2022 H2 Maths Paper 2 Q4 — Worked Solution

Part (a)

2 marks

We factorise the denominator first. Solving $9r^{2} + 3r - 2 = 0$ gives $r = \frac{- 3 \pm \sqrt{9 + 72}}{18} = \frac{- 3 \pm 9}{18}$ , i.e. $r = \frac{1}{3}$ or $r = - \frac{2}{3}$ , so

9r^{2} + 3r - 2 = (3r - 1)(3r + 2)

Using partial fractions,

\frac{1}{(3r - 1)(3r + 2)} = \frac{A}{3r - 1} + \frac{B}{3r + 2}

\Rightarrow 1 = A(3r + 2) + B(3r - 1)

Setting $r = \frac{1}{3}$ : $1 = A(1 + 2) = 3A \Rightarrow A = \frac{1}{3}$ .

Setting $r = - \frac{2}{3}$ : $1 = B( - 2 - 1) = - 3B \Rightarrow B = - \frac{1}{3}$ .

To check your answer: at

r = 1

,

\frac{1}{9 + 3 - 2} = \frac{1}{10}

, and

\frac{1}{3}\left( \frac{1}{2} - \frac{1}{5} \right) = \frac{1}{3} \times \frac{3}{10} = \frac{1}{10}

✓

$\therefore\ \frac{1}{9r^{2} + 3r - 2} = \frac{1}{3}\left( \frac{1}{3r - 1} - \frac{1}{3r + 2} \right)$

Part (b)

4 marks

Approach

The sum runs from $r = m$ to $r = 3m$ , not from $r = 1$ . We telescope directly over this range, writing out the first and last few brackets so the cancellation pattern is visible, and keep only the surviving first and last terms.

Using the partial fractions from (a),

\sum_{r = m}^{3m}\frac{1}{9r^{2} + 3r - 2} = \frac{1}{3}\sum_{r = m}^{3m}\left( \frac{1}{3r - 1} - \frac{1}{3r + 2} \right)

Writing out the terms:

\frac{1}{3}\left\lbrack \left( \frac{1}{3m - 1} - \frac{1}{3m + 2} \right) + \left( \frac{1}{3m + 2} - \frac{1}{3m + 5} \right) + \cdots + \left( \frac{1}{9m - 1} - \frac{1}{9m + 2} \right) \right\rbrack

Since $- \frac{1}{3r + 2} = - \frac{1}{3(r + 1) - 1}$ cancels with the $\text{+}\frac{1}{3(r + 1) - 1}$ of the next bracket, every middle term cancels, leaving only the first and last term. Combining these into a single fraction,

\sum_{r = m}^{3m}\frac{1}{9r^{2} + 3r - 2} = \frac{1}{3}\left( \frac{1}{3m - 1} - \frac{1}{9m + 2} \right)

= \frac{1}{3} \times \frac{(9m + 2) - (3m - 1)}{(3m - 1)(9m + 2)}

= \frac{1}{3} \times \frac{6m + 3}{(3m - 1)(9m + 2)}

= \frac{2m + 1}{(3m - 1)(9m + 2)}

To check your answer: at

m = 1

, direct summation over

r = 1,2,3

gives

\frac{1}{10} + \frac{1}{40} + \frac{1}{88} = \frac{3}{22}

, matching

\frac{2(1) + 1}{\left( 3(1) - 1 \right)\left( 9(1) + 2 \right)} = \frac{3}{22}

✓

$\therefore\ \sum_{r = m}^{3m}\frac{1}{9r^{2} + 3r - 2} = \frac{2m + 1}{(3m - 1)(9m + 2)}$

Part (c)

1 mark

\sum_{r = 1}^{\infty}\frac{1}{9r^{2} + 3r - 2} = \frac{1}{6}

Explanation: using the same telescoping as in (b), but starting the sum at $r = 1$ instead of $r = m$ ,

\sum_{r = 1}^{N}\frac{1}{9r^{2} + 3r - 2} = \frac{1}{3}\left( \frac{1}{2} - \frac{1}{3N + 2} \right)

As $N \rightarrow \infty$ , $\frac{1}{3N + 2} \rightarrow 0$ , so the sum tends to $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ .

Part (d)

2 marks

Writing $S(n) = \sum_{r = 1}^{n}\frac{1}{9r^{2} + 3r - 2}$ , part (c) gives $S(n) = \frac{1}{3}\left( \frac{1}{2} - \frac{1}{3n + 2} \right) = \frac{n}{2(3n + 2)}$ . The difference between the sum to infinity and the sum to $n$ terms is

\frac{1}{6} - S(n) = \frac{1}{6} - \frac{n}{2(3n + 2)}

= \frac{(3n + 2) - 3n}{6(3n + 2)}

= \frac{2}{6(3n + 2)}

= \frac{1}{9n + 6}

We require this to be less than $0.004$ . Since $\frac{1}{0.004} = 250$ ,

\frac{1}{9n + 6} < 0.004

\Rightarrow 9n + 6 > 250

\Rightarrow n > \frac{244}{9}

Since $\frac{244}{9} = 27.11\ldots$ and $n$ is an integer, the smallest value is $n = 28$ .

To check your answer: at

n = 27

, the difference is

\frac{1}{9(27) + 6} = \frac{1}{249} \approx 0.00402

, which is not less than

0.004

; at

n = 28

, it is

\frac{1}{258} \approx 0.00388

, which is ✓

$\therefore$ the smallest value of $n$ is $28$ .

2022 H2 Maths Paper 2, Question 4

Part (a)

Part (b)

Part (c)

Part (d)

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