FP2 2016 1 Edexcel

(a)

Write:

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

for constants A,B.

So that, for \displaystyle r \ne \pm \frac 1 2:

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

We have A - B = 1 and 2A + 2B = 0 so A = -B, giving A = \dfrac 1 2 and B = -\dfrac 1 2, so:

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

Considering the mark distribution, the expectation would be that you’d do this by inspection, noticing that \dfrac 1 {2r - 1} > \dfrac 1 {2r + 1} and 2r + 1 - (2r - 1) = 2.

(b)

We have:

\begin{align*}\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 - \cancel{\frac 1 3}\right) + \left(\cancel{\frac 1 3} - \cancel{\frac 1 5}\right) + \ldots + \left(\cancel{\frac 1 {2n - 3}} - \cancel{\frac 1 {2n - 1}}\right) + \left(\cancel{\frac 1 {2n - 1}} - \frac 1 {2n + 1}\right)\right) \\ & = \frac 1 2 \left(1 - \frac 1 {2n + 1}\right) \\ & = \frac 1 2 \times \frac {2n + 1 - 1} {2n + 1} \\ & = \frac n {2n + 1}\end{align*}