55) = -2$$

(b)

$$u_{9} = 55+(9-1)(-2) = 39$$

(c)

$$S_{14} = \frac{14}{2}(2(55)+(14-1)(-2)) = 588$$

QUESTION 53 6 marks

Easy

A geometric sequence has $u_1 = 12$, $u_2 = 18$, $u_3 = 27$.

a. Find the common ratio, $r$.

[2]

b. Find $u_{9}$.

[2]

c. Find $S_{7}$.

[2]

Show complete worked solution

(a)

$$r = \frac{u_2}{u_1} = \frac{18}{12} = \frac{3}{2}$$

(b)

$$u_{9} = u_1 r^{n-1} = 12 \times \left(\frac{3}{2}\right)^{8} = 307.5$$

(c)

$$S_{7} = \frac{u_1(r^n-1)}{r-1} = 386.1$$

QUESTION 54 6 marks

Medium

A ball is dropped from a height of 20 m. Every time it hits the ground it bounces back to 81\% of its previous height.

a. Find the maximum height it reaches after the 10th bounce.

[3]

b. Find the total distance travelled by the ball until it comes to rest. (Assume no friction and no loss of elasticity.)

[3]

Show complete worked solution

(a)

$$h_{10} = 20 \times (0.81)^{10} = 2.432\text{ m}$$

(b)

$$\text{Total} = 20 + \frac{2(20)(0.81)}{1-0.81} = 190.5\text{ m}$$

QUESTION 55 6 marks

Medium

Leo cycles to the gym each morning. During the first minute, he travels 190 metres. In each subsequent minute, he travels 85% of the distance travelled during the previous minute. The distance from home to the gym is 1700 metres. Leo leaves home at 6:40 am and must arrive by 6:52 am. Will Leo arrive on time? Justify your answer.

Show complete worked solution

This is a geometric series: the distance travelled in minute $n$ is $u_n = 190(0.85)^{n-1}$, so the first term is $u_1 = 190$ and the common ratio is $r = 0.85$. Leo has 12 minutes available (6:40 am to 6:52 am), so find the total distance travelled in 12 minutes using the sum of a finite geometric series: $$S_n = \frac{u_1(1-r^n)}{1-r}$$ Substituting $u_1 = 190$, $r = 0.85$, $n = 12$: $$S_{12} = \frac{190(1-0.85^{12})}{1-0.85}$$ Evaluating $0.85^{12} \approx 0.142238$: $$S_{12} = \frac{190(1-0.142238)}{0.15} = \frac{190(0.857762)}{0.15}$$ $$S_{12} \approx \frac{162.975}{0.15} \approx 1086.49\text{ m}$$ Comparing with the distance to the gym: $$1086.49 \text{ m} < 1700 \text{ m}$$ Since the total distance Leo can cover in the 12 minutes available is less than the distance to the gym, he does not travel far enough in time. $$\boxed{\text{No, Leo will NOT arrive on time (only about 1086 m of the 1700 m is covered).}}$$

QUESTION 56 6 marks

Medium

Consider the sum $$S = \sum_{k=4}^{l} (2k-1), \text{ where } l \text{ is a positive integer greater than } 4.$$

a. Write down the first three terms of the series.

[2]

b. Write down the number of terms in the series, in terms of $l$.

[1]

c. Given that $S = 315$, find the value of $l$.

[3]

Show complete worked solution

(a)

$$7, 9, 11, \ldots$$

(b)

$$\text{number of terms} = l - 4 + 1$$

(c)

Using $S_n = \frac{n}{2}(u_1+u_n)$ with $u_1=7$, and solving for $l$ gives $$l = 18$$

QUESTION 57 9 marks

Medium

An experiment was undertaken to investigate the relationship between the length of a spring and the mass hanging from it. The table shows the extension of the spring (cm) for each mass (g).

Mass (g)	12	24	36	48	60	72	84	96
Extension (cm)	3.5	6.91	10.39	13.87	17.45	20.83	24.44	27.92

It is believed the data can be modelled by an arithmetic sequence, according to Hooke's Law.

a. Find an estimate of the common difference, by using an average of the differences.

[3]

b. Find a model for the $n$th term $u_n$, where $n = \dfrac{\text{mass}}{12}$.

[3]

c. (i) Use your model to estimate the extension for a mass of 72 g. (ii) Calculate the percentage error in this estimate.

[3]

Show complete worked solution

(a)

Consecutive differences: $3.41, 3.48, 3.48, 3.58, 3.38, 3.61, 3.48$. Average $= 3.489$ cm.

(b)

$$u_n = 3.5 + (n-1)(3.489)$$

(c)

(i) $n = 72/12 = 6$. $$u_{6} = 3.5 + (6-1)(3.489) = 20.945\text{ cm}$$ (ii) Actual measured value at 72 g is 20.83 cm. Percentage error $= \left|\dfrac{20.945-20.83}{20.83}\right| \times 100 = 0.552\%$

QUESTION 58 3 marks

Easy

An arithmetic sequence has first term $u_1 = 7$ and common difference $d = 4$. Find the value of $u_{10}$.

Show complete worked solution

Using the formula for the general term of an arithmetic sequence, $u_n = u_1 + (n-1)d$. $u_{10} = 7 + (10-1)(4)$ $u_{10} = 7 + (9)(4)$ $u_{10} = 7 + 36$ Final answer: $\boxed{u_{10} = 43}$

QUESTION 59 3 marks

Easy

The first term of an arithmetic sequence is $5$ and the common difference is $3$. Find the sum of the first $20$ terms of the sequence.

Show complete worked solution

Using the formula for the sum of an arithmetic series, $S_n = \dfrac{n}{2}\left[2u_1 + (n-1)d\right]$. $S_{20} = \dfrac{20}{2}\left[2(5) + (20-1)(3)\right]$ $S_{20} = 10\left[10 + 57\right]$ $S_{20} = 10 \times 67$ Final answer: $\boxed{S_{20} = 670}$

QUESTION 60 4 marks

Medium

In an arithmetic sequence, $u_3 = 11$ and $u_7 = 27$. (a) Find the common difference $d$ and the first term $u_1$. (b) Find $u_{15}$.

Show complete worked solution

Part (a): Using the general term formula, $u_n = u_1 + (n-1)d$, form two equations: $u_3 = u_1 + 2d = 11$ $u_7 = u_1 + 6d = 27$ Subtracting the first equation from the second (elimination): $(u_1+6d) - (u_1+2d) = 27 - 11$ $4d = 16$ $d = 4$ Substitute $d=4$ into $u_1 + 2d = 11$: $u_1 + 8 = 11$ $u_1 = 3$ Final answer: $\boxed{d = 4,\ u_1 = 3}$ Part (b): Using $u_n = u_1 + (n-1)d$: $u_{15} = 3 + (14)(4) = 3 + 56$ Final answer: $\boxed{u_{15} = 59}$

QUESTION 61 5 marks

Medium

The seats in a theatre are arranged in rows that form an arithmetic sequence. The first row has $18$ seats, and each subsequent row has $3$ more seats than the row before it. There are $25$ rows in total. Find the total number of seats in the theatre.

Show complete worked solution

Identify the arithmetic sequence: $u_1 = 18$, $d = 3$, $n = 25$. Using the formula for the sum of an arithmetic series, $S_n = \dfrac{n}{2}\left[2u_1 + (n-1)d\right]$. $S_{25} = \dfrac{25}{2}\left[2(18) + (24)(3)\right]$ $S_{25} = 12.5\left[36 + 72\right]$ $S_{25} = 12.5 \times 108$ Final answer: $\boxed{1350 \text{ seats}}$

QUESTION 62 6 marks

Hard

An arithmetic sequence has first term $u_1 = -5$ and common difference $d = 6$. Find the least value of $n$ for which the sum of the first $n$ terms, $S_n$, exceeds $500$.

Show complete worked solution

Using the formula for the sum of an arithmetic series, $S_n = \dfrac{n}{2}\left[2u_1 + (n-1)d\right]$. $S_n = \dfrac{n}{2}\left[2(-5) + (n-1)(6)\right] = \dfrac{n}{2}\left[-10 + 6n - 6\right] = \dfrac{n}{2}(6n-16)$ $S_n = 3n^2 - 8n$ Set up the inequality $S_n > 500$: $3n^2 - 8n > 500$ $3n^2 - 8n - 500 > 0$ Using the quadratic formula to solve $3n^2-8n-500=0$: $n = \dfrac{8 \pm \sqrt{(-8)^2 - 4(3)(-500)}}{2(3)} = \dfrac{8 \pm \sqrt{64+6000}}{6} = \dfrac{8 \pm \sqrt{6064}}{6}$ $\sqrt{6064} \approx 77.87$ $n \approx \dfrac{8+77.87}{6} \approx 14.31$ (taking the positive root, since $n>0$) Since the coefficient of $n^2$ is positive, $S_n>500$ for $n > 14.31$, so test integer values near this boundary: Check $n=14$: $S_{14} = 3(14)^2 - 8(14) = 588 - 112 = 476$ (not greater than 500) Check $n=15$: $S_{15} = 3(15)^2 - 8(15) = 675 - 120 = 555$ (greater than 500) Final answer: $\boxed{n = 15}$

QUESTION 63 3 marks

Easy

A geometric sequence has first term $u_1 = 3$ and common ratio $r = 2$. Find the value of $u_6$.

Show complete worked solution

Using the formula for the general term of a geometric sequence, $u_n = u_1 r^{\,n-1}$. $u_6 = 3 \times 2^{5}$ $2^5 = 32$ $u_6 = 3 \times 32$ Final answer: $\boxed{u_6 = 96}$

QUESTION 64 3 marks

Easy

A geometric sequence has first term $5$ and common ratio $3$. Find the sum of the first $6$ terms of the sequence.

Show complete worked solution

Using the formula for the sum of a geometric series, $S_n = \dfrac{u_1(r^n - 1)}{r-1}$ (valid since $r e 1$). $S_6 = \dfrac{5(3^6 - 1)}{3-1}$ $3^6 = 729$ $S_6 = \dfrac{5(729-1)}{2} = \dfrac{5(728)}{2} = \dfrac{3640}{2}$ Final answer: $\boxed{S_6 = 1820}$

QUESTION 65 4 marks

Medium

In a geometric sequence, $u_2 = 6$ and $u_5 = 48$. (a) Find the common ratio $r$ and the first term $u_1$. (b) Find $u_8$.

Show complete worked solution

Part (a): Using the general term formula, $u_n = u_1 r^{\,n-1}$, note that $u_5 = u_2 \times r^{3}$ (since the index increases by 3). $r^3 = \dfrac{u_5}{u_2} = \dfrac{48}{6} = 8$ $r = \sqrt[3]{8} = 2$ Since $u_2 = u_1 r$: $u_1 = \dfrac{u_2}{r} = \dfrac{6}{2} = 3$ Final answer: $\boxed{r = 2,\ u_1 = 3}$ Part (b): Using $u_n = u_1 r^{\,n-1}$: $u_8 = 3 \times 2^{7} = 3 \times 128$ Final answer: $\boxed{u_8 = 384}$

QUESTION 66 5 marks

Medium

A geometric series has first term $u_1 = 12$ and common ratio $r = 0.6$. (a) Find the sum to infinity of the series. (b) Find the sum of the first 15 terms, $S_{15}$, giving your answer correct to 3 significant figures.

Show complete worked solution

Part (a): Since $|r| = 0.6 < 1$, the sum to infinity exists. Using the formula $S_\infty = \dfrac{u_1}{1-r}$: $S_\infty = \dfrac{12}{1-0.6} = \dfrac{12}{0.4}$ Final answer: $\boxed{S_\infty = 30}$ Part (b): Using the formula for the sum of a finite geometric series, $S_n = \dfrac{u_1(1-r^n)}{1-r}$. $S_{15} = \dfrac{12(1-0.6^{15})}{1-0.6}$ Using a GDC: $0.6^{15} \approx 0.00047018$ $S_{15} = \dfrac{12(1-0.00047018)}{0.4} = \dfrac{12(0.99952982)}{0.4} = \dfrac{11.994358}{0.4}$ $S_{15} \approx 29.9859$ Rounding to 3 significant figures: Final answer: $\boxed{S_{15} \approx 30.0}$ (Note: this is very close to $S_\infty = 30$, illustrating convergence of the series.)

QUESTION 67 7 marks

Hard

A geometric series has first term $u_1 = 10$ and sum to infinity $S_\infty = 40$. (a) Find the common ratio $r$. (b) Find the least value of $n$ for which $S_n$ exceeds $39.5$.

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Part (a): Using the formula for the sum to infinity, $S_\infty = \dfrac{u_1}{1-r}$: $40 = \dfrac{10}{1-r}$ $1-r = \dfrac{10}{40} = 0.25$ Final answer: $\boxed{r = 0.75}$ Part (b): Using the formula for the sum of a finite geometric series, $S_n = \dfrac{u_1(1-r^n)}{1-r} = \dfrac{10(1-0.75^n)}{0.25} = 40(1-0.75^n)$ Set up the inequality $S_n > 39.5$: $40(1-0.75^n) > 39.5$ $1 - 0.75^n > 0.9875$ $0.75^n < 0.0125$ Taking logarithms of both sides (note the inequality reverses because $\ln(0.75) < 0$): $n \ln(0.75) < \ln(0.0125)$ $n > \dfrac{\ln(0.0125)}{\ln(0.75)}$ Using a GDC: $\ln(0.0125) \approx -4.38203$, $\ln(0.75) \approx -0.28768$ $n > \dfrac{-4.38203}{-0.28768} \approx 15.233$ Since $n$ must be an integer, test values near this boundary: Check $n=15$: $0.75^{15} \approx 0.013363$, so $S_{15} = 40(1-0.013363) \approx 39.4655$ (not greater than 39.5) Check $n=16$: $0.75^{16} \approx 0.010022$, so $S_{16} = 40(1-0.010022) \approx 39.5991$ (greater than 39.5) Final answer: $\boxed{n = 16}$

QUESTION 68 3 marks

Easy

Find the sum to infinity of the geometric series $8 + 4 + 2 + 1 + \ldots$

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Method: using the formula for the sum to infinity of a convergent geometric series, $S_\infty = \dfrac{a}{1-r}$, valid when $|r| < 1$. First term $a = 8$. Common ratio $r = \dfrac{4}{8} = \dfrac{1}{2}$. Since $|r| = 0.5 < 1$, the series converges. $S_\infty = \dfrac{8}{1-\frac{1}{2}} = \dfrac{8}{\frac{1}{2}} = 16$. Final answer: $$\boxed{S_\infty = 16}$$

QUESTION 69 3 marks

Easy

A geometric sequence has first term $u_1 = 5$ and common ratio $r = -\dfrac{1}{3}$. Find the sum to infinity of the sequence.

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Method: using the formula for the sum to infinity, $S_\infty = \dfrac{a}{1-r}$, valid when $|r| < 1$. $a = 5$, $r = -\dfrac{1}{3}$. Since $|r| = \dfrac{1}{3} < 1$, the series converges. $S_\infty = \dfrac{5}{1-(-\frac{1}{3})} = \dfrac{5}{\frac{4}{3}} = 5 \times \dfrac{3}{4} = \dfrac{15}{4}$. Final answer: $$\boxed{S_\infty = \dfrac{15}{4} = 3.75}$$

QUESTION 70 4 marks

Medium

The sum to infinity of a geometric sequence is 20, and its first term is 8. Find the common ratio, $r$, and hence find the value of the second term, $u_2$.

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Method: substituting the given values into $S_\infty = \dfrac{a}{1-r}$ and solving for $r$, then using $u_2 = u_1 \times r$. $20 = \dfrac{8}{1-r}$ $1-r = \dfrac{8}{20} = 0.4$ $r = 1 - 0.4 = 0.6$ Check: $|r| = 0.6 < 1$, so this is a valid common ratio for a convergent series. $u_2 = u_1 \times r = 8 \times 0.6 = 4.8$ Final answer: $$\boxed{r = 0.6, \ u_2 = 4.8}$$

QUESTION 71 5 marks

Medium

In a geometric sequence, the second term is $u_2 = 12$ and the fifth term is $u_5 = 1.5$. Find the sum to infinity of the sequence.

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Method: using $u_n = u_1 r^{n-1}$ to form two equations, dividing them to find $r$, then finding $u_1$ and applying the sum to infinity formula. $u_2 = u_1 r = 12$ ... (1) $u_5 = u_1 r^4 = 1.5$ ... (2) Divide (2) by (1): $\dfrac{u_1 r^4}{u_1 r} = \dfrac{1.5}{12}$ $r^3 = 0.125$ $r = \sqrt[3]{0.125} = 0.5$ From (1): $u_1 = \dfrac{12}{r} = \dfrac{12}{0.5} = 24$ Check: $|r| = 0.5 < 1$, so the series converges. $S_\infty = \dfrac{u_1}{1-r} = \dfrac{24}{1-0.5} = \dfrac{24}{0.5} = 48$ Final answer: $$\boxed{S_\infty = 48}$$

QUESTION 72 6 marks

Hard

Consider the infinite geometric series $1 + (x-1) + (x-1)^2 + (x-1)^3 + \ldots$ (a) Find the set of values of $x$ for which the series converges. (b) Given that the sum to infinity of the series is 5, find the value of $x$.

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(a) Method: identifying the common ratio and applying the convergence condition $|r| < 1$. Here $a = 1$, and the common ratio is $r = x - 1$. The series converges when $|r| < 1$: $|x-1| < 1$ $-1 < x - 1 < 1$ $$\boxed{0 < x < 2}$$ (b) Method: substituting into $S_\infty = \dfrac{a}{1-r}$ and solving the resulting equation for $x$. $S_\infty = \dfrac{1}{1-(x-1)} = \dfrac{1}{2-x} = 5$ $2 - x = \dfrac{1}{5}$ $x = 2 - \dfrac{1}{5} = \dfrac{9}{5} = 1.8$ Check: $0 < 1.8 < 2$, so this value lies within the set found in part (a), confirming it is valid. (3 marks) Final answer: $$\boxed{x = 1.8 = \dfrac{9}{5}}$$

QUESTION 1 5 marks

Easy

Ananya invests \$3000 in an account that pays 5.8\% compound interest per year. She does not invest any further amount or withdraw any money from the account.

a. Find the amount in her account after 8 years.

[2]

b. Find the number of years until the amount in her account exceeds \$5500.

[3]

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(a)

$$A = 3000(1+5.8/100)^{8} = 4,709.84$$

(b)

$$3000(1+5.8/100)^n > 5500 \Rightarrow n > \dfrac{\ln(5500/3000)}{\ln(1+5.8/100)} = 10.75$$ So after 11 years.

QUESTION 2 3 marks

Medium

An investment of \$650 is made at the beginning of every year for 10 years. Interest accumulates at a rate of 3.2\% per year, compounded annually. Find the value of the investment at the end of the 10 years.

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This is an annuity-due (a deposit is made at the beginning of each year, so each of the 10 deposits earns interest for one extra year compared with an ordinary annuity). Using the future value of an annuity-due formula: $$FV = R \times \frac{(1+i)^n-1}{i} \times (1+i)$$ Substituting the payment $R = 650$, the annual rate $i = 0.032$, and $n = 10$ years: $$FV = 650 \times \frac{(1+0.032)^{10}-1}{0.032} \times (1+0.032)$$ Evaluating $(1.032)^{10} \approx 1.36973$: $$FV = 650 \times \frac{1.36973-1}{0.032} \times 1.032 = 650 \times \frac{0.36973}{0.032} \times 1.032$$ $$FV = 650 \times 11.5540 \times 1.032$$ $$FV \approx 7513.09 \times 1.032$$ $$\boxed{FV \approx \$7761.18}$$

QUESTION 3 6 marks

Hard

A family takes out a loan of \$180000 to purchase a house. They agree to pay the bank \$1450 at the end of every month to amortise the loan, and interest accumulates on the balance at a rate of 1.1\% per year, compounded monthly.

a. Find how long it takes to pay back the loan, in years and months.

[4]

b. Calculate how much they have paid in total in amortising the loan.

[2]

Show complete worked solution

(a)

$$n = \dfrac{-\ln\left(1-\dfrac{Pi}{PMT}\right)}{\ln(1+i)} = 131.8 \text{ months} \approx 11\text{ years, } 0\text{ months}$$

(b)

$$\text{Total} = 1450\times 132 = 191400$$

QUESTION 4 4 marks

Easy

At her son's birth, Meera put \$800 into a savings account. The interest earned is 3.6\% per year, compounded quarterly. How much money will he have on his 18th birthday?

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Using the compound interest formula for interest compounded $k$ times per year: $$A = P\left(1+\frac{r}{100k}\right)^{kn}$$ Substituting the principal $P = 800$, annual rate $r = 3.6$, $k = 4$ (quarterly), and $n = 18$ years: $$A = 800\left(1+\frac{3.6}{100\times4}\right)^{4\times18}$$ $$A = 800\left(1+0.009\right)^{72} = 800(1.009)^{72}$$ Evaluating $(1.009)^{72} \approx 1.90618$: $$A = 800 \times 1.90618$$ $$\boxed{A \approx \$1524.94}$$

QUESTION 5 6 marks

Hard

A bank offers loans of \$P at the beginning of a particular month at a monthly interest rate of $I\%$. The interest is calculated at the end of each month and added to the amount outstanding. A repayment of \$R is required at the end of each month. Let $S_n$ denote the amount outstanding, in dollars, immediately after the $n$th monthly repayment.

a. Find an expression for $S_1$.

[2]

b. Find a similar expression for $S_2$, and hence show the general pattern for $S_n$.

[4]

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(a)

After one month, interest adds $\dfrac{I}{100}P$, then a repayment of $R$ is made: $$S_1 = P\left(1+\dfrac{I}{100}\right) - R$$

(b)

$$S_2 = S_1\left(1+\dfrac{I}{100}\right) - R = P\left(1+\dfrac{I}{100}\right)^2 - R\left(1+\left(1+\dfrac{I}{100}\right)\right)$$ Continuing this pattern (the $R$ terms form a geometric series): $$S_n = P\left(1+\dfrac{I}{100}\right)^n - \dfrac{100R}{I}\left(\left(1+\dfrac{I}{100}\right)^n - 1\right)$$

QUESTION 6 5 marks

Hard

Kavya takes out a bank loan of \$200000 to buy a house, at an annual interest rate of 3.8\%, compounded annually and added to the balance. To pay off the loan, she makes annual deposits of \$P at the end of every year, into a savings account paying 2.4\% per year, compounded annually, making her first deposit at the end of year 1. Her goal is to have saved enough to pay off the loan after 20 years.

a. Find the amount Kavya would owe the bank after 20 years.

[2]

b. Find the value of $P$ required, to the nearest dollar.

[3]

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(a)

$$\text{Owed} = 200000(1+3.8/100)^{20} = 421,674.24$$

(b)

Using the ordinary annuity future value formula: $$\text{Owed} = P\times\dfrac{(1+i)^{20}-1}{i}, \quad i=0.024$$ Solving: $$P = 421,674.24 \Big/ \dfrac{(1+0.024)^{20}-1}{0.024} = 16674$$

QUESTION 7 4 marks

Medium

A bank pays interest at a rate of 3.9\% compounded monthly. If the inflation rate is 1.6\%, calculate the real rate of return per year.

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First find the effective annual interest rate, since the nominal rate is compounded monthly. Using the effective annual rate formula: $$i_{\text{eff}} = \left(1+\frac{r}{100k}\right)^k - 1$$ Substituting the nominal rate $r = 3.9$ and $k = 12$ (monthly): $$i_{\text{eff}} = \left(1+\frac{3.9}{1200}\right)^{12} - 1$$ Evaluating $(1.00325)^{12} \approx 1.039704$: $$i_{\text{eff}} \approx 0.039704 = 3.9704\%$$ Now use the Fisher-type real rate of return formula, which adjusts the nominal (effective) rate for inflation: $$i_{\text{real}} = \left(\frac{1+i_{\text{eff}}}{1+i_{\text{inflation}}} - 1\right) \times 100$$ Substituting $i_{\text{eff}} = 3.9704\%$ and the inflation rate $1.6\%$: $$i_{\text{real}} = \left(\frac{1+0.039704}{1+0.016} - 1\right) \times 100$$ $$i_{\text{real}} = \left(\frac{1.039704}{1.016} - 1\right) \times 100$$ $$i_{\text{real}} \approx (1.023331-1)\times100$$ $$\boxed{i_{\text{real}} \approx 2.33\%}$$

QUESTION 8 7 marks

Medium

In 2020, the population growth rate in Country X was 1.4\% and the population was 42.3 million. The population growth rate in Country Y was 0.3\% and the population was 51.8 million. Assume both growth rates remain constant.

a. Find the size of the population in Country X in 2024.

[2]

b. Find the size of the population in Country Y in 2024.

[2]

c. Find when the population of Country X will exceed that of Country Y.

[3]

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(a)

$$42.3(1+1.4/100)^{4} = 44.72 \text{ million}$$

(b)

$$51.8(1+0.3/100)^{4} = 52.42 \text{ million}$$

(c)

Solving $42.3(1.014)^n = 51.8(1.003)^n$ gives $n \approx 18.57$ years after 2020, i.e. during 2039.

QUESTION 9 3 marks

Easy

A bank pays interest at a rate of 4.8\% compounded annually. If the inflation rate is 2.1\%, calculate the real rate of return per year.

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Using the real rate of return formula, which adjusts the nominal annual rate for inflation: $$i_{\text{real}} = \left(\frac{1+i_{\text{nominal}}}{1+i_{\text{inflation}}} - 1\right) \times 100$$ Substituting the nominal annual rate $4.8\%$ (already compounded annually, so no separate effective-rate step is needed) and the inflation rate $2.1\%$: $$i_{\text{real}} = \left(\frac{1+0.048}{1+0.021} - 1\right) \times 100$$ $$i_{\text{real}} = \left(\frac{1.048}{1.021} - 1\right) \times 100$$ $$i_{\text{real}} \approx (1.026445-1)\times100$$ $$\boxed{i_{\text{real}} \approx 2.64\%}$$

QUESTION 10 5 marks

Medium

Farhan invests \$6000 in an account that pays 4.2\% interest compounded monthly. He does not invest any further amount or withdraw any money.

a. Find the amount in his account after 6 years.

[2]

b. Find the number of months until the amount in his account doubles.

[3]

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(a)

$$A = 6000\left(1+\dfrac{4.2/100}{12}\right)^{12\times6} = 7,716.18$$

(b)

$$\left(1+\dfrac{4.2/100}{12}\right)^n = 2 \Rightarrow n = \dfrac{\ln 2}{\ln(1+4.2/1200)} = 198.4$$ So 199 months.

QUESTION 11 6 marks

Hard

A teacher invests \$90 at the beginning of every month into a fund earning 4.5\% per year, compounded monthly.

a. Find the value of the investment at the end of 20 years.

[3]

b. Another teacher wants to have \$35000 at the end of 20 years, investing monthly at the same rate. Calculate how much she must invest per month.

[3]

Show complete worked solution

(a)

$$FV = 90\times\dfrac{(1+i)^{240}-1}{i}\times(1+i), \quad i=\dfrac{4.5/100}{12}$$ $$FV = 35,062.18$$

(b)

Rearranging the same formula for the payment: $$PMT = \dfrac{35000}{\left(\dfrac{(1+i)^{240}-1}{i}\right)(1+i)} = 89.84$$

QUESTION 12 3 marks

Easy

$\$2000$ is invested at a rate of $5\%$ per annum, compounded annually. Find the value of the investment after 3 years.

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Using the compound interest formula, $FV = PV\left(1+\dfrac{r}{100}\right)^n$, with $PV = 2000$, $r=5$, $n=3$. $FV = 2000(1.05)^3$ $(1.05)^2 = 1.1025$ $(1.05)^3 = 1.1025 \times 1.05 = 1.157625$ $FV = 2000 \times 1.157625$ Final answer: $\boxed{FV = \$2315.25}$

QUESTION 13 3 marks

Easy

A car is bought for $\$18\,000$ and depreciates in value at a rate of $12\%$ per year. Find the value of the car after 2 years.

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Using the depreciation formula, $FV = PV\left(1-\dfrac{r}{100}\right)^n$, with $PV = 18000$, $r=12$, $n=2$. $FV = 18000(1-0.12)^2 = 18000(0.88)^2$ $(0.88)^2 = 0.7744$ $FV = 18000 \times 0.7744$ Final answer: $\boxed{FV = \$13\,939.20}$

QUESTION 14 4 marks

Medium

An investment of $\$1000$ grows to $\$1210$ after 2 years, with interest compounded annually. Find the annual interest rate, $r\%$.

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Using the compound interest formula, $FV = PV\left(1+\dfrac{r}{100}\right)^n$, with $PV=1000$, $FV=1210$, $n=2$. $1210 = 1000\left(1+\dfrac{r}{100}\right)^2$ Divide both sides by 1000: $\left(1+\dfrac{r}{100}\right)^2 = 1.21$ Take the square root of both sides (taking the positive root, since a growth factor must be positive): $1+\dfrac{r}{100} = \sqrt{1.21} = 1.1$ $\dfrac{r}{100} = 0.1$ Final answer: $\boxed{r = 10\%}$

QUESTION 15 5 marks

Medium

$\$5000$ is invested at a nominal annual interest rate of $6\%$, compounded monthly, for 4 years. (a) Find the value of the investment at the end of the 4 years, correct to 2 decimal places. (b) Find the total interest earned.

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Part (a): Using the compound interest formula for interest compounded $k$ times per year, $FV = PV\left(1+\dfrac{r}{100k}\right)^{kn}$, with $PV=5000$, $r=6$, $k=12$ (monthly), $n=4$. $FV = 5000\left(1+\dfrac{6}{1200}\right)^{12\times4} = 5000(1.005)^{48}$ Using a GDC: $(1.005)^{48} \approx 1.270489$ $FV = 5000 \times 1.270489 \approx 6352.446$ Final answer: $\boxed{FV \approx \$6352.45}$ Part (b): Interest earned $=$ final value $-$ principal: Interest $= 6352.45 - 5000$ Final answer: $\boxed{\text{Interest} \approx \$1352.45}$

QUESTION 16 7 marks

Hard

Maya invests $\$8000$ in an account paying $4\%$ per annum compound interest. At the same time, she buys a piece of equipment for $\$8000$ that depreciates at a rate of $15\%$ per annum. Find the least number of complete years, $n$, after which the value of Maya's investment is more than twice the value of the equipment.

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Using the compound interest formula, the investment value after $n$ years is: $I(n) = 8000(1.04)^n$ Using the depreciation formula, the equipment value after $n$ years is: $E(n) = 8000(0.85)^n$ We require the investment to exceed twice the equipment value: $8000(1.04)^n > 2 \times 8000(0.85)^n$ Divide both sides by $8000$: $(1.04)^n > 2(0.85)^n$ Divide both sides by $(0.85)^n$: $\left(\dfrac{1.04}{0.85}\right)^n > 2$ $\dfrac{1.04}{0.85} \approx 1.223529$ Taking natural logarithms of both sides (the inequality direction is preserved since $1.223529 > 1$): $n \ln(1.223529) > \ln(2)$ $n > \dfrac{\ln 2}{\ln 1.223529} \approx \dfrac{0.693147}{0.201738} \approx 3.436$ Since $n$ must be a whole number of complete years, test integer values near this boundary: Check $n=3$: $\dfrac{1.04^3}{0.85^3} = \dfrac{1.124864}{0.614125} \approx 1.832$ (not greater than 2) Check $n=4$: $\dfrac{1.04^4}{0.85^4} = \dfrac{1.169859}{0.522006} \approx 2.241$ (greater than 2) Final answer: $\boxed{n = 4 \text{ years}}$

QUESTION 17 4 marks

Easy

Priya invests \$4500 in a savings account that pays $4.5\%$ per year, compounded annually. She makes no further deposits or withdrawals from the account. Find the value of her investment after 6 years, correct to 2 decimal places.

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Using the compound interest formula: $$FV = PV\left(1+\frac{r}{100}\right)^n$$ Substituting the principal $PV = 4500$, annual rate $r = 4.5$, and $n = 6$ years: $$FV = 4500(1.045)^6$$ Working out the power step by step: $(1.045)^2 = 1.092025$ $(1.045)^3 = 1.092025 \times 1.045 = 1.141166$ $(1.045)^6 = (1.141166)^2 = 1.302260$ So: $FV = 4500 \times 1.302260$ Final answer: $\boxed{FV = \$5860.17}$

QUESTION 18 3 marks

Easy

A farmer buys a tractor for \$25\,000. The tractor depreciates in value at a rate of $9\%$ per year. Find the value of the tractor after 5 years, correct to the nearest dollar.

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Using the depreciation formula: $$FV = PV\left(1-\frac{r}{100}\right)^n$$ Substituting $PV = 25000$, $r = 9$, and $n = 5$: $$FV = 25000(0.91)^5$$ $(0.91)^2 = 0.8281$ $(0.91)^5 = 0.91^2 \times 0.91^2 \times 0.91 = 0.8281 \times 0.8281 \times 0.91 = 0.624032$ $FV = 25000 \times 0.624032$ Final answer: $\boxed{FV \approx \$15\,601}$

QUESTION 19 3 marks

Easy

A house is currently valued at \$310\,000. Real estate in the area is appreciating at a rate of $3.5\%$ per year. Assuming this rate stays constant, find the value of the house after 8 years, correct to the nearest dollar.

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Using the compound growth (appreciation) formula: $$FV = PV\left(1+\frac{r}{100}\right)^n$$ Substituting $PV = 310000$, $r = 3.5$, $n = 8$: $$FV = 310000(1.035)^8$$ $(1.035)^2 = 1.071225$ $(1.035)^4 = (1.071225)^2 = 1.147523$ $(1.035)^8 = (1.147523)^2 = 1.316809$ $FV = 310000 \times 1.316809$ Final answer: $\boxed{FV \approx \$408\,211}$

QUESTION 20 4 marks

Easy

An investment grows to \$6000 after 7 years at an interest rate of $3.8\%$ per year, compounded annually. Find the amount that was originally invested, correct to the nearest dollar.

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Using the compound interest formula: $$FV = PV\left(1+\frac{r}{100}\right)^n$$ Substituting $FV = 6000$, $r = 3.8$, $n = 7$, and rearranging for $PV$: $$PV = \frac{FV}{(1+r/100)^n} = \frac{6000}{(1.038)^7}$$ Evaluating the power: $(1.038)^7 = 1.298319$ $$PV = \frac{6000}{1.298319}$$ Final answer: $\boxed{PV \approx \$4621}$

QUESTION 21 4 marks

Easy

\$8000 is invested for 5 years, compounded annually, and grows to \$9733.22. Find the annual interest rate, $r\%$.

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Using the compound interest formula: $$FV = PV\left(1+\frac{r}{100}\right)^n$$ Substituting $PV = 8000$, $FV = 9733.22$, $n = 5$: $$9733.22 = 8000\left(1+\frac{r}{100}\right)^5$$ Dividing both sides by 8000: $$\left(1+\frac{r}{100}\right)^5 = 1.216653$$ Taking the fifth root of both sides: $$1+\frac{r}{100} = 1.216653^{1/5} = 1.04$$ $$\frac{r}{100} = 0.04$$ Final answer: $\boxed{r = 4\%}$

QUESTION 22 4 marks

Easy

An investment earns interest at a rate of $6.5\%$ per year, compounded annually. Find the minimum number of complete years required for the investment to at least double in value.

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We require the smallest integer $n$ such that: $$(1.065)^n \geq 2$$ Taking logarithms of both sides: $$n\ln(1.065) \geq \ln(2)$$ $$n \geq \frac{\ln(2)}{\ln(1.065)} = \frac{0.693147}{0.062975} \approx 11.007$$ Since $n$ must be a whole number of complete years, we round up to $n = 12$. Check: $(1.065)^{11} \approx 1.999$ (just under double), while $(1.065)^{12} \approx 2.129$ (over double). Final answer: $\boxed{n = 12 \text{ years}}$

QUESTION 23 3 marks

Easy

\$3200 is invested at a nominal annual interest rate of $5.2\%$, compounded quarterly, for 3 years. Find the value of the investment at the end of the 3 years, correct to 2 decimal places.

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Using the compound interest formula for interest compounded $k$ times per year: $$FV = PV\left(1+\frac{r}{100k}\right)^{kn}$$ Substituting $PV = 3200$, $r = 5.2$, $k = 4$ (quarterly), $n = 3$, so $kn = 12$: $$FV = 3200\left(1+\frac{5.2}{400}\right)^{12} = 3200(1.013)^{12}$$ Using a GDC: $(1.013)^{12} = 1.167652$ $FV = 3200 \times 1.167652$ Final answer: $\boxed{FV = \$3736.49}$

QUESTION 24 3 marks

Easy

The population of a town is modelled by $P(t) = 12\,000(1.028)^t$, where $t$ is the number of years after 2020. Find the population predicted by this model in 2030.

Show complete worked solution

The year 2030 corresponds to $t = 10$. Substituting $t = 10$ into the model: $$P(10) = 12000(1.028)^{10}$$ Using a GDC: $(1.028)^{10} = 1.318048$ $P(10) = 12000 \times 1.318048$ Final answer: $\boxed{P(10) \approx 15\,817}$ people

QUESTION 25 3 marks

Easy

The value in dollars of a delivery van $t$ years after purchase is modelled by $V(t) = 22\,000(0.87)^t$. Find the value of the van 4 years after purchase, correct to the nearest dollar.

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Substituting $t = 4$ into the model: $$V(4) = 22000(0.87)^4$$ $(0.87)^2 = 0.7569$ $(0.87)^4 = (0.7569)^2 = 0.572898$ $V(4) = 22000 \times 0.572898$ Final answer: $\boxed{V(4) \approx \$12\,604}$

QUESTION 26 4 marks

Easy

A bank advertises a savings account with a nominal annual interest rate of $4.6\%$, compounded monthly. Find the effective annual interest rate, correct to 2 decimal places.

Show complete worked solution

Using the effective annual rate formula: $$i_{\text{eff}} = \left(1+\frac{r}{100k}\right)^k - 1$$ Substituting the nominal rate $r = 4.6$ and $k = 12$ (monthly): $$i_{\text{eff}} = \left(1+\frac{4.6}{1200}\right)^{12} - 1$$ Using a GDC: $\left(1.003833\right)^{12} = 1.046982$ $$i_{\text{eff}} = 1.046982 - 1 = 0.046982$$ Converting to a percentage: Final answer: $\boxed{i_{\text{eff}} \approx 4.70\%}$

QUESTION 27 5 marks

Easy

Rohan invests \$7500 in an account paying $3.3\%$ per year, compounded annually, for 9 years. (a) Find the value of the investment after 9 years, correct to 2 decimal places. [2] (b) Find the total interest earned over the 9 years. [3]

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Part (a): Using the compound interest formula: $$FV = PV\left(1+\frac{r}{100}\right)^n$$ Substituting $PV = 7500$, $r = 3.3$, $n = 9$: $$FV = 7500(1.033)^9$$ $(1.033)^3 = 1.102303$ $(1.033)^9 = (1.102303)^3 = 1.339377$ $FV = 7500 \times 1.339377$ $\boxed{FV = \$10\,045.33}$ Part (b): The interest earned is the final value minus the original principal: $$\text{Interest} = FV - PV = 10045.33 - 7500$$ $\boxed{\text{Interest} = \$2545.33}$

QUESTION 28 5 marks

Easy

\$5000 is invested for 4 years at an annual interest rate of $5\%$. (a) Find the value of the investment if the interest is compounded annually. [2] (b) Find the value of the investment if the interest is instead compounded monthly. [2] (c) State which compounding method gives the greater return, and find the difference between the two final values, correct to 2 decimal places. [1]

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Part (a): Using $FV = PV(1+r/100)^n$ with $PV=5000$, $r=5$, $n=4$: $FV_{\text{annual}} = 5000(1.05)^4 = 5000 \times 1.215506$ $\boxed{FV_{\text{annual}} = \$6077.53}$ Part (b): Using $FV = PV\left(1+\dfrac{r}{100k}\right)^{kn}$ with $k=12$, $kn=48$: $FV_{\text{monthly}} = 5000\left(1+\dfrac{5}{1200}\right)^{48} = 5000(1.004167)^{48} = 5000 \times 1.220895$ $\boxed{FV_{\text{monthly}} = \$6104.48}$ Part (c): Since $6104.48 > 6077.53$, monthly compounding gives the greater return (more frequent compounding means interest is added, and starts earning interest itself, sooner). Difference $= 6104.48 - 6077.53$ $\boxed{\text{Difference} = \$26.95}$

QUESTION 29 4 marks

Easy

\$2000 is invested at an annual interest rate of $7.5\%$, compounded annually. Find the minimum number of complete years required for the investment to exceed \$3500 in value.

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We require the smallest integer $n$ such that: $$2000(1.075)^n > 3500$$ Dividing both sides by 2000: $$(1.075)^n > 1.75$$ Taking logarithms: $$n > \frac{\ln(1.75)}{\ln(1.075)} = \frac{0.559616}{0.072321} \approx 7.738$$ Since $n$ must be a whole number of complete years, round up: $n = 8$. Check: after 7 years, $2000(1.075)^7 = 2000 \times 1.659049 \approx \$3318.10$ (not yet exceeding \$3500); after 8 years, $2000(1.075)^8 = 2000 \times 1.783478 \approx \$3566.96$ (exceeds \$3500). Final answer: $\boxed{n = 8 \text{ years}}$

QUESTION 30 6 marks

Medium

The number of subscribers to a streaming service, $P$, is modelled by $P(t) = P_0 e^{kt}$, where $t$ is measured in years after launch. When the service launched ($t=0$), it had 5000 subscribers. Three years later, it had 6100 subscribers. (a) Find the value of $k$, correct to 4 significant figures. [3] (b) Use your model to predict the number of subscribers 8 years after launch, correct to the nearest hundred. [3]

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Part (a): Since $P_0 = 5000$ (the value at $t=0$), and $P(3) = 6100$: $$6100 = 5000e^{3k}$$ Dividing both sides by 5000: $$e^{3k} = 1.22$$ Taking the natural logarithm of both sides: $$3k = \ln(1.22)$$ $$k = \frac{\ln(1.22)}{3} = \frac{0.198851}{3}$$ $\boxed{k \approx 0.06628}$ Part (b): Substituting $t = 8$ into $P(t) = 5000e^{kt}$: $$P(8) = 5000e^{0.06628 \times 8} = 5000e^{0.530269}$$ Using a GDC: $e^{0.530269} \approx 1.699389$ $P(8) = 5000 \times 1.699389$ $\boxed{P(8) \approx 8500 \text{ subscribers}}$

QUESTION 31 6 marks

Medium

\$4000 is invested at a nominal annual interest rate of $6\%$, compounded monthly. (a) Find the value of the investment after 5 years, correct to 2 decimal places. [3] (b) Find the number of complete months required for the investment to first exceed \$5500. [3]

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Part (a): Using $FV = PV\left(1+\dfrac{r}{100k}\right)^{kn}$ with $PV=4000$, $r=6$, $k=12$, $n=5$ (so $kn=60$): $$FV = 4000\left(1+\frac{6}{1200}\right)^{60} = 4000(1.005)^{60}$$ Using a GDC: $(1.005)^{60} = 1.348850$ $FV = 4000 \times 1.348850$ $\boxed{FV = \$5395.40}$ Part (b): Let $m$ be the number of months. The monthly interest rate is $i = \dfrac{6}{1200} = 0.005$. We require the smallest integer $m$ such that: $$4000(1.005)^m > 5500$$ $$(1.005)^m > 1.375$$ Taking logarithms: $$m > \frac{\ln(1.375)}{\ln(1.005)} = \frac{0.318454}{0.004988} \approx 63.85$$ Since $m$ must be a whole number of complete months, round up. $\boxed{m = 64 \text{ months}}$

QUESTION 32 6 marks

Medium

A savings account pays a nominal annual interest rate of $3.5\%$, compounded annually. The annual rate of inflation is $4.5\%$. (a) Calculate the real rate of return per year, correct to 2 decimal places. [3] (b) Find the number of complete years after which the real value of the savings will have fallen by more than $20\%$. [3]

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Part (a): Using the real rate of return formula, which adjusts the nominal rate for inflation: $$i_{\text{real}} = \left(\frac{1+i_{\text{nominal}}}{1+i_{\text{inflation}}} - 1\right) \times 100$$ Substituting $i_{\text{nominal}} = 0.035$ and $i_{\text{inflation}} = 0.045$: $$i_{\text{real}} = \left(\frac{1.035}{1.045} - 1\right) \times 100$$ $$\frac{1.035}{1.045} = 0.990431$$ $$i_{\text{real}} = (0.990431 - 1) \times 100$$ $\boxed{i_{\text{real}} \approx -0.96\%}$ (the real value of the savings is decreasing, since inflation outpaces the nominal interest rate) Part (b): The real value after $n$ years is scaled by a factor of $(1+i_{\text{real}}/100)^n = (0.990431)^n$. We require the smallest integer $n$ such that the real value has fallen by more than $20\%$, i.e. the remaining value is less than $80\%$ of the original: $$(0.990431)^n < 0.8$$ Taking logarithms of both sides (note both logarithms are negative, and dividing by a negative flips the inequality): $$n > \frac{\ln(0.8)}{\ln(0.990431)} = \frac{-0.223144}{-0.009616} \approx 23.21$$ Since $n$ must be a whole number of complete years, round up. $\boxed{n = 24 \text{ years}}$

QUESTION 33 6 marks

Medium

At the end of every year, Wei deposits \$250 into a savings account that earns $4\%$ per year, compounded annually. She makes her first deposit at the end of year 1. Find the value of the account immediately after her 8th deposit, correct to 2 decimal places.

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This is an ordinary annuity (deposits are made at the end of each year). Using the future value of an ordinary annuity formula: $$FV = R \times \frac{(1+i)^n - 1}{i}$$ where $R$ is the regular payment, $i$ is the interest rate per period as a decimal, and $n$ is the number of payments. Substituting $R = 250$, $i = 0.04$, $n = 8$: $$FV = 250 \times \frac{(1.04)^8 - 1}{0.04}$$ Evaluating the power: $(1.04)^8 = 1.368569$ $$FV = 250 \times \frac{1.368569 - 1}{0.04} = 250 \times \frac{0.368569}{0.04}$$ $$FV = 250 \times 9.214225$$ Final answer: $\boxed{FV = \$2303.56}$

QUESTION 34 6 marks

Medium

A culture of bacteria grows according to the model $N(t) = N_0 \times 2^{t/T}$, where $N_0$ is the initial number of bacteria, $t$ is measured in hours, and $T$ is the doubling period in hours. Initially there are 500 bacteria, and after 6 hours there are 4000 bacteria. (a) Find the value of $T$, the doubling period in hours. [3] (b) Use your model to find the number of bacteria present after 10 hours. [3]

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Part (a): Since $N_0 = 500$ and $N(6) = 4000$: $$4000 = 500 \times 2^{6/T}$$ Dividing both sides by 500: $$2^{6/T} = 8$$ Since $8 = 2^3$: $$2^{6/T} = 2^3$$ Equating exponents (since the bases are equal): $$\frac{6}{T} = 3$$ $\boxed{T = 2 \text{ hours}}$ Part (b): Substituting $t = 10$ and $T = 2$ into $N(t) = 500 \times 2^{t/T}$: $$N(10) = 500 \times 2^{10/2} = 500 \times 2^5 = 500 \times 32$$ $\boxed{N(10) = 16\,000 \text{ bacteria}}$

QUESTION 35 7 marks

Medium

A machine originally worth \$40\,000 can be depreciated using one of two methods. Method S (straight-line): the value decreases by a fixed $8\%$ of the original value every year. Method R (reducing balance): the value decreases by $12\%$ of the current value every year. (a) Find the value of the machine after 5 years under each method, correct to the nearest dollar. [3] (b) Method R starts off lower than Method S, but eventually overtakes it (reducing-balance depreciation slows down as the value shrinks, while straight-line keeps falling at a constant amount every year). Find the first complete year in which the value under Method R exceeds the value under Method S. [4]

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Part (a): Method S: the value falls by a constant $8\%$ of \$40\,000, i.e. \$3200, every year, so after $n$ years: $$V_S(n) = 40000 - 3200n$$ $$V_S(5) = 40000 - 3200(5) = 40000 - 16000$$ $\boxed{V_S(5) = \$24\,000}$ Method R: the value is multiplied by $(1-0.12)$ every year, so: $$V_R(n) = 40000(0.88)^n$$ $(0.88)^5 = 0.527732$ $$V_R(5) = 40000 \times 0.527732$$ $\boxed{V_R(5) \approx \$21\,109}$ Part (b): Tabulating both models year by year: $$\begin{array}{c|c|c} n & V_S(n) & V_R(n) \\ \hline 6 & 20800.00 & 18576.16 \\ 7 & 17600.00 & 16347.02 \\ 8 & 14400.00 & 14385.38 \\ 9 & 11200.00 & 12659.14 \end{array}$$ At $n=8$: $V_R(8) = 14385.38 < V_S(8) = 14400.00$, so Method R is still (barely) below Method S. At $n=9$: $V_R(9) = 12659.14 > V_S(9) = 11200.00$, so Method R has now overtaken Method S. So the crossover occurs between $n=8$ and $n=9$: the first complete year in which Method R's value exceeds Method S's value. $\boxed{n = 9 \text{ years}}$

QUESTION 36 6 marks

Medium

A student takes out a loan of \$15\,000 to buy a car. The loan is charged interest at a nominal annual rate of $7.2\%$, compounded monthly, and is to be fully repaid with equal monthly payments over 3 years, with the first payment made one month after the loan is taken out. Find the value of the monthly repayment, correct to 2 decimal places.

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Using the loan amortisation (present value of an ordinary annuity) formula, where the loan amount equals the present value of all future repayments: $$P = R \times \frac{1-(1+i)^{-n}}{i}$$ where $P$ is the loan principal, $R$ is the monthly repayment, $i$ is the monthly interest rate, and $n$ is the number of monthly payments. Rearranging for $R$: $$R = \frac{P \times i}{1-(1+i)^{-n}}$$ The monthly interest rate is $i = \dfrac{7.2}{1200} = 0.006$, and $n = 3 \times 12 = 36$ monthly payments. $$R = \frac{15000 \times 0.006}{1-(1.006)^{-36}}$$ Using a GDC: $(1.006)^{36} = 1.240302$, so $(1.006)^{-36} = 0.806252$ $$R = \frac{90}{1-0.806252} = \frac{90}{0.193748}$$ Final answer: $\boxed{R = \$464.53}$ per month

QUESTION 37 7 marks

Medium

The population of a coastal town was 2.4 million in 2015 and grew at a constant annual percentage rate to reach 2.75 million in 2023. (a) Find the annual percentage growth rate, correct to 2 decimal places. [3] (b) Assuming the growth rate remains constant, predict the population in 2030, correct to 3 significant figures. [4]

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Part (a): From 2015 to 2023 is $t = 8$ years. Using $P(t) = P_0(1+g)^t$ with $P_0=2.4$ and $P(8)=2.75$: $$2.75 = 2.4(1+g)^8$$ Dividing both sides by 2.4: $$(1+g)^8 = 1.145833$$ Taking the eighth root of both sides: $$1+g = 1.145833^{1/8} = 1.017162$$ $$g = 0.017162$$ $\boxed{g \approx 1.72\%}$ Part (b): From 2015 to 2030 is $t = 15$ years. Using the growth rate found in part (a): $$P(15) = 2.4(1.017162)^{15}$$ Using a GDC: $(1.017162)^{15} \approx 1.290781$ $$P(15) = 2.4 \times 1.290781$$ $\boxed{P(15) \approx 3.10 \text{ million}}$

QUESTION 38 6 marks

Medium

\$3000 is invested at an annual interest rate of $5.5\%$, compounded annually. Find the minimum number of complete years required for the investment to exceed \$4500.

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We require the smallest integer $n$ such that: $$3000(1.055)^n > 4500$$ Dividing both sides by 3000: $$(1.055)^n > 1.5$$ Taking logarithms of both sides: $$n > \frac{\ln(1.5)}{\ln(1.055)} = \frac{0.405465}{0.053541} \approx 7.573$$ Since $n$ must be a whole number of complete years, round up. Check: at $n=7$, $3000(1.055)^7 = 3000 \times 1.454679 \approx \$4364.04$ (not yet exceeding \$4500); at $n=8$, $3000(1.055)^8 = 3000 \times 1.534687 \approx \$4604.06$ (exceeds \$4500). Final answer: $\boxed{n = 8 \text{ years}}$

QUESTION 39 6 marks

Medium

A bank offers an account where interest is compounded continuously, so that the value of an investment after $t$ years is given by $A(t) = Pe^{rt/100}$, where $P$ is the principal and $r$ is the annual interest rate as a percentage. \$6000 is invested in this account at a rate of $4.2\%$ per year. Find the value of the investment after 10 years, correct to the nearest dollar.

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Using the continuous compounding formula: $$A(t) = Pe^{rt/100}$$ Substituting $P = 6000$, $r = 4.2$, $t = 10$: $$A(10) = 6000e^{4.2 \times 10/100} = 6000e^{0.42}$$ Using a GDC: $e^{0.42} \approx 1.521962$ $$A(10) = 6000 \times 1.521962$$ Final answer: $\boxed{A(10) \approx \$9132}$

QUESTION 40 6 marks

Medium

Alicia wants to save \$20\,000 to renovate her kitchen. She plans to make equal deposits at the end of every year, for 6 years, into an account paying $4.5\%$ per year, compounded annually. Find the value of the annual deposit required, correct to 2 decimal places.

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This is a sinking-fund problem: equal deposits are made at the end of each year to accumulate a target future value. Using the future value of an ordinary annuity formula: $$FV = R \times \frac{(1+i)^n - 1}{i}$$ Rearranging for $R$: $$R = \frac{FV \times i}{(1+i)^n - 1}$$ Substituting $FV = 20000$, $i = 0.045$, $n = 6$: $(1.045)^6 = 1.302260$ $$R = \frac{20000 \times 0.045}{1.302260 - 1} = \frac{900}{0.302260}$$ Final answer: $\boxed{R = \$2977.57}$ per year

QUESTION 41 7 marks

Medium

Account A starts with \$5000 and earns $4\%$ per year, compounded annually. Account B starts with \$4500 and earns $5.5\%$ per year, compounded annually. Both accounts receive no further deposits or withdrawals. (a) Find the value of each account after 5 years, correct to 2 decimal places. [3] (b) Find the least number of complete years after which Account B's value first exceeds Account A's value. [4]

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Part (a): $$V_A(5) = 5000(1.04)^5 = 5000 \times 1.216653$$ $\boxed{V_A(5) = \$6083.26}$ $$V_B(5) = 4500(1.055)^5 = 4500 \times 1.306960$$ $\boxed{V_B(5) = \$5881.32}$ (At $n=5$, Account A is still ahead of Account B.) Part (b): We require the smallest integer $n$ such that $4500(1.055)^n > 5000(1.04)^n$. Tabulating values of each account year by year: $$\begin{array}{c|c|c} n & V_A(n) & V_B(n) \\ \hline 6 & 6326.60 & 6204.79 \\ 7 & 6579.66 & 6546.06 \\ 8 & 6842.85 & 6906.09 \end{array}$$ At $n=7$, $V_A(7)=6579.66 > V_B(7)=6546.06$ (A still ahead). At $n=8$, $V_B(8)=6906.09 > V_A(8)=6842.85$ (B has overtaken A). $\boxed{n = 8 \text{ years}}$

QUESTION 42 6 marks

Medium

A company's fleet of vehicles was purchased for a total of \$50\,000. After 5 years of reducing-balance depreciation at a constant annual rate, the fleet is valued at \$28\,800. Find the annual rate of depreciation, correct to 2 decimal places.

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Using the depreciation formula: $$FV = PV\left(1-\frac{r}{100}\right)^n$$ Substituting $PV = 50000$, $FV = 28800$, $n = 5$: $$28800 = 50000\left(1-\frac{r}{100}\right)^5$$ Dividing both sides by 50000: $$\left(1-\frac{r}{100}\right)^5 = 0.576$$ Taking the fifth root of both sides: $$1-\frac{r}{100} = 0.576^{1/5} = 0.895539$$ $$\frac{r}{100} = 1 - 0.895539 = 0.104461$$ Final answer: $\boxed{r \approx 10.45\%}$

QUESTION 43 6 marks

Medium

After a dose of medication, the amount remaining in a patient's bloodstream decreases by $15\%$ every hour. Immediately after the dose, there are 80 mg of the medication in the bloodstream. Find the minimum number of complete hours after the dose for the amount remaining to drop below 10 mg.

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The amount remaining after $t$ hours is modelled by: $$C(t) = 80(0.85)^t$$ We require the smallest integer $t$ such that: $$80(0.85)^t < 10$$ Dividing both sides by 80: $$(0.85)^t < 0.125$$ Taking logarithms of both sides (both logarithms are negative, so the inequality direction flips when dividing): $$t > \frac{\ln(0.125)}{\ln(0.85)} = \frac{-2.079442}{-0.162519} \approx 12.795$$ Since $t$ must be a whole number of complete hours, round up. Check: at $t=12$, $C(12) = 80(0.85)^{12} \approx 80 \times 0.142242 \approx 11.38$ mg (still above 10 mg); at $t=13$, $C(13) = 80(0.85)^{13} \approx 80 \times 0.120905 \approx 9.67$ mg (below 10 mg). Final answer: $\boxed{t = 13 \text{ hours}}$

QUESTION 44 10 marks

Hard

A couple takes out a mortgage of \$250\,000 to buy a house. The loan is charged interest at a nominal annual rate of $4.8\%$, compounded monthly, and is to be fully repaid with equal monthly payments over 25 years, the first payment made one month after the loan begins. (a) Find the value of the monthly repayment, correct to 2 decimal places. [3] (b) Find the amount still owed on the mortgage immediately after the 60th monthly payment (i.e. after 5 years), correct to 2 decimal places. [4] (c) Hence find the total amount of interest paid during the first 5 years of the mortgage. [3]

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Part (a): Using the loan amortisation formula $P = R \times \dfrac{1-(1+i)^{-n}}{i}$, rearranged for $R$: $$R = \frac{P \times i}{1-(1+i)^{-n}}$$ The monthly interest rate is $i = \dfrac{4.8}{1200} = 0.004$, and $n = 25 \times 12 = 300$ monthly payments. $$R = \frac{250000 \times 0.004}{1-(1.004)^{-300}}$$ Using a GDC: $(1.004)^{300} \approx 3.310108$, so $(1.004)^{-300} \approx 0.302103$ $$R = \frac{1000}{1-0.302103} = \frac{1000}{0.697897}$$ $\boxed{R = \$1432.49}$ per month Part (b): The balance outstanding after $m$ payments on a loan of $P$ with monthly rate $i$ and payment $R$ is: $$S_m = P(1+i)^m - R \times \frac{(1+i)^m - 1}{i}$$ Substituting $P = 250000$, $i = 0.004$, $R = 1432.49$, $m = 60$: $(1.004)^{60} = 1.270641$ $$S_{60} = 250000(1.270641) - 1432.49 \times \frac{1.270641-1}{0.004}$$ $$S_{60} = 317660.25 - 1432.49 \times 67.660250$$ $$S_{60} = 317660.25 - 96922.76$$ $\boxed{S_{60} \approx \$220\,737.49}$ Part (c): Total amount paid in 60 monthly payments: $60 \times 1432.49 = \$85\,949.40$ Amount by which the principal was reduced: $250000 - 220737.49 = \$29\,262.51$ The rest of the money paid must have been interest: $$\text{Interest paid} = 85949.40 - 29262.51$$ $\boxed{\text{Interest paid} \approx \$56\,686.89}$ (small roundings from part (a)'s rounded $R$ account for the few cents' difference from the exact value of \$56\,687.03)

QUESTION 45 9 marks

Hard

Nadia borrows \$12\,000 from a bank at a monthly interest rate of $1.5\%$, with interest added to the balance at the end of each month before her repayment is subtracted. She repays \$450 at the end of every month. Let $S_n$ denote the balance owing, in dollars, immediately after the $n$th repayment. (a) Show that $S_n = 12000(1.015)^n - 450 \times \dfrac{(1.015)^n - 1}{0.015}$, and hence find $S_{12}$, correct to 2 decimal places. [3] (b) Find the number of the repayment on which the loan is fully paid off (i.e. the smallest $n$ for which $S_n \leq 0$). [4] (c) Find the value of this final repayment (which will be smaller than \$450, since it only needs to clear the remaining balance). [2]

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Part (a): Each month, the balance is multiplied by $(1.015)$ (interest added), then \$450 is subtracted. Starting from $S_0 = 12000$, this recurrence $S_n = 1.015S_{n-1} - 450$ unwinds (summing the resulting geometric series of the repayments, each carried forward at compound interest for the remaining months) to give the closed form: $$S_n = 12000(1.015)^n - 450 \times \frac{(1.015)^n - 1}{0.015}$$ Substituting $n = 12$: $(1.015)^{12} = 1.195618$ $$S_{12} = 12000(1.195618) - 450 \times \frac{1.195618-1}{0.015}$$ $$S_{12} = 14347.42 - 450 \times 13.041200$$ $$S_{12} = 14347.42 - 5868.54$$ $\boxed{S_{12} = \$8478.87}$ Part (b): Continuing to evaluate $S_n$ using a GDC for increasing $n$, the balance decreases each month until it becomes small: $S_{34} \approx \$138.07$ (still owing), while $S_{35} \approx -\$309.86$ (a full \$450 payment on month 35 would overpay the loan by \$309.86, meaning the loan is actually fully cleared during the 35th payment). $\boxed{n = 35 \text{th repayment}}$ Part (c): Just before the 35th (final) repayment, the balance (after that month's interest is added to $S_{34}$) is: $$S_{34} \times 1.015 = 138.07 \times 1.015$$ This final, reduced repayment exactly clears the remaining balance: $\boxed{\text{Final repayment} \approx \$140.14}$

QUESTION 46 9 marks

Hard

A couple wants to save \$50\,000 for a deposit on a house within 10 years. They plan to make equal deposits at the end of every year into an account paying $5\%$ per year, compounded annually. (a) Find the annual deposit required, correct to 2 decimal places. [5] (b) Find the total interest earned over the 10 years. [4]

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Part (a): This is a sinking-fund problem. Using the future value of an ordinary annuity formula, rearranged for the payment $R$: $$FV = R \times \frac{(1+i)^n - 1}{i} \quad \Longrightarrow \quad R = \frac{FV \times i}{(1+i)^n - 1}$$ Substituting $FV = 50000$, $i = 0.05$, $n = 10$: $(1.05)^{10} = 1.628895$ $$R = \frac{50000 \times 0.05}{1.628895 - 1} = \frac{2500}{0.628895}$$ $\boxed{R = \$3975.23}$ per year Part (b): Total amount deposited over 10 years (before interest): $$\text{Total deposits} = 10 \times 3975.23 = \$39\,752.30$$ The final fund value is \$50\,000 (by design), so the interest earned is the difference: $$\text{Interest earned} = 50000 - 39752.30$$ $\boxed{\text{Interest earned} \approx \$10\,247.70}$

QUESTION 47 8 marks

Hard

Two companies report their annual revenue, in thousands of dollars, using the following models, where $t$ is the number of years after 2020. Company A: $A(t) = 200(1.12)^t$ (revenue grows exponentially, compounding at $12\%$ per year) Company B: $B(t) = 200 + 40t$ (revenue grows linearly, increasing by a fixed \$40\,000 every year) (a) Find $A(5)$ and $B(5)$, and state which company has the greater revenue at $t=5$. [2] (b) Find the least integer value of $t$ for which Company A's revenue first exceeds Company B's revenue. [6]

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Part (a): $$A(5) = 200(1.12)^5 = 200 \times 1.762342$$ $\boxed{A(5) \approx \$352.47 \text{ thousand}}$ $$B(5) = 200 + 40(5) = 200 + 200$$ $\boxed{B(5) = \$400 \text{ thousand}}$ Since $B(5) > A(5)$, Company B has the greater revenue at $t=5$. Part (b): Both models start equal at $t=0$ (both give 200), and initially the linear model B grows faster, since $40 > 200 \times 0.12 = 24$ (B's constant yearly increase exceeds A's yearly increase at $t=0$). However, since A grows by a fixed percentage of an ever-increasing amount, it must eventually overtake the linear model B. There is no simple algebraic way to solve $200(1.12)^t = 200+40t$ for $t$ directly (it mixes an exponential and a linear term), so a GDC table of values is used to find where the crossover occurs: $$\begin{array}{c|c|c} t & A(t) & B(t) \\ \hline 7 & 442.14 & 480 \\ 8 & 495.19 & 520 \\ 9 & 554.62 & 560 \\ 10 & 621.17 & 600 \end{array}$$ At $t=9$: $A(9) = 554.62 < B(9) = 560$ (B still ahead). At $t=10$: $A(10) = 621.17 > B(10) = 600$ (A has now overtaken B). $\boxed{t = 10 \text{ years}}$

QUESTION 48 9 marks

Hard

A small business owns two assets. Asset X (a property) appreciates in value according to $X(t) = 15\,000(1.06)^t$. Asset Y (a piece of equipment) depreciates in value according to $Y(t) = 20\,000(0.82)^t$. Let $C(t) = X(t) + Y(t)$ be the combined value of the two assets, where $t$ is measured in years from today. (a) Complete a table of values of $C(t)$ for $t = 0, 1, 2, \ldots, 8$ (correct to 2 decimal places), and hence state the year in which the combined value is at its lowest. [4] (b) Find the least integer value of $t$, greater than the year found in part (a), for which the combined value $C(t)$ exceeds \$30\,000 again. [5]

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Part (a): Evaluating $C(t) = 15000(1.06)^t + 20000(0.82)^t$ for each integer $t$: $$\begin{array}{c|c|c|c} t & X(t) & Y(t) & C(t) \\ \hline 0 & 15000.00 & 20000.00 & 35000.00 \\ 1 & 15900.00 & 16400.00 & 32300.00 \\ 2 & 16854.00 & 13448.00 & 30302.00 \\ 3 & 17865.24 & 11027.36 & 28892.60 \\ 4 & 18937.15 & 9042.44 & 27979.59 \\ 5 & 20073.38 & 7414.80 & 27488.18 \\ 6 & 21277.79 & 6080.13 & 27357.92 \\ 7 & 22554.45 & 4985.71 & 27540.16 \\ 8 & 23907.72 & 4088.28 & 27996.00 \end{array}$$ The combined value $C(t)$ decreases each year from $t=0$ up to $t=6$ (where $C(6) \approx \$27\,357.92$), then starts increasing again from $t=7$ onward (as the depreciating asset Y becomes small enough that the appreciating asset X's growth dominates the total). $\boxed{\text{Minimum combined value occurs at } t = 6}$ Part (b): Continuing the table beyond $t=8$: $$\begin{array}{c|c} t & C(t) \\ \hline 9 & 28694.58 \\ 10 & 29611.68 \\ 11 & 30728.63 \end{array}$$ At $t=10$: $C(10) = 29611.68 < 30000$ (not yet exceeding \$30\,000 again). At $t=11$: $C(11) = 30728.63 > 30000$ (now exceeds \$30\,000 again). $\boxed{t = 11 \text{ years}}$

QUESTION 49 10 marks

Hard

To save for retirement, Hassan deposits \$4000 into an investment account at the beginning of every year for 20 years, starting today. The account earns $6\%$ per year, compounded annually. (a) Find the value of the account immediately after the last (20th) deposit, correct to 2 decimal places. [4] (b) Hassan then stops making deposits, and instead withdraws a fixed amount $W$ at the end of every year for the next 15 years, with the account now earning $3.5\%$ per year, compounded annually. Find the value of $W$ such that the account balance reaches exactly \$0 immediately after the final (15th) withdrawal. [4] (c) Find the total amount Hassan withdraws over the 15 years, and compare it with the amount in the account at the start of his retirement (found in part (a)). [2]

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Part (a): This is an annuity-due (deposits at the beginning of each year), so each of the 20 deposits earns one extra year of interest compared with an ordinary annuity. Using the future value of an annuity-due formula: $$FV = R \times \frac{(1+i)^n - 1}{i} \times (1+i)$$ Substituting $R = 4000$, $i = 0.06$, $n = 20$: $(1.06)^{20} = 3.207135$ $$FV = 4000 \times \frac{3.207135-1}{0.06} \times 1.06 = 4000 \times 36.785583 \times 1.06$$ $$FV = 147142.33 \times 1.06$$ $\boxed{FV = \$155\,970.91}$ Part (b): This account now behaves as a loan being paid down to zero by fixed withdrawals (an ordinary annuity, since withdrawals occur at the end of each year). Using the present value of an ordinary annuity formula, with the retirement balance from part (a) as the present value: $$FV_{(a)} = W \times \frac{1-(1+i)^{-n}}{i}$$ Rearranging for $W$, with $i = 0.035$, $n = 15$: $$W = \frac{FV_{(a)} \times i}{1-(1+i)^{-n}}$$ $(1.035)^{15} = 1.675349$, so $(1.035)^{-15} = 0.596887$ $$W = \frac{155970.91 \times 0.035}{1-0.596887} = \frac{5458.98}{0.403113}$$ $\boxed{W \approx \$13\,542.18}$ per year Part (c): Total withdrawn over 15 years: $$15 \times 13542.18 = \$203\,132.70$$ This total (\$203\,132.70) is considerably more than the \$155\,970.91 available at the start of retirement, since the remaining balance continues to earn $3.5\%$ interest throughout the 15 years of withdrawals, generating an additional $$203132.70 - 155970.91$$ $\boxed{\approx \$47\,161.79 \text{ in extra interest earned during the withdrawal phase}}$

QUESTION 50 9 marks

Hard

An investment of \$10\,000 grows at an annual rate of $8\%$, compounded annually, for the first 5 years. For the next 4 years, the annual growth rate changes to $5\%$, compounded annually. (a) Find the value of the investment after the first 5 years. [2] (b) Find the value of the investment after the full 9 years. [3] (c) Find the single equivalent annual growth rate that would produce the same overall growth (from \$10\,000 to the value found in part (b)) over the full 9 years, correct to 2 decimal places. [4]

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Part (a): $$V(5) = 10000(1.08)^5 = 10000 \times 1.469328$$ $\boxed{V(5) = \$14\,693.28}$ Part (b): The value at the end of year 5 becomes the principal for the second stage: $$V(9) = V(5)(1.05)^4 = 14693.28 \times 1.215506$$ $\boxed{V(9) \approx \$17\,859.77}$ Part (c): We seek the equivalent constant annual rate $r$ such that: $$10000(1+r)^9 = 17859.77$$ Dividing both sides by 10000: $$(1+r)^9 = 1.785977$$ Taking the ninth root of both sides: $$1+r = 1.785977^{1/9}$$ Using a GDC: $1.785977^{1/9} \approx 1.066562$ $$r \approx 0.066562$$ Final answer: $\boxed{r \approx 6.66\% \text{ per year}}$

QUESTION 1 3 marks

Easy

Write the number $0.0000456$ in the form $a \times 10^{k}$, where $1 \le a < 10$ and $k \in \mathbb{Z}$.

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Using the definition of scientific notation, write the number as a value between 1 and 10 multiplied by a power of 10. Move the decimal point in $0.0000456$ to the right until one non-zero digit remains before the point: $0.0000456 \rightarrow 4.56$ (the decimal point has moved 5 places to the right). Since the original number is less than 1, the exponent is negative: $0.0000456 = 4.56 \times 10^{-5}$ Check: $4.56 \times 10^{-5} = 4.56 \div 100000 = 0.0000456$. Final answer: $\boxed{4.56 \times 10^{-5}}$

QUESTION 2 3 marks

Easy

Simplify $(2.4 \times 10^{6}) \times (5 \times 10^{3})$, giving your answer in the form $a \times 10^{k}$, where $1 \le a < 10$ and $k \in \mathbb{Z}$.

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Using the laws of exponents, multiply the coefficients together and multiply the powers of 10 together separately. Coefficients: $2.4 \times 5 = 12$ Powers of 10: $10^{6} \times 10^{3} = 10^{6+3} = 10^{9}$ So $(2.4\times10^{6})\times(5\times10^{3}) = 12 \times 10^{9}$ Since $12$ is not between 1 and 10, rewrite $12 = 1.2 \times 10^{1}$ and adjust: $12 \times 10^{9} = 1.2 \times 10^{1} \times 10^{9} = 1.2 \times 10^{10}$ Final answer: $\boxed{1.2 \times 10^{10}}$

QUESTION 3 4 marks

Medium

Let $a = 3.2 \times 10^{8}$ and $b = 8 \times 10^{-4}$. Find the value of $\dfrac{a}{b}$, giving your answer in the form $c \times 10^{k}$, where $1 \le c < 10$ and $k \in \mathbb{Z}$.

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Using the laws of exponents for division, divide the coefficients and subtract the exponents. $\dfrac{a}{b} = \dfrac{3.2 \times 10^{8}}{8 \times 10^{-4}} = \left(\dfrac{3.2}{8}\right) \times 10^{8 - (-4)}$ Coefficient: $\dfrac{3.2}{8} = 0.4$ Exponent: $8-(-4) = 12$ So $\dfrac{a}{b} = 0.4 \times 10^{12}$ Since $0.4$ is not between 1 and 10, rewrite $0.4 = 4 \times 10^{-1}$ and adjust: $0.4 \times 10^{12} = 4 \times 10^{-1} \times 10^{12} = 4 \times 10^{11}$ Final answer: $\boxed{4 \times 10^{11}}$

QUESTION 4 4 marks

Medium

A particular star is $4.3$ light years from Earth. Given that one light year is equal to $9.461 \times 10^{12}\text{ km}$, find the distance from Earth to the star in kilometres. Give your answer in the form $a \times 10^{k}$, where $1 \le a < 10$, correct to 3 significant figures.

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Using the definition of a light year, multiply the number of light years by the distance travelled in one light year. Distance $= 4.3 \times (9.461 \times 10^{12})$ $= (4.3 \times 9.461) \times 10^{12}$ $4.3 \times 9.461 = 40.6823$ So distance $= 40.6823 \times 10^{12}\text{ km}$ Rewrite in scientific notation (with $1 \le a < 10$): $40.6823 \times 10^{12} = 4.06823 \times 10^{13}$ Rounding to 3 significant figures: Final answer: $\boxed{4.07 \times 10^{13}\text{ km}}$

QUESTION 5 6 marks

Hard

A company's annual revenue is $\$4.5 \times 10^{7}$ and its annual expenses are $\$6.8 \times 10^{6}$. (a) Find the company's annual profit (profit $=$ revenue $-$ expenses), giving your answer in the form $a \times 10^{k}$, where $1 \le a < 10$. (b) The profit is shared equally as a bonus among $2.5 \times 10^{3}$ employees. Find the bonus received by each employee, giving your answer in the form $a \times 10^{k}$, correct to 3 significant figures.

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Part (a): Using subtraction of numbers in scientific notation, first write both values with the same power of 10. $6.8 \times 10^{6} = 0.68 \times 10^{7}$ Profit $= 4.5 \times 10^{7} - 0.68 \times 10^{7} = (4.5 - 0.68) \times 10^{7} = 3.82 \times 10^{7}$ Since $1 \le 3.82 < 10$, this is already in the correct form. Profit $= \boxed{3.82 \times 10^{7}}$ dollars Part (b): Using division of numbers in scientific notation, divide the profit by the number of employees. Bonus $= \dfrac{3.82 \times 10^{7}}{2.5 \times 10^{3}} = \left(\dfrac{3.82}{2.5}\right) \times 10^{7-3}$ $\dfrac{3.82}{2.5} = 1.528$ Bonus $= 1.528 \times 10^{4}$ Rounding to 3 significant figures: Final answer: $\boxed{1.53 \times 10^{4}\text{ dollars per employee}}$

QUESTION 6 3 marks

Easy

Solve the equation $2^{x} = 32$ for $x$.

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Using the fact that both sides can be written as powers of the same base, express $32$ as a power of $2$. $32 = 2^5$ So the equation becomes: $2^x = 2^5$ Since the bases are equal, the exponents must be equal: Final answer: $\boxed{x = 5}$

QUESTION 7 4 marks

Easy

Find the value of $\log_2 50$, correct to 3 significant figures.

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Using the change of base formula, $\log_2 50 = \dfrac{\ln 50}{\ln 2}$ (natural logarithms are used here, though any base could be chosen). Using a GDC: $\ln 50 \approx 3.912023$ $\ln 2 \approx 0.693147$ $\log_2 50 = \dfrac{3.912023}{0.693147} \approx 5.64386$ Rounding to 3 significant figures: Final answer: $\boxed{\log_2 50 \approx 5.64}$

QUESTION 8 4 marks

Medium

Without using a calculator, find the value of $\log_2 40 - \log_2 5$.

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Using the subtraction law of logarithms, $\log_a x - \log_a y = \log_a\left(\dfrac{x}{y}\right)$. $\log_2 40 - \log_2 5 = \log_2\left(\dfrac{40}{5}\right) = \log_2 8$ Since $8 = 2^3$: $\log_2 8 = \log_2 2^3 = 3$ Final answer: $\boxed{3}$

QUESTION 9 4 marks

Medium

Solve the equation $5^{x} = 130$ for $x$, giving your answer correct to 3 significant figures.

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Taking logarithms of both sides (natural logarithms are used here, though any base could be chosen): $\ln(5^x) = \ln(130)$ Using the power law of logarithms, $\ln(5^x) = x\ln 5$: $x \ln 5 = \ln 130$ $x = \dfrac{\ln 130}{\ln 5}$ Using a GDC: $\ln 130 \approx 4.867534$ $\ln 5 \approx 1.609438$ $x = \dfrac{4.867534}{1.609438} \approx 3.02444$ Rounding to 3 significant figures: Final answer: $\boxed{x \approx 3.02}$

QUESTION 10 5 marks

Hard

Solve the equation $2^{2x+1} - 9(2^{x}) + 4 = 0$ for $x \in \mathbb{R}$, giving your answers as exact values.

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Using the laws of exponents, rewrite $2^{2x+1}$ in terms of $2^x$: $2^{2x+1} = 2 \times 2^{2x} = 2 \times (2^x)^2$ Let $y = 2^x$ (a substitution to form a quadratic equation). Then the equation becomes: $2y^2 - 9y + 4 = 0$ Using the quadratic formula, $y = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$, with $a=2$, $b=-9$, $c=4$: $y = \dfrac{9 \pm \sqrt{(-9)^2 - 4(2)(4)}}{2(2)} = \dfrac{9 \pm \sqrt{81-32}}{4} = \dfrac{9 \pm \sqrt{49}}{4} = \dfrac{9\pm7}{4}$ So $y = \dfrac{16}{4} = 4$ or $y = \dfrac{2}{4} = 0.5$ Now solve for $x$ in each case using $y = 2^x$ (noting $y>0$ is required, and both values satisfy this): Case 1: $2^x = 4 = 2^2 \Rightarrow x = 2$ Case 2: $2^x = 0.5 = 2^{-1} \Rightarrow x = -1$ Final answer: $\boxed{x = 2 \text{ or } x = -1}$

QUESTION 11 3 marks

Easy

Simplify $\dfrac{x^{\frac{3}{2}} \times x^{-\frac{1}{4}}}{x^{\frac{1}{4}}}$, giving your answer in the form $x^{k}$.

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Method: using the laws of exponents - when multiplying powers of the same base, add the exponents; when dividing powers of the same base, subtract the exponents. Numerator: $x^{\frac{3}{2}} \times x^{-\frac{1}{4}} = x^{\frac{3}{2}+(-\frac{1}{4})} = x^{\frac{6}{4}-\frac{1}{4}} = x^{\frac{5}{4}}$. Divide by $x^{\frac{1}{4}}$: $x^{\frac{5}{4}} \div x^{\frac{1}{4}} = x^{\frac{5}{4}-\frac{1}{4}} = x^{\frac{4}{4}} = x^{1}$. Final answer: $$\boxed{x}$$

QUESTION 12 3 marks

Easy

Solve for $x$: $\log_2 x + \log_2 3 = 5$. Give your answer as an exact value.

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Method: using the law of logarithms $\log_a m + \log_a n = \log_a(mn)$, then converting to exponential form. $\log_2 x + \log_2 3 = \log_2(3x) = 5$ Convert to exponential form: $3x = 2^5 = 32$. $x = \dfrac{32}{3}$. Check: $x = \dfrac{32}{3} > 0$, so it is in the domain of $\log_2 x$, and the solution is valid. Final answer: $$\boxed{x = \dfrac{32}{3}}$$

QUESTION 13 4 marks

Medium

Given that $\log_2 5 = a$ and $\log_2 3 = b$, express $\log_2 45$ in terms of $a$ and $b$.

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Method: writing 45 as a product of powers of 3 and 5, then applying the laws of logarithms (log of a product, and log of a power). $45 = 9 \times 5 = 3^2 \times 5$ $\log_2 45 = \log_2(3^2 \times 5) = \log_2(3^2) + \log_2 5$ [log of a product] $= 2\log_2 3 + \log_2 5$ [log of a power: $\log_a(m^k) = k\log_a m$] $= 2b + a$ Final answer: $$\boxed{\log_2 45 = a+2b}$$

QUESTION 14 5 marks

Medium

Solve the equation $3^{x+1} = 20$, giving your answer correct to 3 significant figures.

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Method: taking logarithms of both sides and applying the power law of logarithms, then using change of base to evaluate. $3^{x+1} = 20$ Take $\ln$ of both sides: $\ln(3^{x+1}) = \ln 20$ $(x+1)\ln 3 = \ln 20$ [power law of logarithms] $x + 1 = \dfrac{\ln 20}{\ln 3}$ (this is $\log_3 20$ by change of base) Using a calculator: $\ln 20 = 2.99573\ldots$, $\ln 3 = 1.09861\ldots$ $x + 1 = 2.72683\ldots$ $x = 1.72683\ldots$ Final answer: $$\boxed{x \approx 1.73 \text{ (3 s.f.)}}$$

QUESTION 15 6 marks

Hard

Solve the equation $9^{x} - 4(3^{x}) - 5 = 0$. Give your answer (a) as an exact value, and (b) correct to 3 significant figures.

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Method: using the law of exponents $9^x = (3^2)^x = (3^x)^2$ to rewrite the equation as a quadratic in $y = 3^x$, solving the quadratic, then using change of base to return to $x$. Let $y = 3^x$ (note $y > 0$ for all real $x$). $9^x = (3^x)^2 = y^2$ The equation becomes: $y^2 - 4y - 5 = 0$ Factorise: $(y-5)(y+1) = 0$ $y = 5$ or $y = -1$. Since $y = 3^x > 0$ for all real $x$, reject $y = -1$. So $3^x = 5$. (a) Exact value: $$\boxed{x = \log_3 5 = \dfrac{\ln 5}{\ln 3}}$$ (b) Using a calculator: $\ln 5 = 1.60944\ldots$, $\ln 3 = 1.09861\ldots$ $x = 1.46497\ldots$ (b) Final answer: $$\boxed{x \approx 1.46 \text{ (3 s.f.)}}$$

QUESTION 16 3 marks

Easy

Simplify $a^{5} \times a^{-2}$, giving your answer in the form $a^{k}$.

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Using the law of exponents for multiplication, when multiplying powers of the same base, add the exponents: $a^{m} \times a^{n} = a^{m+n}$. $a^{5} \times a^{-2} = a^{5+(-2)} = a^{3}$ Final answer: $\boxed{a^{3}}$

QUESTION 17 3 marks

Easy

Write the number $0.00302$ in the form $a \times 10^{k}$, where $1 \le a < 10$ and $k \in \mathbb{Z}$.

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Using the definition of scientific notation, write the number as a value between 1 and 10 multiplied by a power of 10. Move the decimal point in $0.00302$ to the right until one non-zero digit remains before the point: $0.00302 \rightarrow 3.02$ (the decimal point has moved 3 places to the right). Since the original number is less than 1, the exponent is negative: $0.00302 = 3.02 \times 10^{-3}$ Check: $3.02 \times 10^{-3} = 3.02 \div 1000 = 0.00302$. Final answer: $\boxed{3.02 \times 10^{-3}}$

QUESTION 18 3 marks

Easy

Without using a calculator, find the value of $\log_5 25$.

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Using the definition of a logarithm, $\log_a x = y$ means $a^y = x$. Let $\log_5 25 = y$, so $5^y = 25$. Since $25 = 5^2$: $5^y = 5^2$ Since the bases are equal, the exponents must be equal, so $y = 2$. Final answer: $\boxed{\log_5 25 = 2}$

QUESTION 19 3 marks

Easy

Solve the equation $3^{x} = 81$ for $x$.

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Using the fact that both sides can be written as powers of the same base, express $81$ as a power of $3$. $81 = 3^4$ So the equation becomes: $3^x = 3^4$ Since the bases are equal, the exponents must be equal: Final answer: $\boxed{x = 4}$

QUESTION 20 3 marks

Easy

Simplify $(2x^{3})^{4}$, giving your answer in the form $kx^{n}$.

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Using the law of exponents for a power of a product, $(pq)^{n} = p^{n}q^{n}$, and the law for a power of a power, $(x^{m})^{n} = x^{mn}$. $(2x^{3})^{4} = 2^{4} \times (x^{3})^{4}$ $2^{4} = 16$ $(x^{3})^{4} = x^{3 \times 4} = x^{12}$ So $(2x^{3})^{4} = 16x^{12}$ Final answer: $\boxed{16x^{12}}$

QUESTION 21 3 marks

Easy

Without using a calculator, find the value of $\log_{10} 1000 - \log_{10} 10$.

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Using the definition of a logarithm, evaluate each term separately. $\log_{10} 1000$: since $1000 = 10^3$, $\log_{10} 1000 = 3$. $\log_{10} 10$: since $10 = 10^1$, $\log_{10} 10 = 1$. So $\log_{10} 1000 - \log_{10} 10 = 3 - 1 = 2$ Final answer: $\boxed{2}$

QUESTION 22 3 marks

Easy

Simplify $\dfrac{y^{7}}{y^{3}}$, giving your answer in the form $y^{k}$.

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Using the law of exponents for division, when dividing powers of the same base, subtract the exponents: $\dfrac{a^{m}}{a^{n}} = a^{m-n}$. $\dfrac{y^{7}}{y^{3}} = y^{7-3} = y^{4}$ Final answer: $\boxed{y^{4}}$

QUESTION 23 3 marks

Easy

Find the exact value of $7^{0} + 7^{-1}$.

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Using the zero exponent law, $a^{0} = 1$ for any $a \ne 0$, and the negative exponent law, $a^{-n} = \dfrac{1}{a^{n}}$. $7^{0} = 1$ $7^{-1} = \dfrac{1}{7}$ So $7^{0} + 7^{-1} = 1 + \dfrac{1}{7} = \dfrac{7}{7} + \dfrac{1}{7} = \dfrac{8}{7}$ Final answer: $\boxed{\dfrac{8}{7}}$

QUESTION 24 3 marks

Easy

Simplify $\sqrt[3]{x^{6}}$, giving your answer in the form $x^{k}$.

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Using the connection between roots and rational exponents, $\sqrt[n]{x^{m}} = x^{\frac{m}{n}}$. $\sqrt[3]{x^{6}} = x^{\frac{6}{3}} = x^{2}$ Final answer: $\boxed{x^{2}}$

QUESTION 25 4 marks

Easy

Find the value of $(2.3 \times 10^{5}) + (4.1 \times 10^{4})$, giving your answer in the form $a \times 10^{k}$, where $1 \le a < 10$ and $k \in \mathbb{Z}$.

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Using the method for adding numbers in scientific notation, first write both values with the same power of 10. $4.1 \times 10^{4} = 0.41 \times 10^{5}$ $(2.3 \times 10^{5}) + (0.41 \times 10^{5}) = (2.3 + 0.41) \times 10^{5} = 2.71 \times 10^{5}$ Since $1 \le 2.71 < 10$, this is already in the correct form. Final answer: $\boxed{2.71 \times 10^{5}}$

QUESTION 26 3 marks

Easy

Solve for $x$: $\log_4 x = 3$. Give your answer as an exact value.

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Using the definition of a logarithm, $\log_a x = y$ means $x = a^y$. $\log_4 x = 3$ means $x = 4^3$ $x = 64$ Final answer: $\boxed{x = 64}$

QUESTION 27 4 marks

Easy

Without using a calculator, find the value of $\log_2 8 + \log_2 4$.

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Using the addition law of logarithms, $\log_a m + \log_a n = \log_a(mn)$. $\log_2 8 + \log_2 4 = \log_2(8 \times 4) = \log_2 32$ Since $32 = 2^5$: $\log_2 32 = \log_2 2^5 = 5$ Final answer: $\boxed{5}$

QUESTION 28 3 marks

Easy

Simplify $x^{\frac{1}{2}} \times x^{\frac{3}{2}}$, giving your answer in the form $x^{k}$.

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Using the law of exponents for multiplication, when multiplying powers of the same base, add the exponents. $x^{\frac{1}{2}} \times x^{\frac{3}{2}} = x^{\frac{1}{2}+\frac{3}{2}} = x^{\frac{4}{2}} = x^{2}$ Final answer: $\boxed{x^{2}}$

QUESTION 29 4 marks

Easy

Find the value of $\log_7 200$, correct to 3 significant figures.

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Using the change of base formula, $\log_7 200 = \dfrac{\ln 200}{\ln 7}$ (natural logarithms are used here, though any base could be chosen). Using a GDC: $\ln 200 \approx 5.29832$ $\ln 7 \approx 1.94591$ $\log_7 200 = \dfrac{5.29832}{1.94591} \approx 2.72280$ Rounding to 3 significant figures: Final answer: $\boxed{\log_7 200 \approx 2.72}$

QUESTION 30 4 marks

Medium

Solve the equation $\log_3(x+2) = 4$ for $x$.

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Using the definition of a logarithm, $\log_a m = n$ means $m = a^n$. $\log_3(x+2) = 4$ means $x + 2 = 3^4$ $x + 2 = 81$ $x = 79$ Check: $x+2 = 81 > 0$, so this is in the domain of $\log_3(x+2)$, and the solution is valid. Final answer: $\boxed{x = 79}$

QUESTION 31 4 marks

Medium

Express $\log_a 20 + \log_a 5 - \log_a 4$ as a single logarithm in the form $\log_a k$, where $k \in \mathbb{Z}$.

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Using the addition and subtraction laws of logarithms, $\log_a m + \log_a n = \log_a(mn)$ and $\log_a m - \log_a n = \log_a\left(\dfrac{m}{n}\right)$. $\log_a 20 + \log_a 5 - \log_a 4 = \log_a(20 \times 5) - \log_a 4$ $= \log_a 100 - \log_a 4$ $= \log_a\left(\dfrac{100}{4}\right)$ $= \log_a 25$ Final answer: $\boxed{\log_a 25}$

QUESTION 32 5 marks

Medium

Given that $\log_5 2 = p$, express $\log_5 500$ in terms of $p$.

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Using the fact that $500$ can be written using a factor of $2$ and a factor of $5$, write $500 = 4 \times 125 = 2^{2} \times 5^{3}$. $\log_5 500 = \log_5(2^{2} \times 5^{3})$ Using the addition law of logarithms: $= \log_5(2^{2}) + \log_5(5^{3})$ Using the power law of logarithms, $\log_a(m^{k}) = k \log_a m$: $= 2\log_5 2 + 3\log_5 5$ Since $\log_5 5 = 1$ and $\log_5 2 = p$: $= 2p + 3(1) = 2p + 3$ Final answer: $\boxed{\log_5 500 = 2p + 3}$

QUESTION 33 5 marks

Medium

Solve the equation $4^{x} = 3^{x+1}$, giving your answer correct to 3 significant figures.

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Taking logarithms of both sides (natural logarithms are used here, though any base could be chosen): $\ln(4^{x}) = \ln(3^{x+1})$ Using the power law of logarithms: $x \ln 4 = (x+1)\ln 3$ Expand the right-hand side: $x \ln 4 = x \ln 3 + \ln 3$ Collect the $x$ terms on one side: $x \ln 4 - x \ln 3 = \ln 3$ $x(\ln 4 - \ln 3) = \ln 3$ $x = \dfrac{\ln 3}{\ln 4 - \ln 3}$ Using a GDC: $\ln 3 \approx 1.098612$, $\ln 4 \approx 1.386294$ $x = \dfrac{1.098612}{1.386294 - 1.098612} = \dfrac{1.098612}{0.287682} \approx 3.81884$ Rounding to 3 significant figures: Final answer: $\boxed{x \approx 3.82}$

QUESTION 34 5 marks

Medium

Solve the equation $\log_2 x + \log_2(x-2) = 3$ for $x$.

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Using the addition law of logarithms, $\log_a m + \log_a n = \log_a(mn)$. $\log_2 x + \log_2(x-2) = \log_2[x(x-2)] = 3$ Convert to exponential form: $x(x-2) = 2^3 = 8$ Expand and rearrange into a quadratic equation: $x^2 - 2x - 8 = 0$ Factorise: $(x-4)(x+2) = 0$ $x = 4$ or $x = -2$ Check the domain: both $x > 0$ and $x - 2 > 0$ are required, so $x > 2$. $x = 4$ satisfies $x > 2$, so it is valid. $x = -2$ does not satisfy $x > 2$ (it makes both logarithms undefined), so it is rejected. Final answer: $\boxed{x = 4}$

QUESTION 35 4 marks

Medium

Simplify $\dfrac{\sqrt{x} \cdot x^{3}}{x^{-\frac{1}{2}}}$, giving your answer in the form $x^{k}$.

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Using the connection between roots and rational exponents, $\sqrt{x} = x^{\frac{1}{2}}$. Numerator: $x^{\frac{1}{2}} \times x^{3} = x^{\frac{1}{2}+3} = x^{\frac{7}{2}}$ Now divide by $x^{-\frac{1}{2}}$, using the law of exponents for division: $x^{\frac{7}{2}} \div x^{-\frac{1}{2}} = x^{\frac{7}{2}-\left(-\frac{1}{2}\right)} = x^{\frac{7}{2}+\frac{1}{2}} = x^{\frac{8}{2}} = x^{4}$ Final answer: $\boxed{x^{4}}$

QUESTION 36 4 marks

Medium

Without using a calculator, find the exact value of $\log_{27} 9$.

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Let $\log_{27} 9 = y$. By the definition of a logarithm, this means $27^{y} = 9$. Write both $27$ and $9$ as powers of $3$: $27 = 3^{3}$ and $9 = 3^{2}$. $(3^{3})^{y} = 3^{2}$ $3^{3y} = 3^{2}$ Since the bases are equal, the exponents must be equal: $3y = 2$ $y = \dfrac{2}{3}$ Final answer: $\boxed{\log_{27} 9 = \dfrac{2}{3}}$

QUESTION 37 3 marks

Medium

Solve the equation $2^{-x} = 0.125$ for $x$.

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Using the fact that both sides can be written as powers of the same base, express $0.125$ as a power of $2$. $0.125 = \dfrac{1}{8} = \dfrac{1}{2^3} = 2^{-3}$ So the equation becomes: $2^{-x} = 2^{-3}$ Since the bases are equal, the exponents must be equal: $-x = -3$ Final answer: $\boxed{x = 3}$

QUESTION 38 5 marks

Medium

The pH of a solution is defined by $\text{pH} = -\log_{10}[\text{H}^{+}]$, where $[\text{H}^{+}]$ is the hydrogen ion concentration measured in mol dm$^{-3}$. A solution has $\text{pH} = 4.5$. Find $[\text{H}^{+}]$, giving your answer in the form $a \times 10^{k}$, where $1 \le a < 10$, correct to 3 significant figures.

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Using the given formula, substitute $\text{pH} = 4.5$: $4.5 = -\log_{10}[\text{H}^{+}]$ Multiply both sides by $-1$: $\log_{10}[\text{H}^{+}] = -4.5$ Convert to exponential form, using the definition of a logarithm ($\log_a m = n$ means $m = a^n$): $[\text{H}^{+}] = 10^{-4.5}$ Using a GDC: $10^{-4.5} \approx 0.0000316228$ Rewrite in the form $a \times 10^{k}$: $0.0000316228 \approx 3.16228 \times 10^{-5}$ Rounding to 3 significant figures: Final answer: $\boxed{[\text{H}^{+}] \approx 3.16 \times 10^{-5} \text{ mol dm}^{-3}}$

QUESTION 39 4 marks

Medium

Solve the equation $9^{x+1} = 27^{x-1}$ for $x$, giving an exact value.

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Using the fact that both $9$ and $27$ can be written as powers of $3$, rewrite both sides with base $3$. $9 = 3^{2}$, so $9^{x+1} = (3^{2})^{x+1} = 3^{2(x+1)} = 3^{2x+2}$ $27 = 3^{3}$, so $27^{x-1} = (3^{3})^{x-1} = 3^{3(x-1)} = 3^{3x-3}$ The equation becomes: $3^{2x+2} = 3^{3x-3}$ Since the bases are equal, the exponents must be equal: $2x + 2 = 3x - 3$ $2 + 3 = 3x - 2x$ Final answer: $\boxed{x = 5}$

QUESTION 40 5 marks

Medium

Consider the expression $\left(\dfrac{x^{4}}{x^{-2}}\right)^{\frac{1}{3}}$. (a) Simplify the expression, giving your answer in the form $x^{k}$. (b) Hence, or otherwise, solve the equation $\left(\dfrac{x^{4}}{x^{-2}}\right)^{\frac{1}{3}} = 8$ for $x$.

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Part (a): Using the law of exponents for division, $\dfrac{x^{4}}{x^{-2}} = x^{4-(-2)} = x^{6}$. Now apply the outer power of $\dfrac{1}{3}$, using the law for a power of a power: $\left(x^{6}\right)^{\frac{1}{3}} = x^{6 \times \frac{1}{3}} = x^{2}$ Part (a) final answer: $\boxed{x^{2}}$ Part (b): Using the result from part (a), the equation becomes: $x^{2} = 8$ Take the square root of both sides: $x = \pm\sqrt{8} = \pm 2\sqrt{2}$ Both values are valid, since the original expression is defined for all $x \ne 0$. Part (b) final answer: $\boxed{x = 2\sqrt{2} \text{ or } x = -2\sqrt{2}}$

QUESTION 41 4 marks

Medium

Solve the equation $\ln(2x - 1) = 3$ for $x$, giving your answer correct to 3 significant figures.

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Using the definition of the natural logarithm, $\ln m = n$ means $m = e^{n}$. $\ln(2x-1) = 3$ means $2x - 1 = e^{3}$ Using a GDC: $e^3 \approx 20.0855$ $2x - 1 = 20.0855$ $2x = 21.0855$ $x = 10.5428$ Check: $2x-1 = 20.0855 > 0$, so this is in the domain of $\ln(2x-1)$, and the solution is valid. Rounding to 3 significant figures: Final answer: $\boxed{x \approx 10.5}$

QUESTION 42 5 marks

Medium

The magnitude of an earthquake is given by $M = \log_{10}\left(\dfrac{I}{I_0}\right)$, where $I$ is the intensity of the earthquake and $I_0$ is a fixed reference intensity. Earthquake A has magnitude $M_A = 6.2$ and earthquake B has magnitude $M_B = 4.7$. Find the value of $\dfrac{I_A}{I_B}$, correct to 3 significant figures.

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Using the given formula, write an expression for the intensity of each earthquake. For earthquake A: $6.2 = \log_{10}\left(\dfrac{I_A}{I_0}\right)$, so $\dfrac{I_A}{I_0} = 10^{6.2}$ For earthquake B: $4.7 = \log_{10}\left(\dfrac{I_B}{I_0}\right)$, so $\dfrac{I_B}{I_0} = 10^{4.7}$ Now form the ratio $\dfrac{I_A}{I_B}$ by dividing: $\dfrac{I_A}{I_B} = \dfrac{I_A}{I_0} \div \dfrac{I_B}{I_0} = \dfrac{10^{6.2}}{10^{4.7}} = 10^{6.2-4.7} = 10^{1.5}$ Using a GDC: $10^{1.5} \approx 31.6228$ Rounding to 3 significant figures: Final answer: $\boxed{\dfrac{I_A}{I_B} \approx 31.6}$

QUESTION 43 4 marks

Medium

Simplify $\log_2 x^{3} - \log_2 \sqrt{x}$, giving your answer in the form $k\log_2 x$, where $k \in \mathbb{Q}$.

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Using the power law of logarithms, $\log_a(m^{n}) = n\log_a m$. $\log_2 x^{3} = 3\log_2 x$ Since $\sqrt{x} = x^{\frac{1}{2}}$: $\log_2 \sqrt{x} = \log_2 x^{\frac{1}{2}} = \dfrac{1}{2}\log_2 x$ So: $\log_2 x^{3} - \log_2 \sqrt{x} = 3\log_2 x - \dfrac{1}{2}\log_2 x = \dfrac{6}{2}\log_2 x - \dfrac{1}{2}\log_2 x = \dfrac{5}{2}\log_2 x$ Final answer: $\boxed{\dfrac{5}{2}\log_2 x}$

QUESTION 44 6 marks

Hard

Solve the equation $\log_2(x+3) + \log_2(x-3) = 4$ for $x$, rejecting any solution that is not valid.

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Using the addition law of logarithms, $\log_a m + \log_a n = \log_a(mn)$. $\log_2(x+3) + \log_2(x-3) = \log_2[(x+3)(x-3)] = 4$ Convert to exponential form: $(x+3)(x-3) = 2^4 = 16$ Using the difference of two squares: $x^2 - 9 = 16$ $x^2 = 25$ $x = 5$ or $x = -5$ Check the domain: both $x+3 > 0$ and $x-3 > 0$ are required, so $x > 3$. $x = 5$ satisfies $x > 3$, so it is valid. $x = -5$ does not satisfy $x > 3$ (both logarithms would be undefined), so it is rejected. Final answer: $\boxed{x = 5}$

QUESTION 45 7 marks

Hard

Solve simultaneously for $x$ and $y$: $2^{x} \times 4^{y} = 32$ and $5^{x} \div 5^{y} = 25$.

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Using the laws of exponents, rewrite each equation with matching bases. Equation 1: $2^{x} \times 4^{y} = 32$ Since $4 = 2^2$ and $32 = 2^5$: $2^{x} \times 2^{2y} = 2^{5}$ $2^{x+2y} = 2^{5}$ Since the bases are equal, the exponents must be equal: $x + 2y = 5 \quad (1)$ Equation 2: $5^{x} \div 5^{y} = 25$ Since $25 = 5^2$: $5^{x-y} = 5^{2}$ Since the bases are equal, the exponents must be equal: $x - y = 2 \quad (2)$ Solve equations (1) and (2) simultaneously. From (2): $x = y + 2$. Substitute into (1): $(y+2) + 2y = 5$ $3y + 2 = 5$ $3y = 3$ $y = 1$ Substitute back: $x = 1 + 2 = 3$ Check in (1): $3 + 2(1) = 5$. Correct. Final answer: $\boxed{x = 3, \ y = 1}$

QUESTION 46 6 marks

Hard

Solve the equation $4^{x} - 5(2^{x+1}) + 16 = 0$ for $x \in \mathbb{R}$, giving your answers as exact values.

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Using the laws of exponents, rewrite $4^x$ and $2^{x+1}$ in terms of $2^x$. $4^x = (2^2)^x = (2^x)^2$ $2^{x+1} = 2 \times 2^x$ Let $y = 2^x$ (a substitution to form a quadratic equation, noting $y > 0$). The equation becomes: $y^2 - 5(2y) + 16 = 0$ $y^2 - 10y + 16 = 0$ Factorise: $(y-2)(y-8) = 0$ $y = 2$ or $y = 8$ Both values satisfy $y > 0$, so both are valid. Now solve for $x$ in each case using $y = 2^x$: Case 1: $2^x = 2 = 2^1 \Rightarrow x = 1$ Case 2: $2^x = 8 = 2^3 \Rightarrow x = 3$ Final answer: $\boxed{x = 1 \text{ or } x = 3}$

QUESTION 47 6 marks

Hard

Solve the equation $\log_5(x+1) - \log_5(x-1) = \log_5 3$ for $x$, giving an exact value.

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Using the subtraction law of logarithms, $\log_a m - \log_a n = \log_a\left(\dfrac{m}{n}\right)$. $\log_5(x+1) - \log_5(x-1) = \log_5\left(\dfrac{x+1}{x-1}\right) = \log_5 3$ Since the logarithms (with the same base) are equal, their arguments must be equal: $\dfrac{x+1}{x-1} = 3$ Multiply both sides by $(x-1)$: $x + 1 = 3(x-1)$ $x + 1 = 3x - 3$ $1 + 3 = 3x - x$ $4 = 2x$ $x = 2$ Check the domain: both $x+1>0$ and $x-1>0$ are required, so $x>1$. Since $x=2$ satisfies $x>1$, the solution is valid. Final answer: $\boxed{x = 2}$

QUESTION 48 7 marks

Hard

Given that $\log_{10} 2 = m$ and $\log_{10} 3 = n$: (a) Express $\log_{10} 72$ in terms of $m$ and $n$. (b) Hence write down, in terms of $m$ and $n$, the exact solution to the equation $10^{y} = 72$. (c) Given further that $m \approx 0.301$ and $n \approx 0.477$, find the value of $y$ correct to 3 significant figures.

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Part (a): Write $72$ as a product of powers of $2$ and $3$: $72 = 8 \times 9 = 2^{3} \times 3^{2}$. $\log_{10} 72 = \log_{10}(2^{3} \times 3^{2})$ Using the addition law of logarithms: $= \log_{10}(2^{3}) + \log_{10}(3^{2})$ Using the power law of logarithms: $= 3\log_{10} 2 + 2\log_{10} 3$ $= 3m + 2n$ Part (a) final answer: $\boxed{\log_{10} 72 = 3m + 2n}$ Part (b): Using the definition of a logarithm, $10^{y} = 72$ means $y = \log_{10} 72$. Using the result from part (a): Part (b) final answer: $\boxed{y = 3m + 2n}$ Part (c): Substitute $m \approx 0.301$ and $n \approx 0.477$ into $y = 3m + 2n$: $y \approx 3(0.301) + 2(0.477)$ $= 0.903 + 0.954$ $= 1.857$ Rounding to 3 significant figures: Part (c) final answer: $\boxed{y \approx 1.86}$

QUESTION 49 6 marks

Hard

Solve the equation $x^{\frac{2}{3}} - 5x^{\frac{1}{3}} + 6 = 0$ for $x$, giving your answers as exact values.

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Using the laws of exponents, note that $x^{\frac{2}{3}} = \left(x^{\frac{1}{3}}\right)^{2}$. Let $u = x^{\frac{1}{3}}$ (a substitution to form a quadratic equation). The equation becomes: $u^{2} - 5u + 6 = 0$ Factorise: $(u-2)(u-3) = 0$ $u = 2$ or $u = 3$ Now solve for $x$ in each case using $u = x^{\frac{1}{3}}$, so $x = u^{3}$: Case 1: $u = 2 \Rightarrow x = 2^{3} = 8$ Case 2: $u = 3 \Rightarrow x = 3^{3} = 27$ Final answer: $\boxed{x = 8 \text{ or } x = 27}$

QUESTION 50 6 marks

Hard

Solve simultaneously for $x$ and $y$, where $x, y > 0$: $\log_2 x + \log_2 y = 4$ and $x = 4y$.

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Using the addition law of logarithms, $\log_a m + \log_a n = \log_a(mn)$. $\log_2 x + \log_2 y = \log_2(xy) = 4$ Convert to exponential form: $xy = 2^4 = 16 \quad (1)$ The second equation gives: $x = 4y \quad (2)$ Substitute (2) into (1): $(4y)(y) = 16$ $4y^2 = 16$ $y^2 = 4$ $y = 2$ or $y = -2$ Since $y > 0$ is required, reject $y = -2$. $y = 2$ Substitute back into (2): $x = 4(2) = 8$ Check: $x=8>0$ and $y=2>0$, so both are valid, and $\log_2 8 + \log_2 2 = 3 + 1 = 4$. Correct. Final answer: $\boxed{x = 8, \ y = 2}$

QUESTION 1 3 marks

Easy

Use Pascal's triangle to find the binomial expansion of $(x+2)^3$.

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Method: reading the coefficients for row $n=3$ of Pascal\'s triangle: $1, 3, 3, 1$. $(x+2)^3 = 1 \cdot x^3 + 3 \cdot x^2 \cdot (2) + 3 \cdot x \cdot (2)^2 + 1 \cdot (2)^3$ $= x^3 + 6x^2 + 3x(4) + 8$ $= x^3 + 6x^2 + 12x + 8$ Final answer: $$\boxed{(x+2)^3 = x^3+6x^2+12x+8}$$

QUESTION 2 3 marks

Easy

Find the coefficient of $x^2$ in the expansion of $(2x-3)^4$.

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Method: using the general term of the binomial expansion, $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$, for $(a+b)^n = (2x + (-3))^4$. General term: $T_{k+1} = \binom{4}{k}(2x)^{4-k}(-3)^k$ We need the power of $x$ to be 2, so $4-k=2 \Rightarrow k=2$. $T_3 = \binom{4}{2}(2x)^2(-3)^2 = 6 \times 4x^2 \times 9 = 216x^2$ Final answer: $$\boxed{216}$$

QUESTION 3 4 marks

Medium

In the expansion of $(1+x)^n$, where $n \in \mathbb{N}$, the coefficient of $x^3$ is 35. Find the value of $n$.

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Method: using the general binomial coefficient $\binom{n}{k}$ for the expansion of $(1+x)^n$; the coefficient of $x^3$ is $\binom{n}{3}$. $\binom{n}{3} = 35$ $\dfrac{n(n-1)(n-2)}{6} = 35$ $n(n-1)(n-2) = 210$ Testing consecutive integers: $7 \times 6 \times 5 = 210$. This works. Since $n(n-1)(n-2)$ is strictly increasing for $n \geq 3$, this is the only positive integer solution. Final answer: $$\boxed{n = 7}$$

QUESTION 4 5 marks

Medium

Find the term independent of $x$ (the constant term) in the expansion of $\left(x^2 + \dfrac{1}{x}\right)^{6}$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ with $a = x^2$, $b = \dfrac{1}{x}$, $n=6$, then finding the value of $k$ for which the power of $x$ is zero. General term: $T_{k+1} = \binom{6}{k}(x^2)^{6-k}(x^{-1})^{k} = \binom{6}{k}x^{12-2k-k} = \binom{6}{k}x^{12-3k}$ Set the exponent of $x$ to zero: $12-3k=0 \Rightarrow k=4$ Constant term $= \binom{6}{4} = 15$ Final answer: $$\boxed{15}$$

QUESTION 5 6 marks

Hard

In the expansion of $(1+ax)^7$, where $a \neq 0$, the coefficient of $x^3$ is twice the coefficient of $x^2$. Find the value of $a$.

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Method: writing the coefficients of $x^2$ and $x^3$ using $\binom{n}{k}a^k$, forming an equation from the given relationship, and solving for $a$. Coefficient of $x^2$: $\binom{7}{2}a^2 = 21a^2$ Coefficient of $x^3$: $\binom{7}{3}a^3 = 35a^3$ Given: $35a^3 = 2(21a^2)$ $35a^3 = 42a^2$ $35a^3 - 42a^2 = 0$ $7a^2(5a-6) = 0$ $a^2 = 0 \Rightarrow a = 0$ (rejected, since $a eq 0$ is given) or $5a - 6 = 0 \Rightarrow a = \dfrac{6}{5}$ Final answer: $$\boxed{a = \dfrac{6}{5} = 1.2}$$

QUESTION 6 4 marks

Easy

Use Pascal's triangle to find the binomial expansion of $(x+3)^4$.

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Method: reading the coefficients for row $n=4$ of Pascal's triangle: $1, 4, 6, 4, 1$. $(x+3)^4 = 1\cdot x^4 + 4\cdot x^3(3) + 6\cdot x^2(3)^2 + 4\cdot x(3)^3 + 1\cdot(3)^4$ $= x^4 + 12x^3 + 6x^2(9) + 4x(27) + 81$ $= x^4 + 12x^3 + 54x^2 + 108x + 81$ Final answer: $$\boxed{(x+3)^4 = x^4+12x^3+54x^2+108x+81}$$

QUESTION 7 4 marks

Easy

Use Pascal's triangle to find the binomial expansion of $(2x-1)^3$.

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Method: reading the coefficients for row $n=3$ of Pascal's triangle: $1, 3, 3, 1$, applied to $(2x + (-1))^3$. $(2x-1)^3 = 1\cdot(2x)^3 + 3\cdot(2x)^2(-1) + 3\cdot(2x)(-1)^2 + 1\cdot(-1)^3$ $= 8x^3 + 3(4x^2)(-1) + 3(2x)(1) + (-1)$ $= 8x^3 - 12x^2 + 6x - 1$ Final answer: $$\boxed{(2x-1)^3 = 8x^3-12x^2+6x-1}$$

QUESTION 8 4 marks

Easy

Write down the row of Pascal's triangle corresponding to $n=5$, and hence expand $(1+x)^5$.

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Method: each row of Pascal's triangle is formed by adding adjacent entries of the row above, starting from row $n=0$: $1$; $1,1$; $1,2,1$; $1,3,3,1$; $1,4,6,4,1$. Row $n=5$: $1, 5, 10, 10, 5, 1$ Using these as the coefficients of $(1+x)^5 = 1^5 + 5(1)^4x + 10(1)^3x^2 + 10(1)^2x^3 + 5(1)x^4 + x^5$: $(1+x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5$ Final answer: $$\boxed{(1+x)^5 = 1+5x+10x^2+10x^3+5x^4+x^5}$$

QUESTION 9 3 marks

Easy

Find the coefficient of $x^3$ in the expansion of $(x+2)^5$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ for $(a+b)^n = (x+2)^5$. General term: $T_{k+1} = \binom{5}{k}x^{5-k}(2)^k$ We need the power of $x$ to be 3, so $5-k=3 \Rightarrow k=2$. $T_3 = \binom{5}{2}x^3(2)^2 = 10 \times x^3 \times 4 = 40x^3$ Final answer: $$\boxed{40}$$

QUESTION 10 4 marks

Easy

Find the coefficient of $x^4$ in the expansion of $(3x-1)^6$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ for $(a+b)^6 = (3x+(-1))^6$. General term: $T_{k+1} = \binom{6}{k}(3x)^{6-k}(-1)^k$ We need the power of $x$ to be 4, so $6-k=4 \Rightarrow k=2$. $T_3 = \binom{6}{2}(3x)^4(-1)^2 = 15 \times 81x^4 \times 1 = 1215x^4$ Final answer: $$\boxed{1215}$$

QUESTION 11 3 marks

Easy

Find the term in $x^2$ in the expansion of $(2+x)^6$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ for $(a+b)^6 = (2+x)^6$. General term: $T_{k+1} = \binom{6}{k}(2)^{6-k}x^{k}$ We need the power of $x$ to be 2, so $k=2$. $T_3 = \binom{6}{2}(2)^4x^2 = 15 \times 16 \times x^2 = 240x^2$ Final answer: $$\boxed{240x^2}$$

QUESTION 12 4 marks

Easy

Expand and simplify $(a-2b)^4$.

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Method: reading the coefficients for row $n=4$ of Pascal's triangle: $1, 4, 6, 4, 1$, applied to $(a + (-2b))^4$. $(a-2b)^4 = a^4 + 4a^3(-2b) + 6a^2(-2b)^2 + 4a(-2b)^3 + (-2b)^4$ $= a^4 - 8a^3b + 6a^2(4b^2) - 4a(8b^3) + 16b^4$ $= a^4 - 8a^3b + 24a^2b^2 - 32ab^3 + 16b^4$ Final answer: $$\boxed{(a-2b)^4 = a^4-8a^3b+24a^2b^2-32ab^3+16b^4}$$

QUESTION 13 3 marks

Easy

Find the value of $\binom{8}{3}$, and hence state the coefficient of $x^3$ in the expansion of $(1+x)^8$.

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Method: computing $\binom{8}{3} = \dfrac{8!}{3!5!}$ directly, then noting that in $(1+x)^8$ the coefficient of $x^k$ is $\binom{8}{k}$ since $1^{8-k}=1$ for every term. $\binom{8}{3} = \dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} = \dfrac{336}{6} = 56$ In $(1+x)^8$, the general term is $T_{k+1} = \binom{8}{k}(1)^{8-k}x^k = \binom{8}{k}x^k$, so the coefficient of $x^3$ is simply $\binom{8}{3}$. Final answer: $$\boxed{\binom{8}{3} = 56, \text{ so the coefficient of } x^3 \text{ is } 56}$$

QUESTION 14 4 marks

Easy

Find the constant term in the expansion of $\left(x + \dfrac{2}{x}\right)^4$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ with $a=x$, $b=\dfrac{2}{x}$, $n=4$, then finding the value of $k$ for which the power of $x$ is zero. General term: $T_{k+1} = \binom{4}{k}x^{4-k}\left(\dfrac{2}{x}\right)^k = \binom{4}{k}2^k x^{4-k-k} = \binom{4}{k}2^k x^{4-2k}$ Set the exponent of $x$ to zero: $4-2k=0 \Rightarrow k=2$ Constant term $= \binom{4}{2}(2)^2 = 6 \times 4 = 24$ Final answer: $$\boxed{24}$$

QUESTION 15 3 marks

Easy

Determine the number of terms in the expansion of $(2x-5)^7$.

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Method: recalling that the binomial expansion of $(a+b)^n$ always has exactly $n+1$ terms, corresponding to $k=0,1,2,\ldots,n$ in the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^k$. Here $n=7$, so the number of terms is $n+1 = 7+1$. Final answer: $$\boxed{8 \text{ terms}}$$

QUESTION 16 3 marks

Easy

Find the coefficient of $x^5$ in the expansion of $(1+x)^9$.

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Method: in the expansion of $(1+x)^9$, the general term is $T_{k+1} = \binom{9}{k}(1)^{9-k}x^k = \binom{9}{k}x^k$, so the coefficient of $x^5$ is $\binom{9}{5}$. $\binom{9}{5} = \dfrac{9!}{5!4!} = \dfrac{9\times8\times7\times6}{4\times3\times2\times1} = \dfrac{3024}{24} = 126$ Final answer: $$\boxed{126}$$

QUESTION 17 4 marks

Easy

Use the binomial theorem to expand and simplify $(x-1)^5$.

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Method: reading the coefficients for row $n=5$ of Pascal's triangle: $1, 5, 10, 10, 5, 1$, applied to $(x+(-1))^5$. $(x-1)^5 = x^5 + 5x^4(-1) + 10x^3(-1)^2 + 10x^2(-1)^3 + 5x(-1)^4 + (-1)^5$ $= x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1$ Final answer: $$\boxed{(x-1)^5 = x^5-5x^4+10x^3-10x^2+5x-1}$$

QUESTION 18 4 marks

Easy

Find the coefficient of $x$ in the expansion of $(3+x)^4$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ for $(a+b)^4 = (3+x)^4$, with $k=1$ giving the term in $x$. General term: $T_{k+1} = \binom{4}{k}(3)^{4-k}x^{k}$ For the term in $x$, $k=1$. $T_2 = \binom{4}{1}(3)^3 x = 4 \times 27 \times x = 108x$ Final answer: $$\boxed{108}$$

QUESTION 19 4 marks

Easy

Use the binomial theorem to find the first three terms, in ascending powers of $x$, of the expansion of $(1+x)^7$.

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Method: using the general term $T_{k+1} = \binom{7}{k}(1)^{7-k}x^k = \binom{7}{k}x^k$ for $k=0,1,2$, since ascending powers of $x$ correspond to increasing $k$. $k=0$: $T_1 = \binom{7}{0}x^0 = 1$ $k=1$: $T_2 = \binom{7}{1}x = 7x$ $k=2$: $T_3 = \binom{7}{2}x^2 = 21x^2$ Final answer: $$\boxed{(1+x)^7 = 1 + 7x + 21x^2 + \ldots}$$

QUESTION 20 4 marks

Easy

Use the binomial theorem to find the first three terms, in descending powers of $x$, of the expansion of $(x+2)^6$.

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Method: using the general term $T_{k+1} = \binom{6}{k}x^{6-k}(2)^k$ for $k=0,1,2$, since descending powers of $x$ correspond to increasing $k$ starting from $x^6$. $k=0$: $T_1 = \binom{6}{0}x^6(2)^0 = x^6$ $k=1$: $T_2 = \binom{6}{1}x^5(2)^1 = 6 \times x^5 \times 2 = 12x^5$ $k=2$: $T_3 = \binom{6}{2}x^4(2)^2 = 15 \times x^4 \times 4 = 60x^4$ Final answer: $$\boxed{(x+2)^6 = x^6 + 12x^5 + 60x^4 + \ldots}$$

QUESTION 21 4 marks

Easy

Find $\binom{10}{4}$, and hence write down the coefficient of $x^4$ in the expansion of $(1+x)^{10}$.

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Method: computing $\binom{10}{4} = \dfrac{10!}{4!6!}$, then noting that in $(1+x)^{10}$ the coefficient of $x^k$ equals $\binom{10}{k}$. $\binom{10}{4} = \dfrac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \dfrac{5040}{24} = 210$ Since $(1+x)^{10}$ has general term $\binom{10}{k}(1)^{10-k}x^k = \binom{10}{k}x^k$, the coefficient of $x^4$ is $\binom{10}{4}$. Final answer: $$\boxed{\binom{10}{4} = 210, \text{ so the coefficient of } x^4 \text{ is } 210}$$

QUESTION 22 4 marks

Easy

Find the coefficient of $x^2y^3$ in the expansion of $(x+y)^5$.

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Method: using the general term of the two-variable binomial expansion $T_{k+1} = \binom{5}{k}x^{5-k}y^{k}$, and finding the value of $k$ that gives $y^3$ (and hence $x^2$). General term: $T_{k+1} = \binom{5}{k}x^{5-k}y^{k}$ We need $y^k = y^3 \Rightarrow k=3$, which correctly gives $x^{5-3}=x^2$. $T_4 = \binom{5}{3}x^2y^3 = 10x^2y^3$ Final answer: $$\boxed{10}$$

QUESTION 23 5 marks

Easy

Evaluate $\binom{6}{2}$ and $\binom{6}{3}$. Hence, using Pascal's rule $\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}$, state the value of $\binom{7}{3}$.

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Method: computing each binomial coefficient directly, then applying Pascal's rule with $n=6$, $k=2$. $\binom{6}{2} = \dfrac{6\times5}{2\times1} = 15$ $\binom{6}{3} = \dfrac{6\times5\times4}{3\times2\times1} = 20$ Pascal's rule states $\binom{6}{2}+\binom{6}{3} = \binom{7}{3}$, since each entry of Pascal's triangle is the sum of the two entries above it. $\binom{7}{3} = 15 + 20 = 35$ Final answer: $$\boxed{\binom{6}{2}=15,\ \binom{6}{3}=20,\ \binom{7}{3}=35}$$

QUESTION 24 8 marks

Medium

Find the constant term in the expansion of $\left(2x^2 - \dfrac{1}{x}\right)^9$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ with $a=2x^2$, $b=-\dfrac{1}{x}$, $n=9$, then finding the value of $k$ for which the power of $x$ is zero. General term: $T_{k+1} = \binom{9}{k}(2x^2)^{9-k}\left(-\dfrac{1}{x}\right)^k = \binom{9}{k}2^{9-k}(-1)^k x^{2(9-k)}x^{-k}$ Simplify the power of $x$: $x^{18-2k-k} = x^{18-3k}$ Set the exponent of $x$ to zero: $18-3k=0 \Rightarrow k=6$ Now substitute $k=6$ into the coefficient part: $\binom{9}{6} = \binom{9}{3} = \dfrac{9\times8\times7}{3\times2\times1} = 84$ $2^{9-6} = 2^3 = 8$ $(-1)^6 = 1$ Constant term $= 84 \times 8 \times 1 = 672$ Final answer: $$\boxed{672}$$

QUESTION 25 5 marks

Medium

In the expansion of $(1+x)^n$, where $n \in \mathbb{N}$, the coefficient of $x^2$ is 45. Find the value of $n$.

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Method: using the general binomial coefficient $\binom{n}{2}$ for the coefficient of $x^2$ in $(1+x)^n$, forming an equation, and solving for $n$. $\binom{n}{2} = 45$ $\dfrac{n(n-1)}{2} = 45$ $n(n-1) = 90$ $n^2 - n - 90 = 0$ $(n-10)(n+9) = 0$ $n = 10$ or $n = -9$ Since $n \in \mathbb{N}$, reject $n=-9$. Final answer: $$\boxed{n = 10}$$

QUESTION 26 6 marks

Medium

Find the coefficient of $x^5$ in the expansion of $(2-3x)^8$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ for $(a+b)^8 = (2+(-3x))^8$. General term: $T_{k+1} = \binom{8}{k}(2)^{8-k}(-3x)^k$ We need the power of $x$ to be 5, so $k=5$. $T_6 = \binom{8}{5}(2)^3(-3)^5 x^5$ $\binom{8}{5} = \binom{8}{3} = \dfrac{8\times7\times6}{3\times2\times1} = 56$ $(2)^3 = 8$ $(-3)^5 = -243$ $T_6 = 56 \times 8 \times (-243) \, x^5 = -108864x^5$ Final answer: $$\boxed{-108864}$$

QUESTION 27 5 marks

Medium

The coefficient of $x^3$ in the expansion of $(1+kx)^6$ is 160. Find the value of $k$.

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Method: writing the coefficient of $x^3$ using $\binom{n}{j}$ notation, forming an equation, and solving for the unknown constant $k$ (using $j$ for the summation index to avoid confusion with $k$). General term: $T_{j+1} = \binom{6}{j}(1)^{6-j}(kx)^{j} = \binom{6}{j}k^{j}x^{j}$ For the term in $x^3$, $j=3$: coefficient $= \binom{6}{3}k^3$ $\binom{6}{3} = 20$ $20k^3 = 160$ $k^3 = 8$ $k = \sqrt[3]{8} = 2$ Final answer: $$\boxed{k = 2}$$

QUESTION 28 6 marks

Medium

By expanding $(1+x)^6$ and $(1-x)$ separately, find the coefficient of $x^3$ in the expansion of $(1+x)^6(1-x)$.

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Method: expanding $(1+x)^6$ fully using Pascal's triangle (row $n=6$: $1,6,15,20,15,6,1$), then multiplying by $(1-x)$ and collecting the terms that produce $x^3$. $(1+x)^6 = 1 + 6x + 15x^2 + 20x^3 + 15x^4 + 6x^5 + x^6$ Multiplying by $(1-x)$, the coefficient of $x^3$ in the product comes from two sources: (i) the $x^3$ term of $(1+x)^6$ multiplied by the constant term $1$ of $(1-x)$: $20 \times 1 = 20$ (ii) the $x^2$ term of $(1+x)^6$ multiplied by the $-x$ term of $(1-x)$: $15 \times (-1) = -15$ Adding these contributions: $20 + (-15) = 5$ Final answer: $$\boxed{5}$$

QUESTION 29 5 marks

Medium

Find the term independent of $x$ in the expansion of $\left(3x - \dfrac{2}{x^2}\right)^6$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ with $a=3x$, $b=-\dfrac{2}{x^2}$, $n=6$, then finding the value of $k$ for which the power of $x$ is zero. General term: $T_{k+1} = \binom{6}{k}(3x)^{6-k}\left(-\dfrac{2}{x^2}\right)^k = \binom{6}{k}3^{6-k}(-2)^k x^{6-k-2k} = \binom{6}{k}3^{6-k}(-2)^k x^{6-3k}$ Set the exponent of $x$ to zero: $6-3k=0 \Rightarrow k=2$ $\binom{6}{2} = 15$ $3^{6-2} = 3^4 = 81$ $(-2)^2 = 4$ Term independent of $x$ $= 15 \times 81 \times 4 = 4860$ Final answer: $$\boxed{4860}$$

QUESTION 30 8 marks

Medium

Find the coefficient of $x^4$ in the expansion of $(1+2x)^5(1-x)^3$.

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Method: expanding each bracket separately using Pascal's triangle, then multiplying the two expansions and collecting the terms whose powers of $x$ sum to 4. $(1+2x)^5$: using row $n=5$ ($1,5,10,10,5,1$) with $a=1$, $b=2x$: $(1+2x)^5 = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5$ $(1-x)^3$: using row $n=3$ ($1,3,3,1$) with $a=1$, $b=-x$: $(1-x)^3 = 1 - 3x + 3x^2 - x^3$ The coefficient of $x^4$ in the product is found by pairing terms whose exponents add to 4: $x^1 \times x^3$: $10 \times (-1) = -10$ $x^2 \times x^2$: $40 \times 3 = 120$ $x^3 \times x^1$: $80 \times (-3) = -240$ $x^4 \times x^0$: $80 \times 1 = 80$ Summing these contributions: $-10 + 120 - 240 + 80 = -50$ Final answer: $$\boxed{-50}$$

QUESTION 31 8 marks

Medium

The first three terms, in ascending powers of $x$, in the expansion of $(1+ax)^n$ are $1$, $12x$ and $60x^2$. Find the values of $a$ and $n$.

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Method: writing the first three terms of the general expansion in terms of $a$ and $n$, then equating coefficients with the given terms to form two equations and solving simultaneously. General expansion: $(1+ax)^n = \binom{n}{0} + \binom{n}{1}ax + \binom{n}{2}a^2x^2 + \ldots = 1 + nax + \dfrac{n(n-1)}{2}a^2x^2 + \ldots$ Equating the coefficient of $x$: $na = 12 \quad (1)$ Equating the coefficient of $x^2$: $\dfrac{n(n-1)}{2}a^2 = 60 \quad (2)$ From (1): $a = \dfrac{12}{n}$ Substituting into (2): $\dfrac{n(n-1)}{2}\left(\dfrac{12}{n}\right)^2 = 60$ $\dfrac{n(n-1)}{2} \times \dfrac{144}{n^2} = 60$ $\dfrac{72(n-1)}{n} = 60$ $72(n-1) = 60n$ $72n - 72 = 60n$ $12n = 72$ $n = 6$ Substituting back into (1): $6a = 12 \Rightarrow a = 2$ Check with (2): $\binom{6}{2}(2)^2 = 15 \times 4 = 60$. Correct. Final answer: $$\boxed{a = 2, \ n = 6}$$

QUESTION 32 5 marks

Medium

Find the term in $x^3$ in the expansion of $\left(x - \dfrac{2}{x}\right)^7$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ with $a=x$, $b=-\dfrac{2}{x}$, $n=7$, then finding the value of $k$ for which the power of $x$ is 3. General term: $T_{k+1} = \binom{7}{k}x^{7-k}\left(-\dfrac{2}{x}\right)^k = \binom{7}{k}(-2)^k x^{7-2k}$ Set the exponent of $x$ to 3: $7-2k=3 \Rightarrow k=2$ $\binom{7}{2} = 21$ $(-2)^2 = 4$ Term $= 21 \times 4 \, x^3 = 84x^3$ Final answer: $$\boxed{84x^3}$$

QUESTION 33 5 marks

Medium

In the expansion of $(1+2x)^n$, the coefficient of $x^2$ is four times the coefficient of $x$. Find the value of $n$.

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Method: expressing the coefficient of $x$ and the coefficient of $x^2$ using $\binom{n}{k}$, forming an equation from the given relationship, and solving for $n$. Coefficient of $x$: $\binom{n}{1}(2)^1 = 2n$ Coefficient of $x^2$: $\binom{n}{2}(2)^2 = \dfrac{n(n-1)}{2}\times 4 = 2n(n-1)$ Given: coefficient of $x^2$ = 4 $\times$ coefficient of $x$ $2n(n-1) = 4(2n)$ $2n(n-1) = 8n$ Dividing both sides by $2n$ (since $n \neq 0$): $n - 1 = 4$ $n = 5$ Final answer: $$\boxed{n = 5}$$

QUESTION 34 5 marks

Medium

Find the sum of the coefficients in the expansion of $(3x-2)^5$.

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Method: the sum of all coefficients in the expansion of a polynomial in $x$ is found by substituting $x=1$, since every power of $x$ then equals 1. Sum of coefficients $= (3(1)-2)^5 = (1)^5 = 1$ This can be verified using the general term: $(3x-2)^5 = \displaystyle\sum_{k=0}^{5}\binom{5}{k}(3x)^{5-k}(-2)^k$. Substituting $x=1$ gives $\displaystyle\sum_{k=0}^{5}\binom{5}{k}(3)^{5-k}(-2)^k = (3-2)^5 = 1$, confirming the result without needing to expand every term individually. Final answer: $$\boxed{1}$$

QUESTION 35 5 marks

Medium

Given that the coefficient of $x^2$ in the expansion of $(1+2x)^n$ is 84, find the value of $n$.

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Method: writing the coefficient of $x^2$ using $\binom{n}{2}$, forming an equation, and solving for the positive integer $n$. Coefficient of $x^2$: $\binom{n}{2}(2)^2 = 4\binom{n}{2}$ $4\binom{n}{2} = 84$ $\binom{n}{2} = 21$ $\dfrac{n(n-1)}{2} = 21$ $n(n-1) = 42$ $n^2 - n - 42 = 0$ $(n-7)(n+6) = 0$ $n = 7$ or $n = -6$ Since $n$ must be a positive integer, reject $n=-6$. Final answer: $$\boxed{n = 7}$$

QUESTION 36 6 marks

Medium

Find the constant term in the expansion of $\left(x^2 - \dfrac{3}{x}\right)^6$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ with $a=x^2$, $b=-\dfrac{3}{x}$, $n=6$, then finding the value of $k$ for which the power of $x$ is zero. General term: $T_{k+1} = \binom{6}{k}(x^2)^{6-k}\left(-\dfrac{3}{x}\right)^k = \binom{6}{k}(-3)^k x^{12-2k-k} = \binom{6}{k}(-3)^k x^{12-3k}$ Set the exponent of $x$ to zero: $12-3k=0 \Rightarrow k=4$ $\binom{6}{4} = \binom{6}{2} = 15$ $(-3)^4 = 81$ Constant term $= 15 \times 81 = 1215$ Final answer: $$\boxed{1215}$$

QUESTION 37 5 marks

Medium

In the expansion of $(1+x)^n$, the ratio of the coefficient of $x^3$ to the coefficient of $x^2$ is $4:1$. Find the value of $n$.

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Method: writing the two coefficients as binomial coefficients, forming the ratio equation, and solving for $n$. Coefficient of $x^3$: $\binom{n}{3}$ Coefficient of $x^2$: $\binom{n}{2}$ $\dfrac{\binom{n}{3}}{\binom{n}{2}} = 4$ Using $\dfrac{\binom{n}{3}}{\binom{n}{2}} = \dfrac{n-2}{3}$ (a standard simplification, since $\binom{n}{3} = \binom{n}{2}\times\dfrac{n-2}{3}$): $\dfrac{n-2}{3} = 4$ $n-2 = 12$ $n = 14$ Final answer: $$\boxed{n = 14}$$

QUESTION 38 5 marks

Medium

Find the coefficient of $x^7$ in the expansion of $\left(x^3 - \dfrac{2}{x}\right)^5$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ with $a=x^3$, $b=-\dfrac{2}{x}$, $n=5$, then finding the value of $k$ for which the power of $x$ is 7. General term: $T_{k+1} = \binom{5}{k}(x^3)^{5-k}\left(-\dfrac{2}{x}\right)^k = \binom{5}{k}(-2)^k x^{15-3k-k} = \binom{5}{k}(-2)^k x^{15-4k}$ Set the exponent of $x$ to 7: $15-4k=7 \Rightarrow k=2$ $\binom{5}{2} = 10$ $(-2)^2 = 4$ Coefficient $= 10 \times 4 = 40$ Final answer: $$\boxed{40}$$

QUESTION 39 7 marks

Medium

By using the binomial expansion of $(1+x)^5$, find the value of $(1.02)^5$ correct to 4 decimal places.

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Method: expanding $(1+x)^5$ fully using Pascal's triangle, then substituting $x=0.02$ so that $1+x=1.02$, and evaluating each term. Row $n=5$ of Pascal's triangle: $1,5,10,10,5,1$ $(1+x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5$ Substituting $x=0.02$: $5x = 5(0.02) = 0.1$ $10x^2 = 10(0.0004) = 0.004$ $10x^3 = 10(0.000008) = 0.00008$ $5x^4 = 5(0.00000016) = 0.0000008$ $x^5 = 0.0000000032$ (negligible at this precision) Adding all terms: $1 + 0.1 + 0.004 + 0.00008 + 0.0000008 + \ldots = 1.1040808\ldots$ Rounding to 4 decimal places: Final answer: $$\boxed{(1.02)^5 \approx 1.1041}$$

QUESTION 40 5 marks

Medium

Given that the coefficient of $x^3$ in the expansion of $(2+kx)^5$ is 320, find the value of $k$.

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Method: using the general term $T_{j+1} = \binom{5}{j}(2)^{5-j}(kx)^{j}$, setting $j=3$ for the term in $x^3$, forming an equation, and solving for $k$. General term: $T_{j+1} = \binom{5}{j}(2)^{5-j}k^{j}x^{j}$ For the term in $x^3$, $j=3$: coefficient $= \binom{5}{3}(2)^2k^3$ $\binom{5}{3} = 10$ $10 \times 4 \times k^3 = 320$ $40k^3 = 320$ $k^3 = 8$ $k = \sqrt[3]{8} = 2$ Final answer: $$\boxed{k = 2}$$

QUESTION 41 8 marks

Medium

Use the binomial theorem to find the first four terms, in ascending powers of $x$, of the expansion of $(1-2x)^7$. Hence find an approximate value for $(0.98)^7$, correct to 4 decimal places.

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Method: expanding $(1-2x)^7$ using the general term $T_{k+1} = \binom{7}{k}(1)^{7-k}(-2x)^k$ for $k=0,1,2,3$, then substituting a suitable value of $x$ so that $1-2x = 0.98$. $k=0$: $T_1 = \binom{7}{0}(-2x)^0 = 1$ $k=1$: $T_2 = \binom{7}{1}(-2x)^1 = 7(-2x) = -14x$ $k=2$: $T_3 = \binom{7}{2}(-2x)^2 = 21(4x^2) = 84x^2$ $k=3$: $T_4 = \binom{7}{3}(-2x)^3 = 35(-8x^3) = -280x^3$ So $(1-2x)^7 \approx 1 - 14x + 84x^2 - 280x^3$ To find $(0.98)^7$, set $1-2x = 0.98 \Rightarrow x = 0.01$: $-14x = -14(0.01) = -0.14$ $84x^2 = 84(0.0001) = 0.0084$ $-280x^3 = -280(0.000001) = -0.00028$ Adding: $1 - 0.14 + 0.0084 - 0.00028 = 0.86812$ Final answer: $$\boxed{(0.98)^7 \approx 0.8681}$$

QUESTION 42 8 marks

Hard

Consider the expansion of $(2x-1)^7$. (a) Write down the general term $T_{k+1}$ of the expansion. [2] (b) Find the term in $x^4$. [3] (c) By substituting a suitable value of $x$, find the sum of the coefficients of all the terms in the expansion. [3]

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Part (a): Method: using $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ with $a=2x$, $b=-1$, $n=7$. $$T_{k+1} = \binom{7}{k}(2x)^{7-k}(-1)^{k}$$ Part (b): We need the power of $x$ to be 4, so $7-k=4 \Rightarrow k=3$. $T_4 = \binom{7}{3}(2x)^4(-1)^3 = 35 \times 16x^4 \times (-1) = -560x^4$ Final answer to (b): $$\boxed{-560x^4}$$ Part (c): Method: the sum of the coefficients of a polynomial expansion in $x$ is obtained by substituting $x=1$. Sum of coefficients $= (2(1)-1)^7 = (1)^7 = 1$ Final answer to (c): $$\boxed{1}$$

QUESTION 43 8 marks

Hard

In the expansion of $(1+ax)^7$, where $a \neq 0$, the coefficient of $x^2$ is 84. (a) Find the possible values of $a$. [5] (b) For the positive value of $a$ found in part (a), find the coefficient of $x^3$. [3]

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Part (a): Method: writing the coefficient of $x^2$ as $\binom{7}{2}a^2$, forming an equation, and solving for $a$. Coefficient of $x^2$: $\binom{7}{2}a^2 = 21a^2$ $21a^2 = 84$ $a^2 = 4$ $a = \pm 2$ Final answer to (a): $$\boxed{a = 2 \text{ or } a = -2}$$ Part (b): Using the positive value $a=2$: Coefficient of $x^3$: $\binom{7}{3}a^3 = 35 \times (2)^3 = 35 \times 8 = 280$ Final answer to (b): $$\boxed{280}$$

QUESTION 44 7 marks

Hard

Find the constant term in the expansion of $\left(\sqrt{x} - \dfrac{3}{x}\right)^{12}$.

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Method: using the general term $T_{k+1} = \binom{n}{k}a^{n-k}b^{k}$ with $a=\sqrt{x}=x^{1/2}$, $b=-\dfrac{3}{x}=-3x^{-1}$, $n=12$, then finding the value of $k$ for which the total power of $x$ is zero. General term: $T_{k+1} = \binom{12}{k}\left(x^{1/2}\right)^{12-k}\left(-3x^{-1}\right)^{k} = \binom{12}{k}(-3)^{k}x^{\frac{12-k}{2}-k}$ Simplify the exponent of $x$: $\dfrac{12-k}{2} - k = \dfrac{12-k-2k}{2} = \dfrac{12-3k}{2}$ Set the exponent of $x$ to zero: $\dfrac{12-3k}{2}=0 \Rightarrow 12-3k=0 \Rightarrow k=4$ $\binom{12}{4} = \dfrac{12\times11\times10\times9}{4\times3\times2\times1} = \dfrac{11880}{24} = 495$ $(-3)^4 = 81$ Constant term $= 495 \times 81 = 40095$ Final answer: $$\boxed{40095}$$

QUESTION 45 9 marks

Hard

(a) Use the binomial theorem to expand $(1+x)^5$ fully. [3] (b) By substituting a suitable value of $x$ into your expansion, show that $\displaystyle\sum_{k=0}^{5}\binom{5}{k} = 32$. [3] (c) By substituting a different suitable value of $x$, show that $\displaystyle\sum_{k=0}^{5}(-1)^k\binom{5}{k} = 0$, and interpret this result in terms of the number of ways of choosing an even-sized versus an odd-sized subset from a set of 5 elements. [3]

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Part (a): Method: reading the coefficients for row $n=5$ of Pascal's triangle: $1,5,10,10,5,1$. $$(1+x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5$$ Part (b): Substituting $x=1$: every power of $x$ becomes 1, so the right-hand side becomes exactly the sum of the coefficients, which are the binomial coefficients $\binom{5}{k}$. $(1+1)^5 = 1+5+10+10+5+1$ $2^5 = 32$ $$\boxed{\sum_{k=0}^{5}\binom{5}{k} = 32}$$ Part (c): Substituting $x=-1$: $(1+(-1))^5 = 1 - 5 + 10 - 10 + 5 - 1$ $0^5 = 0$ $$\boxed{\sum_{k=0}^{5}(-1)^k\binom{5}{k} = 0}$$ Interpretation: since the terms with even $k$ (giving $+\binom{5}{k}$) and the terms with odd $k$ (giving $-\binom{5}{k}$) sum to zero, the sum of the even-sized subset counts $\binom{5}{0}+\binom{5}{2}+\binom{5}{4}$ must equal the sum of the odd-sized subset counts $\binom{5}{1}+\binom{5}{3}+\binom{5}{5}$: a set of 5 elements has exactly as many even-sized subsets as odd-sized subsets (16 each).

QUESTION 46 11 marks

Hard

In the expansion of $(1+x)^n$, the coefficients of the 2nd, 3rd and 4th terms are consecutive terms of an arithmetic sequence. (a) Write down expressions, in terms of $n$, for the coefficients of the 2nd, 3rd and 4th terms. [3] (b) Form an equation using the arithmetic sequence condition and solve it to find all possible values of $n$. [6] (c) Explain why one of your solutions from part (b) must be rejected, and state the valid value of $n$. [2]

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Part (a): Method: in $(1+x)^n$, the term $T_{r+1} = \binom{n}{r}x^r$, so the 2nd term ($r=1$) has coefficient $\binom{n}{1}$, the 3rd term ($r=2$) has coefficient $\binom{n}{2}$, and the 4th term ($r=3$) has coefficient $\binom{n}{3}$. 2nd term coefficient: $\binom{n}{1} = n$ 3rd term coefficient: $\binom{n}{2} = \dfrac{n(n-1)}{2}$ 4th term coefficient: $\binom{n}{3} = \dfrac{n(n-1)(n-2)}{6}$ Part (b): Method: three quantities $u_1, u_2, u_3$ are consecutive terms of an arithmetic sequence if $2u_2 = u_1+u_3$. Applying this to the coefficients found in part (a): $$2\binom{n}{2} = \binom{n}{1} + \binom{n}{3}$$ $$2 \times \dfrac{n(n-1)}{2} = n + \dfrac{n(n-1)(n-2)}{6}$$ $$n(n-1) = n + \dfrac{n(n-1)(n-2)}{6}$$ Dividing every term by $n$ (since $n \neq 0$): $$n-1 = 1 + \dfrac{(n-1)(n-2)}{6}$$ Multiplying every term by 6: $$6(n-1) = 6 + (n-1)(n-2)$$ $$6n - 6 = 6 + n^2 - 3n + 2$$ $$6n - 6 = n^2 - 3n + 8$$ $$0 = n^2 - 9n + 14$$ Factorising: $$(n-7)(n-2) = 0$$ $$n = 7 \text{ or } n = 2$$ Part (c): For the 4th term to exist in the expansion, we require $n \geq 3$. Since $n=2$ does not give a 4th term (the expansion of $(1+x)^2$ only has 3 terms), it must be rejected. Final answer: $$\boxed{n = 7}$$

QUESTION 47 11 marks

Hard

Find the coefficient of $x^5$ in the expansion of $(1+x)^4(2-x)^6$. (a) Expand $(1+x)^4$ fully. [2] (b) Write down, in terms of $x$, the terms of $(2-x)^6$ needed to combine with your answer to part (a) to produce a term in $x^5$. [4] (c) Hence find the coefficient of $x^5$ in the full product $(1+x)^4(2-x)^6$. [5]

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Part (a): Method: reading the coefficients for row $n=4$ of Pascal's triangle: $1,4,6,4,1$. $$(1+x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$$ Part (b): Since $(1+x)^4$ only contains powers of $x$ from $x^0$ to $x^4$, to obtain $x^5$ in the product we must pair each term $x^i$ from $(1+x)^4$ (for $i=0,1,2,3,4$) with the term $x^{5-i}$ from $(2-x)^6$. General term of $(2-x)^6$: $T_{k+1} = \binom{6}{k}(2)^{6-k}(-x)^k = \binom{6}{k}2^{6-k}(-1)^k x^k$ We therefore need the coefficients of $x^1, x^2, x^3, x^4, x^5$ in $(2-x)^6$: $k=1$: $\binom{6}{1}2^5(-1)^1 = 6\times32\times(-1) = -192$ $k=2$: $\binom{6}{2}2^4(-1)^2 = 15\times16\times1 = 240$ $k=3$: $\binom{6}{3}2^3(-1)^3 = 20\times8\times(-1) = -160$ $k=4$: $\binom{6}{4}2^2(-1)^4 = 15\times4\times1 = 60$ $k=5$: $\binom{6}{5}2^1(-1)^5 = 6\times2\times(-1) = -12$ Part (c): Method: multiply each coefficient of $(1+x)^4$ (from part (a): $1,4,6,4,1$ for $x^0,\ldots,x^4$) by the matching coefficient of $(2-x)^6$ found in part (b), so that the powers of $x$ sum to 5, then add the products. $x^0$(coeff $1$) $\times x^5$(coeff $-12$): $1\times(-12) = -12$ $x^1$(coeff $4$) $\times x^4$(coeff $60$): $4\times60 = 240$ $x^2$(coeff $6$) $\times x^3$(coeff $-160$): $6\times(-160) = -960$ $x^3$(coeff $4$) $\times x^2$(coeff $240$): $4\times240 = 960$ $x^4$(coeff $1$) $\times x^1$(coeff $-192$): $1\times(-192) = -192$ Summing all five contributions: $-12 + 240 - 960 + 960 - 192 = 36$ Final answer: $$\boxed{36}$$

QUESTION 48 10 marks

Hard

(a) Use the binomial theorem to find the first four terms, in ascending powers of $x$, of the expansion of $(1+3x)^{10}$. [4] (b) By substituting $x=0.01$ into your expansion, find an approximate value for $(1.03)^{10}$, correct to 3 decimal places. [4] (c) State one reason why using more terms of the expansion would improve the accuracy of the approximation found in part (b). [2]

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Part (a): Method: using the general term $T_{k+1} = \binom{10}{k}(1)^{10-k}(3x)^{k}$ for $k=0,1,2,3$. $k=0$: $T_1 = \binom{10}{0} = 1$ $k=1$: $T_2 = \binom{10}{1}(3x) = 10(3x) = 30x$ $k=2$: $T_3 = \binom{10}{2}(3x)^2 = 45(9x^2) = 405x^2$ $k=3$: $T_4 = \binom{10}{3}(3x)^3 = 120(27x^3) = 3240x^3$ $$(1+3x)^{10} \approx 1 + 30x + 405x^2 + 3240x^3$$ Part (b): Setting $1+3x = 1.03 \Rightarrow x = 0.01$: $30x = 30(0.01) = 0.3$ $405x^2 = 405(0.0001) = 0.0405$ $3240x^3 = 3240(0.000001) = 0.00324$ Adding: $1 + 0.3 + 0.0405 + 0.00324 = 1.34374$ Final answer to (b): $$\boxed{(1.03)^{10} \approx 1.344}$$ Part (c): The terms of the expansion that were omitted (from $T_5 = \binom{10}{4}(3x)^4$ onwards) are not zero; they contribute additional small positive amounts to the true value. Including more terms captures more of this contribution, reducing the truncation error and giving an approximation closer to the exact value of $(1.03)^{10}$.

QUESTION 49 7 marks

Hard

The coefficient of $x^3$ in the expansion of $(2+x)^n$ is twice the coefficient of $x^2$. Find the value of $n$.

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Method: writing both coefficients using $\binom{n}{k}$, forming an equation from the given relationship, and solving for $n$. Coefficient of $x^3$: $\binom{n}{3}(2)^{n-3}$ Coefficient of $x^2$: $\binom{n}{2}(2)^{n-2}$ Given: coefficient of $x^3$ = 2 $\times$ coefficient of $x^2$ $$\binom{n}{3}(2)^{n-3} = 2\binom{n}{2}(2)^{n-2}$$ $$\binom{n}{3}(2)^{n-3} = \binom{n}{2}(2)^{n-1}$$ Dividing both sides by $2^{n-3}$: $$\binom{n}{3} = \binom{n}{2}\times 2^{2} = 4\binom{n}{2}$$ Using $\dfrac{n(n-1)(n-2)}{6} = 4\times\dfrac{n(n-1)}{2}$: $$\dfrac{n(n-1)(n-2)}{6} = 2n(n-1)$$ Dividing both sides by $n(n-1)$ (since $n \neq 0, 1$): $$\dfrac{n-2}{6} = 2$$ $$n-2 = 12$$ $$n = 14$$ Final answer: $$\boxed{n = 14}$$

QUESTION 50 6 marks

Hard

Given that the coefficient of $x^n$ in the expansion of $(1+x)^{2n}$ is 924, find the value of $n$.

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Method: writing the coefficient of $x^n$ in $(1+x)^{2n}$ as $\binom{2n}{n}$, forming an equation, and identifying the value of $n$ that satisfies it. Coefficient of $x^n$ in $(1+x)^{2n}$: $\binom{2n}{n}$ $$\binom{2n}{n} = 924$$ Testing $2n=12$ (so $n=6$): $$\binom{12}{6} = \dfrac{12!}{6!6!} = \dfrac{12\times11\times10\times9\times8\times7}{6\times5\times4\times3\times2\times1} = \dfrac{665280}{720} = 924$$ This matches, and since $\binom{2n}{n}$ is strictly increasing for $n \geq 1$, this is the only solution. Final answer: $$\boxed{n = 6}$$

QUESTION 1 3 marks

Easy

Prove that $(n+1)^2 - (n-1)^2 = 4n$ for all integers $n$.

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Method: LHS \to RHS proof, expanding each squared bracket on the left-hand side. LHS $= (n+1)^2 - (n-1)^2$ Expand each square: $(n+1)^2 = n^2+2n+1$ and $(n-1)^2 = n^2-2n+1$. LHS $= (n^2+2n+1) - (n^2-2n+1)$ $= n^2+2n+1-n^2+2n-1$ $= 4n$ $=$ RHS. Since LHS $=$ RHS for all integers $n$, the identity is proven. $\blacksquare$ $$\boxed{(n+1)^2-(n-1)^2=4n}$$

QUESTION 2 3 marks

Easy

Prove that $(2n+1)^2 - (2n-1)^2 = 8n$ for all integers $n$.

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Method: LHS \to RHS proof using the difference of two squares, $a^2-b^2=(a-b)(a+b)$, with $a=2n+1$ and $b=2n-1$. LHS $= (2n+1)^2-(2n-1)^2$ $= [(2n+1)-(2n-1)][(2n+1)+(2n-1)]$ $= [2][4n]$ $= 8n$ $=$ RHS. Since LHS $=$ RHS for all integers $n$, the identity is proven. $\blacksquare$ $$\boxed{(2n+1)^2-(2n-1)^2=8n}$$

QUESTION 3 4 marks

Medium

Prove that $\dfrac{1}{n} - \dfrac{1}{n+2} = \dfrac{2}{n(n+2)}$ for $n \in \mathbb{R}$, $n \neq 0, -2$.

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Method: LHS \to RHS proof by combining the two fractions on the left-hand side over a common denominator. LHS $= \dfrac{1}{n} - \dfrac{1}{n+2}$ The common denominator is $n(n+2)$: $= \dfrac{n+2}{n(n+2)} - \dfrac{n}{n(n+2)}$ $= \dfrac{(n+2)-n}{n(n+2)}$ $= \dfrac{2}{n(n+2)}$ $=$ RHS. Since LHS $=$ RHS (for $n eq 0,-2$, so that all fractions are defined), the identity is proven. $\blacksquare$ $$\boxed{\dfrac{1}{n}-\dfrac{1}{n+2}=\dfrac{2}{n(n+2)}}$$

QUESTION 4 5 marks

Medium

(a) Prove that $(x+y)^2 - (x-y)^2 = 4xy$ for all real $x$ and $y$. (b) Hence find the value of $xy$, given that $(x+y)^2 = 50$ and $(x-y)^2 = 18$.

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(a) Method: LHS \to RHS proof by expanding both squared brackets. LHS $= (x+y)^2-(x-y)^2$ $(x+y)^2 = x^2+2xy+y^2$ and $(x-y)^2 = x^2-2xy+y^2$. LHS $= (x^2+2xy+y^2) - (x^2-2xy+y^2)$ $= 4xy$ $=$ RHS. So the identity is proven for all real $x,y$. $$\boxed{(x+y)^2-(x-y)^2=4xy}$$ (b) Method: substituting the given values into the identity proven in part (a). From (a): $4xy = (x+y)^2 - (x-y)^2 = 50 - 18 = 32$. So $xy = \dfrac{32}{4} = 8$. Final answer: $$\boxed{xy = 8}$$

QUESTION 5 6 marks

Hard

(a) Prove that $(n+3)^2 - (n-3)^2 = 12n$ for all real $n$. (b) Hence solve the equation $(x+3)^2 - (x-3)^2 = 180$.

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(a) Method: LHS \to RHS proof using the difference of two squares, $a^2-b^2=(a-b)(a+b)$, with $a=n+3$, $b=n-3$. LHS $= (n+3)^2-(n-3)^2$ $= [(n+3)-(n-3)][(n+3)+(n-3)]$ $= [6][2n]$ $= 12n$ $=$ RHS. So the identity is proven for all real $n$. $$\boxed{(n+3)^2-(n-3)^2=12n}$$ (b) Method: applying the identity from part (a) with $n=x$ to reduce the equation to a linear equation. From (a), $(x+3)^2-(x-3)^2 = 12x$. So the equation becomes $12x = 180$. $x = \dfrac{180}{12} = 15$. Final answer: $$\boxed{x = 15}$$

QUESTION 6 3 marks

Easy

Prove that $(n+4)^2 - (n-4)^2 = 16n$ for all integers $n$.

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Method: LHS \to RHS proof, expanding each squared bracket on the left-hand side. LHS $= (n+4)^2 - (n-4)^2$ Expand each square: $(n+4)^2 = n^2+8n+16$ and $(n-4)^2 = n^2-8n+16$. LHS $= (n^2+8n+16) - (n^2-8n+16)$ $= n^2+8n+16-n^2+8n-16$ $= 16n$ $=$ RHS. Since LHS $=$ RHS for all integers $n$, the identity is proven. $\blacksquare$ $$\boxed{(n+4)^2-(n-4)^2=16n}$$

QUESTION 7 4 marks

Easy

Prove that $(3n+2)^2 - (3n-2)^2 = 24n$ for all integers $n$.

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Method: LHS \to RHS proof using the difference of two squares, $a^2-b^2=(a-b)(a+b)$, with $a=3n+2$ and $b=3n-2$. LHS $= (3n+2)^2-(3n-2)^2$ $= [(3n+2)-(3n-2)][(3n+2)+(3n-2)]$ Simplify each bracket: $(3n+2)-(3n-2) = 4$ and $(3n+2)+(3n-2) = 6n$. LHS $= [4][6n]$ $= 24n$ $=$ RHS. Since LHS $=$ RHS for all integers $n$, the identity is proven. $\blacksquare$ $$\boxed{(3n+2)^2-(3n-2)^2=24n}$$

QUESTION 8 4 marks

Easy

Prove that $n^2 - n$ is always even for every integer $n$.

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Method: direct deductive proof by factorising $n^2-n$ into a product of two consecutive integers. To prove: $n^2-n$ is even for every integer $n$. Factorise: $n^2-n = n(n-1)$. Now $n$ and $n-1$ are two consecutive integers, so one of them must be even and the other odd (consecutive integers always have opposite parity). Since the product $n(n-1)$ contains an even factor, the product $n(n-1)$ must itself be even. That is, $n(n-1) = 2m$ for some integer $m$. Therefore $n^2-n = n(n-1) = 2m$, which is even. Hence $n^2-n$ is even for every integer $n$, as required. $\blacksquare$

QUESTION 9 3 marks

Easy

Prove that the sum of any two consecutive integers is always odd.

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Method: direct deductive proof using the general algebraic representation of consecutive integers. Let the two consecutive integers be $n$ and $n+1$, where $n$ is any integer. Their sum is $n + (n+1) = 2n+1$. Since $2n$ is even (it is $2$ times an integer), $2n+1$ is one more than an even number, so $2n+1$ is odd by definition. Hence the sum of any two consecutive integers is always odd, as required. $\blacksquare$ $$\boxed{n+(n+1)=2n+1 \text{ (odd)}}$$

QUESTION 10 4 marks

Easy

Prove that the sum of any three consecutive integers is always a multiple of 3.

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Method: direct deductive proof, letting the middle integer be $n$. Let the three consecutive integers be $n-1$, $n$, $n+1$, where $n$ is any integer. Their sum is: $(n-1) + n + (n+1) = 3n$ Since $3n = 3 \times n$ with $n$ an integer, $3n$ is by definition a multiple of $3$. Hence the sum of any three consecutive integers is always a multiple of $3$, as required. $\blacksquare$ $$\boxed{(n-1)+n+(n+1)=3n}$$

QUESTION 11 3 marks

Easy

Prove that if $n$ is an even integer, then $n^2$ is even.

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Method: direct deductive proof, writing an even integer in the general form $2k$. Since $n$ is even, $n$ can be written as $n = 2k$ for some integer $k$. Then $n^2 = (2k)^2 = 4k^2 = 2(2k^2)$. Since $2k^2$ is an integer, $n^2 = 2(2k^2)$ is of the form $2 \times \text{(an integer)}$, so $n^2$ is even. Hence if $n$ is an even integer, $n^2$ is also even, as required. $\blacksquare$

QUESTION 12 4 marks

Easy

Prove that if $n$ is an odd integer, then $n^2$ is odd.

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Method: direct deductive proof, writing an odd integer in the general form $2k+1$. Since $n$ is odd, $n$ can be written as $n = 2k+1$ for some integer $k$. Then: $n^2 = (2k+1)^2 = 4k^2+4k+1$ $= 2(2k^2+2k)+1$ Since $2k^2+2k$ is an integer, $n^2 = 2(2k^2+2k)+1$ is of the form $2 \times \text{(an integer)} + 1$, so $n^2$ is odd. Hence if $n$ is an odd integer, $n^2$ is also odd, as required. $\blacksquare$

QUESTION 13 3 marks

Easy

Prove the identity $x^2 - y^2 = (x-y)(x+y)$ for all real numbers $x$ and $y$, by expanding the right-hand side.

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Method: RHS \to LHS proof, expanding the product on the right-hand side. RHS $= (x-y)(x+y)$ Expand using the distributive law: $= x(x+y) - y(x+y)$ $= x^2+xy-xy-y^2$ $= x^2-y^2$ $=$ LHS. Since RHS $=$ LHS for all real $x,y$, the identity is proven. $\blacksquare$ $$\boxed{x^2-y^2=(x-y)(x+y)}$$

QUESTION 14 4 marks

Easy

Prove that $\dfrac{1}{n-1} + \dfrac{1}{n+1} = \dfrac{2n}{n^2-1}$ for $n \in \mathbb{R}$, $n \neq 1, -1$.

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Method: LHS \to RHS proof by combining the two fractions on the left-hand side over a common denominator. LHS $= \dfrac{1}{n-1} + \dfrac{1}{n+1}$ The common denominator is $(n-1)(n+1) = n^2-1$: $= \dfrac{n+1}{(n-1)(n+1)} + \dfrac{n-1}{(n-1)(n+1)}$ $= \dfrac{(n+1)+(n-1)}{n^2-1}$ $= \dfrac{2n}{n^2-1}$ $=$ RHS. Since LHS $=$ RHS (for $n \neq 1,-1$, so that all fractions are defined), the identity is proven. $\blacksquare$ $$\boxed{\dfrac{1}{n-1}+\dfrac{1}{n+1}=\dfrac{2n}{n^2-1}}$$

QUESTION 15 4 marks

Easy

Prove that $(x+1)^3 - x^3 = 3x^2+3x+1$ for all real $x$.

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Method: LHS \to RHS proof by expanding the cube on the left-hand side. LHS $= (x+1)^3 - x^3$ Expand $(x+1)^3$ using the binomial expansion: $(x+1)^3 = x^3+3x^2+3x+1$ LHS $= (x^3+3x^2+3x+1) - x^3$ $= 3x^2+3x+1$ $=$ RHS. Since LHS $=$ RHS for all real $x$, the identity is proven. $\blacksquare$ $$\boxed{(x+1)^3-x^3=3x^2+3x+1}$$

QUESTION 16 3 marks

Easy

Prove that the product of any two consecutive integers is always even.

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Method: direct deductive proof, considering the parity of two consecutive integers. Let the two consecutive integers be $n$ and $n+1$, where $n$ is any integer. Among any two consecutive integers, exactly one is even and the other is odd, since consecutive integers alternate in parity. Therefore the product $n(n+1)$ contains at least one even factor. A product with at least one even factor is itself even, since if the even factor equals $2k$, then $n(n+1) = 2k \times (\text{the other factor}) = 2 \times \text{(an integer)}$. Hence the product of any two consecutive integers is always even, as required. $\blacksquare$

QUESTION 17 4 marks

Easy

Prove the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ for all real numbers $a$ and $b$, by expanding the right-hand side.

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Method: RHS \to LHS proof, expanding the product on the right-hand side. RHS $= (a-b)(a^2+ab+b^2)$ Expand using the distributive law: $= a(a^2+ab+b^2) - b(a^2+ab+b^2)$ $= a^3+a^2b+ab^2 - a^2b-ab^2-b^3$ The middle terms cancel: $a^2b-a^2b=0$ and $ab^2-ab^2=0$. $= a^3-b^3$ $=$ LHS. Since RHS $=$ LHS for all real $a,b$, the identity is proven. $\blacksquare$ $$\boxed{a^3-b^3=(a-b)(a^2+ab+b^2)}$$

QUESTION 18 4 marks

Easy

Prove that $\sin\theta\tan\theta + \cos\theta = \dfrac{1}{\cos\theta}$ for $\cos\theta \neq 0$.

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Method: LHS \to RHS proof, using $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$. LHS $= \sin\theta\tan\theta + \cos\theta$ Substitute $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$: $= \sin\theta \cdot \dfrac{\sin\theta}{\cos\theta} + \cos\theta$ $= \dfrac{\sin^2\theta}{\cos\theta} + \cos\theta$ Write $\cos\theta$ with denominator $\cos\theta$: $= \dfrac{\sin^2\theta}{\cos\theta} + \dfrac{\cos^2\theta}{\cos\theta}$ $= \dfrac{\sin^2\theta+\cos^2\theta}{\cos\theta}$ By the Pythagorean identity, $\sin^2\theta+\cos^2\theta = 1$: $= \dfrac{1}{\cos\theta}$ $=$ RHS. Since LHS $=$ RHS for all $\theta$ with $\cos\theta \neq 0$, the identity is proven. $\blacksquare$ $$\boxed{\sin\theta\tan\theta+\cos\theta=\dfrac{1}{\cos\theta}}$$

QUESTION 19 3 marks

Easy

Prove that $(2n)^2 - (2n-2)^2 = 8n-4$ for all integers $n$.

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Method: LHS \to RHS proof, expanding each squared bracket on the left-hand side. LHS $= (2n)^2 - (2n-2)^2$ Expand each square: $(2n)^2 = 4n^2$ and $(2n-2)^2 = 4n^2-8n+4$. LHS $= 4n^2 - (4n^2-8n+4)$ $= 4n^2-4n^2+8n-4$ $= 8n-4$ $=$ RHS. Since LHS $=$ RHS for all integers $n$, the identity is proven. $\blacksquare$ $$\boxed{(2n)^2-(2n-2)^2=8n-4}$$

QUESTION 20 4 marks

Easy

Prove that the difference between the squares of any two consecutive odd integers is always a multiple of 8.

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Method: direct deductive proof, representing consecutive odd integers algebraically as $2k+1$ and $2k+3$. Let the two consecutive odd integers be $2k+1$ and $2k+3$, where $k$ is any integer (each is odd, and they differ by $2$). Their difference of squares is: $(2k+3)^2 - (2k+1)^2$ Use the difference of two squares, $a^2-b^2=(a-b)(a+b)$, with $a=2k+3$, $b=2k+1$: $= [(2k+3)-(2k+1)][(2k+3)+(2k+1)]$ $= [2][4k+4]$ $= 8k+8$ $= 8(k+1)$ Since $k+1$ is an integer, $8(k+1)$ is a multiple of $8$. Hence the difference between the squares of any two consecutive odd integers is always a multiple of $8$, as required. $\blacksquare$ $$\boxed{(2k+3)^2-(2k+1)^2=8(k+1)}$$

QUESTION 21 3 marks

Easy

Prove that $\dfrac{n^2-1}{n+1} = n-1$ for $n \in \mathbb{R}$, $n \neq -1$.

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Method: LHS \to RHS proof by factorising the numerator as a difference of two squares. LHS $= \dfrac{n^2-1}{n+1}$ Factorise the numerator using $a^2-b^2=(a-b)(a+b)$ with $a=n$, $b=1$: $n^2-1 = (n-1)(n+1)$ So LHS $= \dfrac{(n-1)(n+1)}{n+1}$ Since $n \neq -1$, the factor $(n+1)$ is non-zero and can be cancelled: $= n-1$ $=$ RHS. Since LHS $=$ RHS for all $n \neq -1$, the identity is proven. $\blacksquare$ $$\boxed{\dfrac{n^2-1}{n+1}=n-1}$$

QUESTION 22 3 marks

Easy

Prove that $(x-1)^2 + (x+1)^2 = 2x^2+2$ for all real $x$.

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Method: LHS \to RHS proof, expanding each squared bracket on the left-hand side. LHS $= (x-1)^2 + (x+1)^2$ Expand each square: $(x-1)^2 = x^2-2x+1$ and $(x+1)^2 = x^2+2x+1$. LHS $= (x^2-2x+1) + (x^2+2x+1)$ $= 2x^2+2$ $=$ RHS. Since LHS $=$ RHS for all real $x$, the identity is proven. $\blacksquare$ $$\boxed{(x-1)^2+(x+1)^2=2x^2+2}$$

QUESTION 23 3 marks

Easy

Prove that $\tan\theta \cdot \cos\theta = \sin\theta$ for $\cos\theta \neq 0$.

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Method: LHS \to RHS proof, using the quotient identity $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$. LHS $= \tan\theta \cdot \cos\theta$ Substitute $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$: $= \dfrac{\sin\theta}{\cos\theta} \cdot \cos\theta$ Since $\cos\theta \neq 0$, the $\cos\theta$ factors cancel: $= \sin\theta$ $=$ RHS. Since LHS $=$ RHS for all $\theta$ with $\cos\theta \neq 0$, the identity is proven. $\blacksquare$ $$\boxed{\tan\theta\cos\theta=\sin\theta}$$

QUESTION 24 6 marks

Medium

(a) Prove that $(n+4)^2 + (n-4)^2 = 2n^2+32$ for all real $n$. [3] (b) Hence solve the equation $(x+4)^2 + (x-4)^2 = 66$. [3]

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(a) Method: LHS \to RHS proof by expanding each squared bracket. LHS $= (n+4)^2+(n-4)^2$ $(n+4)^2 = n^2+8n+16$ and $(n-4)^2 = n^2-8n+16$. LHS $= (n^2+8n+16) + (n^2-8n+16)$ $= 2n^2+32$ $=$ RHS. So the identity is proven for all real $n$. $$\boxed{(n+4)^2+(n-4)^2=2n^2+32}$$ (b) Method: applying the identity from part (a) with $n=x$ to reduce the equation to a simple quadratic. From (a), $(x+4)^2+(x-4)^2 = 2x^2+32$. So the equation becomes: $2x^2+32 = 66$ $2x^2 = 34$ $x^2 = 17$ $x = \pm\sqrt{17}$ Final answer: $$\boxed{x = \pm\sqrt{17}}$$

QUESTION 25 5 marks

Medium

Prove that $\dfrac{\sin\theta}{1-\cos\theta} = \dfrac{1+\cos\theta}{\sin\theta}$ for $\sin\theta \neq 0$.

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Method: proof by cross-multiplication, showing that the product of the extremes equals the product of the means, then using the Pythagorean identity. To prove: $\dfrac{\sin\theta}{1-\cos\theta} = \dfrac{1+\cos\theta}{\sin\theta}$. Since $\sin\theta \neq 0$, this holds if and only if, after cross-multiplying, $\sin\theta \times \sin\theta = (1-\cos\theta)(1+\cos\theta)$. Consider the right-hand side of this cross-multiplied equation: $(1-\cos\theta)(1+\cos\theta) = 1-\cos^2\theta$ By the Pythagorean identity, $\sin^2\theta+\cos^2\theta=1$, so $1-\cos^2\theta = \sin^2\theta$. Therefore $(1-\cos\theta)(1+\cos\theta) = \sin^2\theta = \sin\theta \times \sin\theta$. This confirms that the cross-multiplied equation holds identically, so the original identity $\dfrac{\sin\theta}{1-\cos\theta} = \dfrac{1+\cos\theta}{\sin\theta}$ is proven for all $\theta$ with $\sin\theta \neq 0$. $\blacksquare$ $$\boxed{\dfrac{\sin\theta}{1-\cos\theta}=\dfrac{1+\cos\theta}{\sin\theta}}$$

QUESTION 26 5 marks

Medium

Prove that $\dfrac{1}{n} + \dfrac{1}{n+1} + \dfrac{1}{n(n+1)} = \dfrac{2}{n}$ for $n \in \mathbb{R}$, $n \neq 0, -1$.

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Method: LHS \to RHS proof by combining all three fractions on the left-hand side over the common denominator $n(n+1)$. LHS $= \dfrac{1}{n} + \dfrac{1}{n+1} + \dfrac{1}{n(n+1)}$ Write each fraction with common denominator $n(n+1)$: $= \dfrac{n+1}{n(n+1)} + \dfrac{n}{n(n+1)} + \dfrac{1}{n(n+1)}$ $= \dfrac{(n+1)+n+1}{n(n+1)}$ $= \dfrac{2n+2}{n(n+1)}$ Factorise the numerator: $2n+2 = 2(n+1)$. $= \dfrac{2(n+1)}{n(n+1)}$ Since $n \neq -1$, the factor $(n+1)$ is non-zero and cancels: $= \dfrac{2}{n}$ $=$ RHS. Since LHS $=$ RHS (for $n \neq 0,-1$, so that all fractions are defined), the identity is proven. $\blacksquare$ $$\boxed{\dfrac{1}{n}+\dfrac{1}{n+1}+\dfrac{1}{n(n+1)}=\dfrac{2}{n}}$$

QUESTION 27 6 marks

Medium

(a) Prove that $(n+1)^3 - n^3 = 3n^2+3n+1$ for all integers $n$. [4] (b) Hence prove that $(n+1)^3-n^3$ is always odd. [2]

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(a) Method: LHS \to RHS proof by expanding the cube. LHS $= (n+1)^3-n^3$ Expand $(n+1)^3 = n^3+3n^2+3n+1$. LHS $= (n^3+3n^2+3n+1) - n^3$ $= 3n^2+3n+1$ $=$ RHS. So the identity is proven for all integers $n$. $$\boxed{(n+1)^3-n^3=3n^2+3n+1}$$ (b) Method: using the result of (a) and the fact that the product of two consecutive integers is even. From (a), $(n+1)^3-n^3 = 3n^2+3n+1 = 3n(n+1)+1$. Now $n$ and $n+1$ are consecutive integers, so their product $n(n+1)$ is even (one of the two must be even). Write $n(n+1) = 2m$ for some integer $m$. Then $3n(n+1)+1 = 3(2m)+1 = 6m+1$, which is of the form $2 \times \text{(an integer)} + 1$, i.e. odd. Hence $(n+1)^3-n^3$ is always odd, as required. $\blacksquare$

QUESTION 28 5 marks

Medium

Prove that the sum of the squares of any two consecutive integers is always odd.

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Method: direct deductive proof, letting the two consecutive integers be $n$ and $n+1$. To prove: $n^2+(n+1)^2$ is odd for every integer $n$. Expand: $n^2+(n+1)^2 = n^2 + n^2+2n+1 = 2n^2+2n+1$. Factorise the even part: $2n^2+2n+1 = 2(n^2+n)+1$. Since $n^2+n$ is an integer (the sum of two integers), $2(n^2+n)$ is even, and $2(n^2+n)+1$ is one more than an even number, so it is odd. Hence the sum of the squares of any two consecutive integers is always odd, as required. $\blacksquare$ $$\boxed{n^2+(n+1)^2=2(n^2+n)+1}$$

QUESTION 29 5 marks

Medium

(a) Prove that $\cos^2\theta - \sin^2\theta = 2\cos^2\theta-1$ for all $\theta$. [3] (b) Hence find the value of $\cos^2\theta-\sin^2\theta$ given that $\cos\theta = \dfrac{2}{3}$. [2]

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(a) Method: LHS \to RHS proof, using the Pythagorean identity to replace $\sin^2\theta$. LHS $= \cos^2\theta-\sin^2\theta$ By the Pythagorean identity, $\sin^2\theta+\cos^2\theta=1$, so $\sin^2\theta = 1-\cos^2\theta$. LHS $= \cos^2\theta - (1-\cos^2\theta)$ $= \cos^2\theta-1+\cos^2\theta$ $= 2\cos^2\theta-1$ $=$ RHS. So the identity is proven for all $\theta$. $$\boxed{\cos^2\theta-\sin^2\theta=2\cos^2\theta-1}$$ (b) Method: substituting $\cos\theta=\dfrac{2}{3}$ into the identity proven in part (a). From (a): $\cos^2\theta-\sin^2\theta = 2\cos^2\theta-1 = 2\left(\dfrac{2}{3}\right)^2-1 = 2\left(\dfrac{4}{9}\right)-1 = \dfrac{8}{9}-1 = -\dfrac{1}{9}$. Final answer: $$\boxed{\cos^2\theta-\sin^2\theta = -\dfrac{1}{9}}$$

QUESTION 30 5 marks

Medium

Prove that $\sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta$ for all $\theta$.

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Method: LHS \to RHS proof, factorising the left-hand side as a difference of two squares and applying the Pythagorean identity. LHS $= \sin^4\theta-\cos^4\theta$ Write this as $(\sin^2\theta)^2-(\cos^2\theta)^2$ and factorise using $a^2-b^2=(a-b)(a+b)$ with $a=\sin^2\theta$, $b=\cos^2\theta$: $= (\sin^2\theta-\cos^2\theta)(\sin^2\theta+\cos^2\theta)$ By the Pythagorean identity, $\sin^2\theta+\cos^2\theta=1$: $= (\sin^2\theta-\cos^2\theta)(1)$ $= \sin^2\theta-\cos^2\theta$ $=$ RHS. Since LHS $=$ RHS for all $\theta$, the identity is proven. $\blacksquare$ $$\boxed{\sin^4\theta-\cos^4\theta=\sin^2\theta-\cos^2\theta}$$

QUESTION 31 4 marks

Medium

Prove that $(1-\sin\theta)(1+\sin\theta) = \cos^2\theta$ for all $\theta$.

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Method: LHS \to RHS proof, expanding the product as a difference of two squares and applying the Pythagorean identity. LHS $= (1-\sin\theta)(1+\sin\theta)$ Expand using $a^2-b^2=(a-b)(a+b)$ with $a=1$, $b=\sin\theta$: $= 1^2-\sin^2\theta$ $= 1-\sin^2\theta$ By the Pythagorean identity, $\sin^2\theta+\cos^2\theta=1$, so $1-\sin^2\theta = \cos^2\theta$. $= \cos^2\theta$ $=$ RHS. Since LHS $=$ RHS for all $\theta$, the identity is proven. $\blacksquare$ $$\boxed{(1-\sin\theta)(1+\sin\theta)=\cos^2\theta}$$

QUESTION 32 5 marks

Medium

Prove that the product of any two consecutive even integers is always divisible by 8.

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Method: direct deductive proof, representing consecutive even integers algebraically as $2k$ and $2k+2$. Let the two consecutive even integers be $2k$ and $2k+2$, where $k$ is any integer. Their product is: $2k(2k+2) = 4k(k+1)$ Now $k$ and $k+1$ are consecutive integers, so one of them must be even, meaning their product $k(k+1)$ is always even. Write $k(k+1) = 2m$ for some integer $m$. Then $4k(k+1) = 4(2m) = 8m$. Since $m$ is an integer, $8m$ is a multiple of $8$. Hence the product of any two consecutive even integers is always divisible by $8$, as required. $\blacksquare$ $$\boxed{2k(2k+2)=8m}$$

QUESTION 33 6 marks

Medium

(a) Prove that $x^3+y^3=(x+y)(x^2-xy+y^2)$ for all real $x$ and $y$, by expanding the right-hand side. [4] (b) Hence factorise $x^3+8$ fully. [2]

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(a) Method: RHS \to LHS proof, expanding the product on the right-hand side. RHS $= (x+y)(x^2-xy+y^2)$ Expand using the distributive law: $= x(x^2-xy+y^2) + y(x^2-xy+y^2)$ $= x^3-x^2y+xy^2 + x^2y-xy^2+y^3$ The middle terms cancel: $-x^2y+x^2y=0$ and $xy^2-xy^2=0$. $= x^3+y^3$ $=$ LHS. So the identity is proven for all real $x,y$. $$\boxed{x^3+y^3=(x+y)(x^2-xy+y^2)}$$ (b) Method: applying the identity from part (a) with $x=x$ and $y=2$, since $8=2^3$. Write $x^3+8 = x^3+2^3$. Using the identity from (a) with $y=2$: $x^3+2^3 = (x+2)(x^2-2x+4)$ Final answer: $$\boxed{x^3+8=(x+2)(x^2-2x+4)}$$

QUESTION 34 4 marks

Medium

Prove that $n^2+n+1$ is always odd for every integer $n$.

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Method: direct deductive proof, factorising the even part of the expression. To prove: $n^2+n+1$ is odd for every integer $n$. Write $n^2+n+1 = n(n+1)+1$. Now $n$ and $n+1$ are consecutive integers, so their product $n(n+1)$ is always even (one of the two consecutive integers must be even). Write $n(n+1) = 2k$ for some integer $k$. Then $n^2+n+1 = n(n+1)+1 = 2k+1$, which is of the form $2 \times \text{(an integer)}+1$, i.e. odd. Hence $n^2+n+1$ is always odd for every integer $n$, as required. $\blacksquare$

QUESTION 35 5 marks

Medium

Prove that $\dfrac{n+1}{n} - \dfrac{n}{n+1} = \dfrac{2n+1}{n(n+1)}$ for $n \in \mathbb{R}$, $n \neq 0, -1$.

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Method: LHS \to RHS proof by combining the two fractions over the common denominator $n(n+1)$. LHS $= \dfrac{n+1}{n} - \dfrac{n}{n+1}$ Write each fraction with common denominator $n(n+1)$: $= \dfrac{(n+1)^2}{n(n+1)} - \dfrac{n^2}{n(n+1)}$ $= \dfrac{(n+1)^2-n^2}{n(n+1)}$ Expand the numerator: $(n+1)^2-n^2 = (n^2+2n+1)-n^2 = 2n+1$. $= \dfrac{2n+1}{n(n+1)}$ $=$ RHS. Since LHS $=$ RHS (for $n \neq 0,-1$, so that all fractions are defined), the identity is proven. $\blacksquare$ $$\boxed{\dfrac{n+1}{n}-\dfrac{n}{n+1}=\dfrac{2n+1}{n(n+1)}}$$

QUESTION 36 5 marks

Medium

(a) Prove that $(a-b)^2 \geq 0$ for all real numbers $a$ and $b$. [2] (b) Hence prove that $a^2+b^2 \geq 2ab$ for all real numbers $a$ and $b$. [3]

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(a) Method: direct deductive proof, using the fact that the square of any real number is non-negative. $a-b$ is a real number (since $a,b$ are real). The square of any real number is never negative. Hence $(a-b)^2 \geq 0$ for all real $a,b$, as required. $\blacksquare$ (b) Method: expanding the result of part (a) and rearranging. From (a): $(a-b)^2 \geq 0$. Expand the left-hand side: $a^2-2ab+b^2 \geq 0$ Add $2ab$ to both sides of the inequality (this does not change the direction of the inequality): $a^2-2ab+b^2+2ab \geq 0+2ab$ $a^2+b^2 \geq 2ab$ Hence $a^2+b^2 \geq 2ab$ for all real numbers $a$ and $b$, as required. $\blacksquare$ $$\boxed{a^2+b^2\geq 2ab}$$

QUESTION 37 4 marks

Medium

Prove that the sum of any five consecutive integers is always a multiple of 5.

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Method: direct deductive proof, letting the middle integer be $n$. Let the five consecutive integers be $n-2$, $n-1$, $n$, $n+1$, $n+2$, where $n$ is any integer. Their sum is: $(n-2)+(n-1)+n+(n+1)+(n+2)$ $= 5n + (-2-1+0+1+2)$ $= 5n+0$ $= 5n$ Since $5n = 5 \times n$ with $n$ an integer, $5n$ is by definition a multiple of $5$. Hence the sum of any five consecutive integers is always a multiple of $5$, as required. $\blacksquare$ $$\boxed{(n-2)+(n-1)+n+(n+1)+(n+2)=5n}$$

QUESTION 38 5 marks

Medium

Prove that $\tan\theta + \dfrac{1}{\tan\theta} = \dfrac{1}{\sin\theta\cos\theta}$ for $\sin\theta\cos\theta \neq 0$.

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Method: LHS \to RHS proof, writing $\tan\theta$ in terms of $\sin\theta$ and $\cos\theta$ and combining fractions. LHS $= \tan\theta + \dfrac{1}{\tan\theta}$ Substitute $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, so $\dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$: $= \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta}$ Combine over the common denominator $\sin\theta\cos\theta$: $= \dfrac{\sin^2\theta}{\sin\theta\cos\theta} + \dfrac{\cos^2\theta}{\sin\theta\cos\theta}$ $= \dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$ By the Pythagorean identity, $\sin^2\theta+\cos^2\theta=1$: $= \dfrac{1}{\sin\theta\cos\theta}$ $=$ RHS. Since LHS $=$ RHS for all $\theta$ with $\sin\theta\cos\theta \neq 0$, the identity is proven. $\blacksquare$ $$\boxed{\tan\theta+\dfrac{1}{\tan\theta}=\dfrac{1}{\sin\theta\cos\theta}}$$

QUESTION 39 5 marks

Medium

Prove that the difference between the squares of any two consecutive multiples of 3 is always a multiple of 9.

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Method: direct deductive proof, representing consecutive multiples of 3 algebraically as $3k$ and $3k+3$. Let the two consecutive multiples of $3$ be $3k$ and $3k+3$, where $k$ is any integer. Their difference of squares is: $(3k+3)^2-(3k)^2$ Use the difference of two squares, $a^2-b^2=(a-b)(a+b)$, with $a=3k+3$, $b=3k$: $= [(3k+3)-3k][(3k+3)+3k]$ $= [3][6k+3]$ $= 18k+9$ $= 9(2k+1)$ Since $2k+1$ is an integer, $9(2k+1)$ is a multiple of $9$. Hence the difference between the squares of any two consecutive multiples of $3$ is always a multiple of $9$, as required. $\blacksquare$ $$\boxed{(3k+3)^2-(3k)^2=9(2k+1)}$$

QUESTION 40 5 marks

Medium

Prove that $\dfrac{1}{(n-1)(n+1)} = \dfrac{1}{2}\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)$ for $n \in \mathbb{R}$, $n \neq 1,-1$.

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Method: RHS \to LHS proof, combining the two fractions inside the bracket on the right-hand side. RHS $= \dfrac{1}{2}\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)$ First combine the fractions inside the bracket, using common denominator $(n-1)(n+1)$: $\dfrac{1}{n-1}-\dfrac{1}{n+1} = \dfrac{(n+1)-(n-1)}{(n-1)(n+1)} = \dfrac{2}{(n-1)(n+1)}$ Substitute back: RHS $= \dfrac{1}{2} \times \dfrac{2}{(n-1)(n+1)}$ $= \dfrac{1}{(n-1)(n+1)}$ $=$ LHS. Since RHS $=$ LHS (for $n \neq 1,-1$, so that all fractions are defined), the identity is proven. $\blacksquare$ $$\boxed{\dfrac{1}{(n-1)(n+1)}=\dfrac{1}{2}\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)}$$

QUESTION 41 6 marks

Medium

(a) Prove that $(2n-1)^2+(2n+1)^2 = 8n^2+2$ for all integers $n$. [4] (b) Hence prove that the sum of the squares of any two consecutive odd integers is always even. [2]

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(a) Method: LHS \to RHS proof by expanding each squared bracket, noting that $2n-1$ and $2n+1$ are consecutive odd integers. LHS $= (2n-1)^2+(2n+1)^2$ Expand each square: $(2n-1)^2 = 4n^2-4n+1$ and $(2n+1)^2 = 4n^2+4n+1$. LHS $= (4n^2-4n+1)+(4n^2+4n+1)$ $= 8n^2+2$ $=$ RHS. So the identity is proven for all integers $n$. $$\boxed{(2n-1)^2+(2n+1)^2=8n^2+2}$$ (b) Method: using the result of (a) to show the expression is always a multiple of 2. The integers $2n-1$ and $2n+1$ differ by $2$ and are both odd, so they represent any pair of consecutive odd integers as $n$ ranges over all integers. From (a), the sum of their squares is $8n^2+2 = 2(4n^2+1)$. Since $4n^2+1$ is an integer, $2(4n^2+1)$ is of the form $2 \times \text{(an integer)}$, so it is even. Hence the sum of the squares of any two consecutive odd integers is always even, as required. $\blacksquare$

QUESTION 42 8 marks

Hard

(a) Prove that $(n+1)^4-(n-1)^4 = 8n^3+8n$ for all integers $n$. [5] (b) Hence prove that $(n+1)^4-(n-1)^4$ is always a multiple of 8. [3]

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(a) Method: LHS \to RHS proof by expanding each fourth power using the binomial expansion. LHS $= (n+1)^4-(n-1)^4$ Expand each power: $(n+1)^4 = n^4+4n^3+6n^2+4n+1$ $(n-1)^4 = n^4-4n^3+6n^2-4n+1$ LHS $= (n^4+4n^3+6n^2+4n+1) - (n^4-4n^3+6n^2-4n+1)$ $= n^4+4n^3+6n^2+4n+1-n^4+4n^3-6n^2+4n-1$ Collect like terms: the $n^4$ terms cancel, the $6n^2$ terms cancel, and the constant terms cancel. $= 8n^3+8n$ $=$ RHS. So the identity is proven for all integers $n$. $$\boxed{(n+1)^4-(n-1)^4=8n^3+8n}$$ (b) Method: factorising the result of (a) to expose an explicit factor of 8. From (a), $(n+1)^4-(n-1)^4 = 8n^3+8n = 8(n^3+n)$. Since $n$ is an integer, $n^3+n$ is also an integer. Therefore $(n+1)^4-(n-1)^4 = 8 \times \text{(an integer)}$, which is by definition a multiple of $8$. Hence $(n+1)^4-(n-1)^4$ is always a multiple of $8$ for every integer $n$, as required. $\blacksquare$

QUESTION 43 7 marks

Hard

(a) Prove that $(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$ for all real numbers $a$, $b$, $c$. [2] (b) Hence prove that $a^2+b^2+c^2 \geq ab+bc+ca$ for all real numbers $a$, $b$, $c$. [5]

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(a) Method: direct deductive proof, using the fact that each squared term is non-negative. Since $a,b,c$ are real, each of $a-b$, $b-c$, $c-a$ is a real number, and the square of any real number is non-negative. So $(a-b)^2 \geq 0$, $(b-c)^2 \geq 0$, $(c-a)^2 \geq 0$. Adding three non-negative quantities gives a non-negative result: $(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$ as required. $\blacksquare$ (b) Method: expanding the result of part (a) and rearranging to isolate $a^2+b^2+c^2$ and $ab+bc+ca$. From (a): $(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$. Expand each square: $(a-b)^2 = a^2-2ab+b^2$ $(b-c)^2 = b^2-2bc+c^2$ $(c-a)^2 = c^2-2ca+a^2$ Sum these: $(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2) \geq 0$ $2a^2+2b^2+2c^2-2ab-2bc-2ca \geq 0$ Divide both sides by $2$ (a positive number, so the inequality direction is unchanged): $a^2+b^2+c^2-ab-bc-ca \geq 0$ Add $ab+bc+ca$ to both sides: $a^2+b^2+c^2 \geq ab+bc+ca$ Hence $a^2+b^2+c^2 \geq ab+bc+ca$ for all real numbers $a,b,c$, as required. $\blacksquare$ $$\boxed{a^2+b^2+c^2\geq ab+bc+ca}$$

QUESTION 44 7 marks

Hard

(a) Show that $n^3-n = (n-1)n(n+1)$ for every integer $n$. [2] (b) Explain why the product of any three consecutive integers is always divisible by 6. [3] (c) Hence prove that $n^3-n$ is divisible by 6 for every integer $n$. [2]

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(a) Method: factorising $n^3-n$ directly. $n^3-n = n(n^2-1)$ Factorise $n^2-1$ as a difference of two squares: $n^2-1=(n-1)(n+1)$. $n^3-n = n(n-1)(n+1) = (n-1)n(n+1)$ as required. $\blacksquare$ $$\boxed{n^3-n=(n-1)n(n+1)}$$ (b) Method: explaining divisibility by 2 and by 3 separately among any three consecutive integers. Consider any three consecutive integers $(n-1),n,(n+1)$. Divisibility by 2: among any two consecutive integers, at least one is even; since $(n-1),n,(n+1)$ contains at least two consecutive integers (e.g. $n-1,n$), at least one of the three is even. So the product $(n-1)n(n+1)$ is divisible by $2$. Divisibility by 3: among any three consecutive integers, exactly one must be a multiple of $3$. This is because multiples of $3$ occur every third integer, so any run of three consecutive integers must contain exactly one of them. So the product $(n-1)n(n+1)$ is divisible by $3$. Since the product is divisible by both $2$ and $3$, and $2$ and $3$ share no common factor other than $1$, the product must be divisible by $2 \times 3 = 6$. Hence the product of any three consecutive integers is always divisible by $6$, as required. $\blacksquare$ (c) Method: combining the factorisation from (a) with the divisibility result from (b). From (a), $n^3-n = (n-1)n(n+1)$, the product of three consecutive integers. From (b), the product of any three consecutive integers is divisible by $6$. Therefore $n^3-n$ is divisible by $6$ for every integer $n$, as required. $\blacksquare$

QUESTION 45 7 marks

Hard

(a) Prove that $\cos^4\theta-\sin^4\theta = \cos2\theta$ for all $\theta$. [4] (b) Hence prove that $\cos^4\theta-\sin^4\theta+2\sin^2\theta = 1$ for all $\theta$. [3]

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(a) Method: LHS \to RHS proof, factorising as a difference of two squares and applying the Pythagorean and double-angle identities. LHS $= \cos^4\theta-\sin^4\theta$ Write this as $(\cos^2\theta)^2-(\sin^2\theta)^2$ and factorise using $a^2-b^2=(a-b)(a+b)$ with $a=\cos^2\theta$, $b=\sin^2\theta$: $= (\cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^2\theta)$ By the Pythagorean identity, $\cos^2\theta+\sin^2\theta=1$: $= (\cos^2\theta-\sin^2\theta)(1)$ $= \cos^2\theta-\sin^2\theta$ By the double angle identity, $\cos2\theta = \cos^2\theta-\sin^2\theta$: $= \cos2\theta$ $=$ RHS. Since LHS $=$ RHS for all $\theta$, the identity is proven. $$\boxed{\cos^4\theta-\sin^4\theta=\cos2\theta}$$ (b) Method: substituting the result of (a) and using the double angle identity $\cos2\theta=1-2\sin^2\theta$. From (a), $\cos^4\theta-\sin^4\theta = \cos2\theta$. So LHS $= \cos^4\theta-\sin^4\theta+2\sin^2\theta = \cos2\theta+2\sin^2\theta$. By the double angle identity, $\cos2\theta = 1-2\sin^2\theta$: $= (1-2\sin^2\theta)+2\sin^2\theta$ $= 1$ $=$ RHS. Hence $\cos^4\theta-\sin^4\theta+2\sin^2\theta = 1$ for all $\theta$, as required. $\blacksquare$

QUESTION 46 7 marks

Hard

(a) Prove that $\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}$ for $n \in \mathbb{R}$, $n \neq 0,-1$. [3] (b) Hence, using the result of part (a), prove that $\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4} = \dfrac{3}{4}$. [4]

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(a) Method: RHS \to LHS proof, combining the two fractions on the right-hand side. RHS $= \dfrac{1}{n} - \dfrac{1}{n+1}$ Combine over the common denominator $n(n+1)$: $= \dfrac{n+1}{n(n+1)} - \dfrac{n}{n(n+1)}$ $= \dfrac{(n+1)-n}{n(n+1)}$ $= \dfrac{1}{n(n+1)}$ $=$ LHS. Since RHS $=$ LHS (for $n\neq0,-1$), the identity is proven. $$\boxed{\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}}$$ (b) Method: applying the identity from (a) with $n=1,2,3$ in turn and observing the telescoping cancellation. Using the identity from (a): For $n=1$: $\dfrac{1}{1\times2} = \dfrac{1}{1}-\dfrac{1}{2}$ For $n=2$: $\dfrac{1}{2\times3} = \dfrac{1}{2}-\dfrac{1}{3}$ For $n=3$: $\dfrac{1}{3\times4} = \dfrac{1}{3}-\dfrac{1}{4}$ Adding these three results: $\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4} = \left(\dfrac{1}{1}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)$ The intermediate terms cancel in pairs ($-\dfrac{1}{2}$ with $+\dfrac{1}{2}$, and $-\dfrac{1}{3}$ with $+\dfrac{1}{3}$), leaving only the first and last terms: $= \dfrac{1}{1} - \dfrac{1}{4}$ $= 1-\dfrac{1}{4}$ $= \dfrac{3}{4}$ Hence $\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4} = \dfrac{3}{4}$, as required. $\blacksquare$

QUESTION 47 6 marks

Hard

(a) Prove that $(x+y+z)^2 = x^2+y^2+z^2+2xy+2yz+2zx$ for all real $x,y,z$. [3] (b) Hence find the value of $xy+yz+zx$, given that $x+y+z=10$ and $x^2+y^2+z^2=58$. [3]

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(a) Method: LHS \to RHS proof, expanding the square of the trinomial by repeated use of the distributive law. LHS $= (x+y+z)^2 = (x+y+z)(x+y+z)$ Expand by distributing each term of the first bracket over the second: $= x(x+y+z)+y(x+y+z)+z(x+y+z)$ $= x^2+xy+xz+xy+y^2+yz+xz+yz+z^2$ Collect like terms ($xy$ appears twice, $xz$ appears twice, $yz$ appears twice): $= x^2+y^2+z^2+2xy+2yz+2zx$ $=$ RHS. So the identity is proven for all real $x,y,z$. $$\boxed{(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx}$$ (b) Method: substituting the given values into the identity proven in part (a) and solving for $xy+yz+zx$. From (a): $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)$. Substitute $x+y+z=10$ and $x^2+y^2+z^2=58$: $10^2 = 58+2(xy+yz+zx)$ $100 = 58+2(xy+yz+zx)$ $2(xy+yz+zx) = 42$ $xy+yz+zx = 21$ Final answer: $$\boxed{xy+yz+zx = 21}$$

QUESTION 48 6 marks

Hard

Prove algebraically that the square of any odd integer can be written in the form $8k+1$ for some integer $k$.

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Method: direct deductive proof, writing a general odd integer as $2n+1$ and analysing the parity of the resulting coefficient. To prove: for any odd integer, its square has the form $8k+1$ for some integer $k$. Let the odd integer be $2n+1$, where $n$ is any integer (this represents every odd integer as $n$ ranges over all integers). Then: $(2n+1)^2 = 4n^2+4n+1$ Factorise the first two terms: $= 4n(n+1)+1$ Now $n$ and $n+1$ are consecutive integers, so their product $n(n+1)$ is always even (one of any two consecutive integers must be even). Write $n(n+1) = 2k$ for some integer $k$. Substituting: $(2n+1)^2 = 4(2k)+1 = 8k+1$ Since $k$ is an integer, this shows that the square of any odd integer, $(2n+1)^2$, can always be written in the form $8k+1$. Hence the result is proven for every odd integer, as required. $\blacksquare$ $$\boxed{(2n+1)^2=8k+1, \text{ where } k=\tfrac{n(n+1)}{2} \text{ is an integer}}$$

QUESTION 49 8 marks

Hard

(a) Prove that $\dfrac{1}{\sin\theta}-\sin\theta = \dfrac{\cos^2\theta}{\sin\theta}$ for $\sin\theta \neq 0$. [3] (b) Hence prove that $\left(\dfrac{1}{\sin\theta}-\sin\theta\right)\left(\dfrac{1}{\cos\theta}-\cos\theta\right) = \sin\theta\cos\theta$ for $\sin\theta\cos\theta \neq 0$. [5]

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(a) Method: LHS \to RHS proof, writing $\sin\theta$ with denominator $\sin\theta$ and applying the Pythagorean identity. LHS $= \dfrac{1}{\sin\theta}-\sin\theta$ $= \dfrac{1}{\sin\theta}-\dfrac{\sin^2\theta}{\sin\theta}$ $= \dfrac{1-\sin^2\theta}{\sin\theta}$ By the Pythagorean identity, $1-\sin^2\theta = \cos^2\theta$: $= \dfrac{\cos^2\theta}{\sin\theta}$ $=$ RHS. So the identity is proven for $\sin\theta \neq 0$. $$\boxed{\dfrac{1}{\sin\theta}-\sin\theta=\dfrac{\cos^2\theta}{\sin\theta}}$$ (b) Method: applying the result of (a), and the corresponding identity for cosine obtained by the same method, then multiplying and simplifying. From (a): $\dfrac{1}{\sin\theta}-\sin\theta = \dfrac{\cos^2\theta}{\sin\theta}$. By an identical argument with $\sin\theta$ and $\cos\theta$ interchanged (using $1-\cos^2\theta=\sin^2\theta$): $\dfrac{1}{\cos\theta}-\cos\theta = \dfrac{1-\cos^2\theta}{\cos\theta} = \dfrac{\sin^2\theta}{\cos\theta}$ Now consider the product: $\left(\dfrac{1}{\sin\theta}-\sin\theta\right)\left(\dfrac{1}{\cos\theta}-\cos\theta\right) = \dfrac{\cos^2\theta}{\sin\theta} \times \dfrac{\sin^2\theta}{\cos\theta}$ Multiply the fractions and cancel one factor of $\sin\theta$ and one factor of $\cos\theta$ (valid since $\sin\theta\cos\theta \neq 0$): $= \dfrac{\cos^2\theta\sin^2\theta}{\sin\theta\cos\theta}$ $= \sin\theta\cos\theta$ Hence $\left(\dfrac{1}{\sin\theta}-\sin\theta\right)\left(\dfrac{1}{\cos\theta}-\cos\theta\right) = \sin\theta\cos\theta$, as required. $\blacksquare$

QUESTION 50 7 marks

Hard

(a) Prove that $(2n+1)^3-(2n-1)^3 = 24n^2+2$ for all integers $n$. [5] (b) Hence prove that $(2n+1)^3-(2n-1)^3$ is always even. [2]

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(a) Method: LHS \to RHS proof by expanding each cube using the binomial expansion. LHS $= (2n+1)^3-(2n-1)^3$ Expand each cube: $(2n+1)^3 = 8n^3+12n^2+6n+1$ $(2n-1)^3 = 8n^3-12n^2+6n-1$ LHS $= (8n^3+12n^2+6n+1) - (8n^3-12n^2+6n-1)$ $= 8n^3+12n^2+6n+1-8n^3+12n^2-6n+1$ Collect like terms: the $8n^3$ terms cancel and the $6n$ terms cancel. $= 24n^2+2$ $=$ RHS. So the identity is proven for all integers $n$. $$\boxed{(2n+1)^3-(2n-1)^3=24n^2+2}$$ (b) Method: factorising the result of (a) to expose an explicit factor of 2. From (a), $(2n+1)^3-(2n-1)^3 = 24n^2+2 = 2(12n^2+1)$. Since $n$ is an integer, $12n^2+1$ is also an integer. Therefore $(2n+1)^3-(2n-1)^3 = 2 \times \text{(an integer)}$, which is by definition even. Hence $(2n+1)^3-(2n-1)^3$ is always even for every integer $n$, as required. $\blacksquare$

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