A2 June 2025 Paper 1 Q5
5. Use the method of differences to prove that for \(n \gt 2\)
\[\sum_{r=2}^{n} \frac{4}{r^2 - 1} = \frac{(pn + q)(n - 1)}{n(n + 1)}\]where \(p\) and \(q\) are constants to be determined. (5)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{4}{r^2 - 1} \equiv \dfrac{4}{(r - 1)(r + 1)} \equiv \dfrac{A}{r - 1} + \dfrac{B}{r + 1} \Rightarrow A = \ldots,\ B = \ldots\) \((\text{NB } A = 2,\ B = -2)\) | M1 | 3.1a |
| \(\begin{aligned} r &= 2 &&\Rightarrow \dfrac{2}{1} - \dfrac{2}{3}\\[6pt] r &= 3 &&\Rightarrow \dfrac{2}{2} - \dfrac{2}{4}\\[6pt] r &= 4 &&\Rightarrow \dfrac{2}{3} - \dfrac{2}{5}\\[6pt] r &= n - 1 &&\Rightarrow \dfrac{2}{n - 2} - \dfrac{2}{n}\\[6pt] r &= n &&\Rightarrow \dfrac{2}{n - 1} - \dfrac{2}{n + 1}\end{aligned}\) | dM1 | 2.1 |
| \(2 + 1 - \dfrac{2}{n} - \dfrac{2}{n + 1}\) | A1 | 1.1b |
| \(\dfrac{3n(n + 1) - 2(n + 1) - 2n}{n(n + 1)} = \dfrac{3n^2 + 3n - 2n - 2 - 2n}{n(n + 1)} = \dfrac{3n^2 - n - 2}{n(n + 1)}\) | M1 | 1.1b |
| \(\dfrac{(3n + 2)(n - 1)}{n(n + 1)}\) cso | A1 | 2.2a |
| (5) | ||
| (5 marks) |
Notes
M1: A complete strategy to write as partial fractions
dM1: Dependent on previous method mark. Completes the method of differences process, must have a minimum substitution of \(r = 2,\ r = 3,\ r = n - 1\) and \(r = n\). Condone the use of \(r\) for \(n\) for this mark.
If pairs of terms are not seen in each substitution for \(r\), this mark can be implied by sight of \(2 + 1 - \dfrac{2}{n} - \dfrac{2}{n + 1}\) following correct partial fractions.
A1: Correct ‘fractions’ from the beginning and end that do not cancel. Accept exact equivalents. Must be in terms of \(n\)
M1: Combines all their ‘fractions’ (at least two algebraic fractions) over a correct common denominator, leading to a quadratic in the numerator, where brackets may not be expanded. Condone the use of \(r\) for \(n\) for this mark.
A1: Since the form of the solution is given, only accept \(\dfrac{(3n + 2)(n - 1)}{n(n + 1)}\) and not equivalents, but isw
Do not accept values for \(p\) and \(q\) written alone.
Note: If they start with \(r = 0,\ r = 1\) or \(r = 3\), the maximum they can score is M1M0A0M1A0
Similarly, if they use \(r = n + 1\) the maximum they can score is M1M0A0M1A0