A2 June 2023 Paper 1 Q3
3
(a) Using partial fractions and the method of differences, show that\[\dfrac{1}{1 \times 3} + \dfrac{1}{2 \times 4} + \dfrac{1}{3 \times 5} + \ldots + \dfrac{1}{n(n + 2)} = \dfrac{3}{4} - \dfrac{an + b}{2(n + 1)(n + 2)},\]where \(a\) and \(b\) are integers to be determined. [5]
(b) Deduce the sum to infinity of the series.\[\dfrac{1}{1 \times 3} + \dfrac{1}{2 \times 4} + \dfrac{1}{3 \times 5} + \ldots.\] [1]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{r(r + 2)} = \dfrac{A}{r} + \dfrac{B}{(r + 2)}\) \(1 = A(r + 2) + Br\) | M1 | 1.1 |
| \(r = 0 \Rightarrow A = \frac{1}{2}, \quad r = -2 \Rightarrow B = -\frac{1}{2}\) | A1 | 1.1 |
| \(\displaystyle\sum_{r=1}^{n}\frac{1}{r(r + 2)} = \frac{1}{2}\sum_{r=1}^{n}\left(\frac{1}{r} - \frac{1}{r + 2}\right)\) \(= k\left[1 - \dfrac{1}{3} + \dfrac{1}{2} - \dfrac{1}{4} + \dfrac{1}{3} - \dfrac{1}{5} + \ldots + \dfrac{1}{n} - \dfrac{1}{n + 2}\right]\) | M1 | 1.1 |
| \(= \dfrac{1}{2}\left[1 + \dfrac{1}{2} - \dfrac{1}{n + 1} - \dfrac{1}{n + 2}\right]\) | A1 | 1.1 |
| \(= \dfrac{3}{4} - \dfrac{2n + 3}{2(n + 1)(n + 2)}\) | A1 | 1.1 |
| [5] |
Notes
M1: (2nd) enough terms to show consistent cancellation in their series
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{3}{4}\) | B1 | 2.2a |
| [1] |