2015 H2 Maths Paper 2 Q4 — Worked Solution

Part (a)

6 marks

Let $P(n)$ be the statement

1 \times 3 \times 6 + 2 \times 4 \times 7 + 3 \times 5 \times 8 + \ldots + n(n + 2)(n + 5) = \frac{1}{12}n(n + 1)\left( 3n^{2} + 31n + 74 \right)

for $n \in \mathbb{Z}^{\text{+}}$ .

When $n = 1$ : LHS $\text{=}1 \times 3 \times 6 = 18$ . RHS $\text{=}\frac{1}{12}(1)(2)(3 + 31 + 74) = \frac{1}{12}(2)(108) = 18$ . Since LHS $\text{=}$ RHS, $P(1)$ is true.

Assume $P(k)$ is true for some $k \in \mathbb{Z}^{\text{+}}$ , i.e.

\sum_{r = 1}^{k}r(r + 2)(r + 5) = \frac{1}{12}k(k + 1)\left( 3k^{2} + 31k + 74 \right)

We want to show $P(k + 1)$ is true, i.e.

\sum_{r = 1}^{k + 1}r(r + 2)(r + 5) = \frac{1}{12}(k + 1)(k + 2)\left( 3(k + 1)^{2} + 31(k + 1) + 74 \right)

Now,

\sum_{r = 1}^{k + 1}r(r + 2)(r + 5) = \sum_{r = 1}^{k}r(r + 2)(r + 5) + (k + 1)(k + 3)(k + 6)

= \frac{1}{12}k(k + 1)\left( 3k^{2} + 31k + 74 \right) + (k + 1)(k + 3)(k + 6)

= \frac{1}{12}(k + 1)\left\lbrack k\left( 3k^{2} + 31k + 74 \right) + 12(k + 3)(k + 6) \right\rbrack

= \frac{1}{12}(k + 1)\left\lbrack 3k^{3} + 43k^{2} + 182k + 216 \right\rbrack

= \frac{1}{12}(k + 1)(k + 2)\left( 3k^{2} + 37k + 108 \right)

Since $3(k + 1)^{2} + 31(k + 1) + 74 = 3k^{2} + 37k + 108$ , this is

\sum_{r = 1}^{k + 1}r(r + 2)(r + 5) = \frac{1}{12}(k + 1)(k + 2)\left( 3(k + 1)^{2} + 31(k + 1) + 74 \right)

so $P(k + 1)$ is true whenever $P(k)$ is true.

Since $P(1)$ is true, and $P(k)$ true $\Rightarrow$ $P(k + 1)$ true, by mathematical induction $P(n)$ is true for all $n \in \mathbb{Z}^{\text{+}}$ . (shown)

Part (b)(i)

1 mark

\frac{2}{4r^{2} + 8r + 3} = \frac{2}{(2r + 1)(2r + 3)}

Approach

write $\frac{2}{(2r + 1)(2r + 3)} = \frac{A}{2r + 1} + \frac{B}{2r + 3}$ and compare coefficients after multiplying through.

2 = A(2r + 3) + B(2r + 1)

Comparing coefficients of $r$ : $2A + 2B = 0$ , so $B = - A$ . Comparing constants: $3A + B = 2$ , so $3A - A = 2$ , giving $A = 1$ , $B = - 1$ .

$\therefore$ $A = 1$ , $B = - 1$ , so $\frac{2}{4r^{2} + 8r + 3} = \frac{1}{2r + 1} - \frac{1}{2r + 3}$ . (shown)

Part (b)(ii)

4 marks

Approach

the partial-fraction form telescopes; we write out the first few and last few terms to see which cancel.

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

S_{n} = \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{7} \right) + \left( \frac{1}{7} - \frac{1}{9} \right) + \ldots + \left( \frac{1}{2n + 1} - \frac{1}{2n + 3} \right)

All the middle terms cancel, leaving

S_{n} = \frac{1}{3} - \frac{1}{2n + 3}

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

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

$\therefore$ $S_{n} = \frac{1}{3} - \frac{1}{2n + 3} = \frac{2n}{3(2n + 3)}$ .

Part (b)(iii)

3 marks

Approach

as $n \rightarrow \infty$ , $\frac{1}{2n + 3} \rightarrow 0$ , so the sum to infinity is $\frac{1}{3}$ . We need the smallest $n$ for which $S_{n}$ is within $10^{- 3}$ of this value, i.e. for which $S_{\infty} - S_{n} < 10^{- 3}$ .

S_{\infty} = \frac{1}{3}

S_{\infty} - S_{n} = \frac{1}{2n + 3}

We require

\frac{1}{2n + 3} < 10^{- 3}

2n + 3 > 1000

n > 498.5

By GC, checking the integers near this boundary: at $n = 498$ , $\frac{1}{2(498) + 3} = \frac{1}{999} \approx 0.001001 > 0.001$ ; at $n = 499$ , $\frac{1}{2(499) + 3} = \frac{1}{1001} \approx 0.000999 < 0.001$ .

$\therefore$ the smallest value of $n$ is $499$ .

2015 H2 Maths Paper 2, Question 4

Part (a)

Part (b)(i)

Part (b)(ii)

Part (b)(iii)

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