A2 June 2019 Paper 2 Q1
1 In this question you must show detailed reasoning.
(a) By using partial fractions show that \(\displaystyle\sum_{r=1}^{n} \frac{1}{r^2 + 3r + 2} = \frac{1}{2} - \frac{1}{n + 2}\). [5]
(b) Hence determine the value of \(\displaystyle\sum_{r=1}^{\infty} \frac{1}{r^2 + 3r + 2}\). [2]
| Scheme | Marks | AO |
|---|---|---|
| DR \((r + 2)(r + 1)\) | B1 | 1.1a |
| \(\dfrac{A}{r + 1} + \dfrac{B}{r + 2}\) | M1 | 1.1 |
| \(A = 1,\ B = -1\) | A1 | 1.1 |
| \(\Sigma = \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4}\ \ldots - \dfrac{1}{n + 1} + \dfrac{1}{n + 1} - \dfrac{1}{n + 2}\) | M1 | 2.1 |
| \(= \dfrac{1}{2} - \dfrac{1}{n + 2}\) | A1 | 1.1 |
| [5] |
Notes
B1: Correct factorisation of denominator soi
M1: Correct general form for partial fractions soi by correct answer. Could be incorrect if recovered. eg \(\frac{A}{r + 1} + \frac{B}{r + 2} + C\) and \(C = 0\)
A1: Both: \(\dfrac{1}{r + 1} - \dfrac{1}{r + 2}\)
M1: Condone omission of \(+\frac{1}{4}\) and \(-\frac{1}{n + 1}\) for M1 only
A1: AG. Cancellation must be evident
| Scheme | Marks | AO |
|---|---|---|
| DR \(\displaystyle\sum^{\infty} = \frac{1}{2}\) | B1 | 2.2a |
| since \(\dfrac{1}{n + 2} \to 0\) as \(n \to \infty\) | B1 | 2.4 |
| [2] |
Notes
B1: Or \(\displaystyle\lim_{n \to \infty}\left(\frac{1}{2} - \frac{1}{n + 2}\right) = \frac{1}{2} - 0 = \frac{1}{2}\)
B1: Indication that \(1/(n + 2)\) is close to zero (accept “small”) when \(n\) is large