AS June 2025 Paper 1 Q6
6
(a) Express \(\dfrac{1}{(r - 1)^2} - \dfrac{1}{(r + 1)^2}\) as a single simplified fraction. [2]
(b) Hence determine the limit which \(\displaystyle\sum_{r=2}^{n} \frac{r}{(r - 1)^2(r + 1)^2}\) converges to as \(n \to \infty\). [5]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{(r - 1)^2} - \dfrac{1}{(r + 1)^2} = \dfrac{(r + 1)^2 - (r - 1)^2}{(r + 1)^2(r - 1)^2}\) | M1 | 1.1 |
| \(= \dfrac{4r}{(r + 1)^2(r - 1)^2}\) | A1 | 1.1 |
| [2] |
Notes
M1: Correctly combine both given fractions
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle 4\sum_{r=2}^{n} \frac{r}{(r + 1)^2(r - 1)^2} = \sum_{r=2}^{n}\left(\frac{1}{(r - 1)^2} - \frac{1}{(r + 1)^2}\right)\) | M1 | 2.1 |
| \(= \left(\begin{aligned} &\frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{2^2} - \frac{1}{4^2} + \frac{1}{3^2} - \frac{1}{5^2} + \ldots \\ &\left[\ldots + \frac{1}{(n - 3)^2} - \frac{1}{(n - 1)^2} + \frac{1}{(n - 2)^2} - \frac{1}{n^2} + \frac{1}{(n - 1)^2} - \frac{1}{(n + 1)^2}\right] \end{aligned}\right)\) | M1 A1 | 2.5 2.1 |
| \(= 1 + \dfrac{1}{4} - \dfrac{1}{n^2} - \dfrac{1}{(n + 1)^2}\) | B1* | 1.1 |
| As \(n \to \infty\), this tends to \(1\frac{1}{4}\) oe, so \(\displaystyle\sum_{r=2}^{n} \frac{r}{(r + 1)^2(r - 1)^2}\) tends to \(\dfrac{5}{16}\) | B1 | 2.2a |
| [5] |
Notes
M1: splitting fractions using above result
M1: method of differences attempted
A1: at least first three terms (for \(n = 2\), 3 and 4) shown. The last three for \(n - 2\), \(n - 1\) and \(n\) can be omitted, but if shown must be correct
B1*: or \(= 1 + \dfrac{1}{4}\) (if without the terms in \(n\))
B1: 0.3125 dep B1*