2021 H2 Maths Paper 1 Q5 — Worked Solution

Part (a)

3 marks

\frac{x}{(x + 2)(x + 3)(x + 4)} = \frac{A}{x + 2} + \frac{B}{x + 3} + \frac{C}{x + 4}

Multiplying both sides by $(x + 2)(x + 3)(x + 4)$ ,

x = A(x + 3)(x + 4) + B(x + 2)(x + 4) + C(x + 2)(x + 3)

Substituting $x = - 2,\, - 3,\, - 4$ in turn to isolate each constant:

$x = - 2$ : $\ - 2 = A(1)(2) \Rightarrow A = - 1$

$x = - 3$ : $\ - 3 = B( - 1)(1) \Rightarrow B = 3$

$x = - 4$ : $\ - 4 = C( - 2)( - 1) \Rightarrow C = - 2$

To check your answer: at

x = 1

,

\frac{1}{(3)(4)(5)} = \frac{1}{60}

, and

\frac{- 1}{3} + \frac{3}{4} - \frac{2}{5} = \frac{- 20 + 45 - 24}{60} = \frac{1}{60}

✓.

$\therefore$ $\frac{x}{(x + 2)(x + 3)(x + 4)} = \frac{3}{x + 3} - \frac{1}{x + 2} - \frac{2}{x + 4}$ .

Part (b)

3 marks

Approach

Grouping the three partial-fraction terms as they stand does not telescope, since $\frac{- 1}{r + 2}$ and $\frac{- 2}{r + 4}$ sit two steps apart in $r$ . Splitting $\frac{3}{r + 3}$ into $\frac{1}{r + 3} + \frac{2}{r + 3}$ pairs each term with a neighbour exactly one step away, giving two ordinary telescoping sums.

\sum_{r = 1}^{n}\frac{r}{(r + 2)(r + 3)(r + 4)} = \sum_{r = 1}^{n}\left( \frac{- 1}{r + 2} + \frac{3}{r + 3} - \frac{2}{r + 4} \right)

= \sum_{r = 1}^{n}\left( \frac{1}{r + 3} - \frac{1}{r + 2} \right) + 2\sum_{r = 1}^{n}\left( \frac{1}{r + 3} - \frac{1}{r + 4} \right)

Writing out the terms of the first sum,

\sum_{r = 1}^{n}\left( \frac{1}{r + 3} - \frac{1}{r + 2} \right) = \left( \frac{1}{4} - \frac{1}{3} \right) + \left( \frac{1}{5} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{n + 3} - \frac{1}{n + 2} \right)

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

and similarly, with a gap of one extra step, the second sum telescopes to

\sum_{r = 1}^{n}\left( \frac{1}{r + 3} - \frac{1}{r + 4} \right) = \frac{1}{4} - \frac{1}{n + 4}

Combining,

\sum_{r = 1}^{n}\frac{r}{(r + 2)(r + 3)(r + 4)} = \left( \frac{1}{n + 3} - \frac{1}{3} \right) + 2\left( \frac{1}{4} - \frac{1}{n + 4} \right)

= \frac{1}{n + 3} - \frac{2}{n + 4} + \frac{1}{6}

= \frac{1}{6} - \frac{n + 2}{(n + 3)(n + 4)}

= \frac{(n + 3)(n + 4) - 6(n + 2)}{6(n + 3)(n + 4)}

= \frac{n(n + 1)}{6(n + 3)(n + 4)}

To check your answer: at

n = 3

, direct summation gives

\frac{1}{60} + \frac{1}{60} + \frac{1}{70} \approx 0.0476

, matching

\frac{3(4)}{6(6)(7)} \approx 0.0476

✓.

$\therefore$ $\sum_{r = 1}^{n}\frac{r}{(r + 2)(r + 3)(r + 4)} = \frac{n(n + 1)}{6(n + 3)(n + 4)}$ .

Part (c)

1 mark

\sum_{r = 1}^{\infty}\frac{r}{(r + 2)(r + 3)(r + 4)} = \frac{1}{6}

Explanation: dividing the numerator and denominator of the result in (b) by $n^{2}$ gives $\frac{1 + 1/n}{6(1 + 3/n)(1 + 4/n)}$ , which tends to $\frac{1}{6}$ as $n \rightarrow \infty$ .

2021 H2 Maths Paper 1, Question 5

Part (a)

Part (b)

Part (c)

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