A2 October 2021 Paper 1 Q1
1
(a) Express \(\dfrac{1}{(2r - 1)(2r + 1)}\) in partial fractions. [3]
(b) Hence find \(\displaystyle\sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)}\), expressing the result as a single fraction. [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{(2r - 1)(2r + 1)} = \dfrac{A}{2r - 1} + \dfrac{B}{2r + 1}\) \(\Rightarrow 1 = A(2r + 1) + B(2r - 1)\) | M1 | 1.1a |
| \(r = \frac{1}{2} \Rightarrow A = \frac{1}{2}\) | A1 | 1.1 |
| \(r = -\frac{1}{2} \Rightarrow B = -\frac{1}{2}\) | A1 | 1.1 |
| [3] |
Notes
M1: or cover-up method
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)} = \frac{1}{2}\sum_{r=1}^{n}\left(\frac{1}{2r - 1} - \frac{1}{2r + 1}\right)\) | M1* | 1.1 |
| \(= \dfrac{1}{2}\left(1 - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + \ldots + \dfrac{1}{2n - 3} - \dfrac{1}{2n - 1} + \dfrac{1}{2n - 1} - \dfrac{1}{2n + 1}\right)\) | M1dep* | 1.2 |
| \(= \dfrac{1}{2}\left(1 - \dfrac{1}{2n + 1}\right)\) | A1 | 1.1 |
| \(= \dfrac{n}{2n + 1}\) | A1 | 1.1 |
| [4] |
Notes
M1dep*: showing cancellation clearly
SC: B3 For fully correct summation with A = \(-\frac{1}{2}\) and B = \(\frac{1}{2}\)