A2 June 2022 Paper 2 Q8
8 In this question you must show detailed reasoning.
It is given that \(\displaystyle\sum_{r=k}^{98}\frac{5r + 2}{r(r + 1)(r + 2)} = \frac{20539}{34650}\) for some \(k\).
Determine the value of \(k\). [7]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{5r + 2}{r(r + 1)(r + 2)} = \dfrac{A}{r} + \dfrac{B}{r + 1} + \dfrac{C}{r + 2}\) | M1 | 3.1a |
| \(= \dfrac{1}{r} + \dfrac{3}{r + 1} - \dfrac{4}{r + 2}\) | A1 | 1.1 |
| \(\displaystyle\sum_{r=k}^{98}\frac{5r + 2}{r(r + 1)(r + 2)} = \sum_{r=k}^{98}\left(\frac{1}{r} + \frac{3}{r + 1} - \frac{4}{r + 2}\right)\) \(\displaystyle \begin{aligned} &= \frac{1}{k} + \frac{3}{k + 1} - \frac{4}{k + 2} \\[6pt] &+ \frac{1}{k + 1} + \frac{3}{k + 2} - \frac{4}{k + 3} \\[6pt] &+ \frac{1}{k + 2} + \frac{3}{k + 3} - \frac{4}{k + 4} \\[6pt] &+ \ldots \\[6pt] &+ \frac{1}{96} + \frac{3}{97} - \frac{4}{98} \\[6pt] &+ \frac{1}{97} + \frac{3}{98} - \frac{4}{99} \\[6pt] &+ \frac{1}{98} + \frac{3}{99} - \frac{4}{100} \end{aligned}\) | M1 | 3.1a |
| \(= \dfrac{1}{k} + \dfrac{4}{k + 1} - \dfrac{124}{2475}\) or \(\dfrac{1}{k} + \dfrac{4}{k + 1} - \dfrac{1}{99} - \dfrac{4}{100}\) oe | A1 | 2.2b |
| \(\dfrac{1}{k} + \dfrac{4}{k + 1} - \dfrac{124}{2475} = \dfrac{20539}{34650} \Rightarrow \dfrac{1}{k} + \dfrac{4}{k + 1} = \dfrac{9}{14}\) \(\Rightarrow 14(k + 1) + 56k = 9k(k + 1)\) \(\Rightarrow 9k^2 - 61k - 14 = 0\) | M1 | 1.1 |
| \((9k + 2)(k - 7) = 0 \Rightarrow k = -\dfrac{2}{9}\) or \(7\) | A1 | 1.1 |
| But \(k\) is an integer so \(k \neq -\dfrac{2}{9}\) so \(k = 7\) | A1 | 3.2a |
| [7] |
Notes
M1: Correct form for partial fractions
A1: or \(A = 1,\ B = 3,\ C = -4\)
M1: Writing out sufficient terms so that cancellation pattern becomes evident. Could see \(N\) rather than 98.
If use \(r = a\) to \(N\) then M1 awarded when \(N = 98\) and \(N = k - 1\) both considered.
A1: Could see \(\dfrac{5k + 1}{k(k + 1)} - \dfrac{124}{2475}\)
M1: Equating their expression for LHS to 20539/34650 and rearranging to 3 term quadratic
A1: \(\dfrac{--61 \pm \sqrt{(-61)^2 - 4 \times 9 \times (-14)}}{2 \times 9} = \dfrac{61 \pm \sqrt{4225}}{18} = \dfrac{61 \pm 65}{18}\)
or \(9\left(\left(k - \dfrac{61}{18}\right)^2 - \left(\dfrac{61}{18}\right)^2\right) - 14 = 0\)
\(9\left(k - \dfrac{61}{18}\right)^2 - \dfrac{4225}{36} = 0\)
A1: Explicitly rejecting non-integer value, with reason, and deducing correct value… …or stating that the original function is undefined if \(r\) is allowed to take the value 0.
NB \(k \geqslant 0\) (or \(k \gt 0\)) alone is not a valid reason to reject \(-2/9\)