A2 June 2023 Paper 2 Q7
7 In this question you must show detailed reasoning.
You are given that \(\displaystyle\sum_{r=1}^{\infty}\frac{1}{r^2}\) exists and is equal to \(\dfrac{1}{6}\pi^2\).
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{5r + 6}{r^3 + r^2} = \dfrac{A}{r} + \dfrac{B}{r^2} + \dfrac{C}{r + 1}\) | *M1 | 3.1a |
| \(= \dfrac{Ar(r + 1) + B(r + 1) + Cr^2}{r^2(r + 1)}\) Consider \(r = -1\) | *M1 | 1.1 |
| \(C = 1\) | A1 | 1.1 |
| \(r = 0 \Rightarrow B = 6\) and \(A + C = 0 \Rightarrow A = -1\) | A1 | 1.1 |
| \(\therefore \displaystyle\sum_{r=1}^{n}\frac{5r + 6}{r^3 + r^2} = \sum_{r=1}^{n}\left(\frac{-1}{r} + \frac{6}{r^2} + \frac{1}{r + 1}\right) = \sum_{r=1}^{n}\frac{6}{r^2} + \sum_{r=1}^{n}\left(\frac{1}{r + 1} - \frac{1}{r}\right)\) \(= \displaystyle\sum_{r=1}^{n}\frac{6}{r^2} + \frac{1}{2} - \frac{1}{1} + \frac{1}{3} - \frac{1}{2} + \frac{1}{4} - \frac{1}{3} + \ldots + \frac{1}{n} - \frac{1}{n - 1} + \frac{1}{n + 1} - \frac{1}{n}\) | dep*M1 | 1.1 |
| \(= \dfrac{1}{n + 1} - 1 + 6\displaystyle\sum_{r=1}^{n}\frac{1}{r^2}\) so \(a = 1,\ b = -1,\ c = 6\) | A1 | 2.2a |
| [6] |
Notes
*M1: (1st) Correct PF expansion used in solution. Allow extraneous terms only if stated/evaluated as 0. Accept \(\frac{Ar + B}{r^2} + \frac{C}{r + 1}\) but not with additional \(\frac{D}{r}\) unless recovered later
*M1: (2nd) Recombining and use valid method to find coefficients (e.g. appropriate choice of \(r\) or comparing coefficients in \(r^2\), \(r\) or \(r^0\)). Indep of 1st M1 if from PF terms involving 3+ unknowns and their factors seen in a denominator
A1: (1st) Any one correct non-zero coefficient
A1: (2nd) Other two coefficients by valid method
dep*M1: Separating and expressing sum in form in which cancellation pattern is clear
\(r^{\prime} = r + 1\): \(\therefore \displaystyle\sum_{r=1}^{n}\frac{1}{r + 1} - \sum_{r=1}^{n}\frac{1}{r} = \sum_{r^{\prime}=2}^{n+1}\frac{1}{r^{\prime}} - \sum_{r=1}^{n}\frac{1}{r} = \frac{1}{n + 1} + \sum_{r=2}^{n}\left(\frac{1}{r} - \frac{1}{r}\right) - \frac{1}{1} = \frac{1}{n + 1} - 1\)
A1: (3rd) Complete argument with all detail. Allow embedded answers. Minimum of first and last cancellation terms shown (If algebraic approach, must see full argument)
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\lim_{n \to \infty}\frac{1}{n + 1} = 0\) | M1 | 3.1a |
| \(\displaystyle\sum_{r=1}^{\infty}\frac{5r + 6}{r^3 + r^2} = \lim_{n \to \infty}\left\{\sum_{r=1}^{n}\frac{5r + 6}{r^3 + r^2}\right\}\) \(= \displaystyle\lim_{n \to \infty}\left\{\frac{1}{n + 1} - 1 + 6\sum_{r=1}^{n}\frac{1}{r^2}\right\} = \pi^2 - 1\) \(= (\pi - 1)(\pi + 1)\) | A1 | 2.2a |
| [2] |
Notes
M1: AG. Considering the limit as \(n\) tends to infinity of \(\frac{1}{n + 1}\) term; not \(\frac{1}{\text{infinity}}\)
A1: AG. Joined up argument. Some intermediate working must be shown. Condone poor, but clear limit notation