2019 H2 Maths Paper 1 Q6 — Worked Solution

Part (i)

4 marks

Approach

$4r^{2} - 1$ factorises as a difference of two squares, $(2r - 1)(2r + 1)$ , so we expect the partial fractions to telescope when summed — each term should cancel with a term two steps away.

Since $4r^{2} - 1 = (2r - 1)(2r + 1)$ , we write

\frac{1}{4r^{2} - 1} = \frac{A}{2r - 1} + \frac{B}{2r + 1}

for constants $A$ and $B$ . Multiplying through by $(2r - 1)(2r + 1)$ ,

1 = A(2r + 1) + B(2r - 1).

Substituting $r = \frac{1}{2}$ : $1 = 2A$ , so $A = \frac{1}{2}$ . Substituting $r = - \frac{1}{2}$ : $1 = - 2B$ , so $B = - \frac{1}{2}$ .

\therefore\frac{1}{4r^{2} - 1} = \frac{1}{2}\left( \frac{1}{2r - 1} - \frac{1}{2r + 1} \right).

To check your answer: at

r = 1

\frac{1}{4(1)^{2} - 1} = \frac{1}{3}

, and

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

. ✓

Now we sum, writing out enough terms to see the cancellation:

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

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

Every term except the first and the last cancels, leaving

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

= \frac{1}{2} \cdot \frac{2n}{2n + 1}

= \frac{n}{2n + 1}.

$\therefore$ $\sum_{r = 1}^{n}\frac{1}{4r^{2} - 1} = \frac{n}{2n + 1}$ .

Part (ii)

2 marks

Approach

The given sum starts at $r = 11$ instead of $r = 1$ . We get it from the result of (i) by taking $n \rightarrow \infty$ to get the sum to infinity, then subtracting off the first $10$ terms (i.e. the sum up to $r = 10$ ), since $\sum_{r = 11}^{\infty}u_{r} = \sum_{r = 1}^{\infty}u_{r} - \sum_{r = 1}^{10}u_{r}$ , where $u_{r} = \frac{1}{4r^{2} - 1}$ .

As $n \rightarrow \infty$ , $\frac{n}{2n + 1} = \frac{1}{2 + \frac{1}{n}} \rightarrow \frac{1}{2}$ , so

\sum_{r = 1}^{\infty}\frac{1}{4r^{2} - 1} = \frac{1}{2}.

Using the result of (i) with $n = 10$ ,

\sum_{r = 1}^{10}\frac{1}{4r^{2} - 1} = \frac{10}{21}.

Hence

\sum_{r = 11}^{\infty}\frac{1}{4r^{2} - 1} = \frac{1}{2} - \frac{10}{21}

= \frac{21 - 20}{42}

= \frac{1}{42}.

$\therefore$ $\sum_{r = 11}^{\infty}\frac{1}{4r^{2} - 1} = \frac{1}{42}$ .

2019 H2 Maths Paper 1, Question 6

Part (i)

Part (ii)

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