2023 H2 Maths Paper 1 Q5 — Worked Solution

Part (a)

3 marks

We split each logarithm using $\ln(r - 1) + \ln(r + 1) - 2\ln r$ , then regroup it into a difference. Let $u_{r} = \ln r - \ln(r + 1)$ .

\ln\left\lbrack \frac{(r - 1)(r + 1)}{r^{2}} \right\rbrack

= \ln(r - 1) + \ln(r + 1) - 2\ln r

= \left\lbrack \ln(r - 1) - \ln r \right\rbrack - \left\lbrack \ln r - \ln(r + 1) \right\rbrack

= u_{r - 1} - u_{r}

Summing from $r = 2$ to $n$ , this telescopes:

\sum_{r = 2}^{n}\ln\left\lbrack \frac{(r - 1)(r + 1)}{r^{2}} \right\rbrack

= \sum_{r = 2}^{n}\left( u_{r - 1} - u_{r} \right)

= u_{1} - u_{n}

= \left\lbrack \ln 1 - \ln 2 \right\rbrack - \left\lbrack \ln n - \ln(n + 1) \right\rbrack

= \ln(n + 1) - \ln n - \ln 2

= \ln\left( \frac{n + 1}{n} \right) - \ln 2\ \ \left( \text{shown} \right)

To check your answer: for

n = 2

, LHS

\text{=}\ln\frac{3}{4}

and RHS

\text{=}\ln\frac{3}{2} - \ln 2 = \ln\frac{3}{4}

✓

Part (b)

2 marks

Letting $n \rightarrow \infty$ in the result of (a): since $\frac{n + 1}{n} = 1 + \frac{1}{n} \rightarrow 1$ , $\ln\left( \frac{n + 1}{n} \right) \rightarrow \ln 1 = 0$ , so the partial sum

\sum_{r = 2}^{n}\ln\left\lbrack \frac{(r - 1)(r + 1)}{r^{2}} \right\rbrack \rightarrow 0 - \ln 2 = - \ln 2\ \text{as }n \rightarrow \infty.

Since the partial sums tend to a finite limit, the series is convergent.

$\therefore$ the series converges, with sum to infinity $- \ln 2$ .

Part (c)

2 marks

Approach

writing $S_{n}$ for the partial sum found in (a), the terms $r = 2,\ldots,9$ cancel between $S_{20}$ and $S_{9}$ , so the required sum is simply $S_{20} - S_{9}$ .

Using $S_{n} = \sum_{r = 2}^{n}\ln\left\lbrack \frac{(r - 1)(r + 1)}{r^{2}} \right\rbrack = \ln\left( \frac{n + 1}{n} \right) - \ln 2$ from (a),

\sum_{r = 10}^{20}\ln\left\lbrack \frac{(r - 1)(r + 1)}{r^{2}} \right\rbrack = S_{20} - S_{9}

= \left\lbrack \ln\frac{21}{20} - \ln 2 \right\rbrack - \left\lbrack \ln\frac{10}{9} - \ln 2 \right\rbrack

= \ln\frac{21}{20} - \ln\frac{10}{9}

= \ln\left( \frac{21/20}{10/9} \right)

= \ln\left( \frac{189}{200} \right)

To check your answer: direct numerical summation of

\sum_{r = 10}^{20}\ln\left\lbrack \frac{(r - 1)(r + 1)}{r^{2}} \right\rbrack

gives

\text{≈} - 0.0566

, matching

\ln(189/200) \approx - 0.0566

✓

$\therefore a = 189,\ b = 200$ .

2023 H2 Maths Paper 1, Question 5

Part (a)

Part (b)

Part (c)

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