AS June 2025 Paper 1 Q8
8
(a) Show that, for all positive integers \(r\),\[\frac{1}{r^2} - \frac{1}{(r + 1)^2} = \frac{2r + 1}{r^2(r + 1)^2}\] [1 mark]
(b) Hence, using the method of differences, show that\[\sum_{r=1}^{n} \frac{2r + 1}{r^2(r + 1)^2} = \frac{an^2 + bn}{(n + 1)^2}\]
where \(a\) and \(b\) are integers to be found. [3 marks]
(c) Hence show that, for all positive integers \(c\),\[\sum_{r=c}^{2c} \frac{2r + 1}{r^2(r + 1)^2}\]
can be written in the form
\[\frac{(pc + 1)(c + 1)}{c^2(qc + 1)^2}\]where \(p\) and \(q\) are integers to be found. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Completes a rigorous argument to prove the required result. Must include the LHS, at least one intermediate stage, and the RHS. Accept the abbreviation LHS. | R1 | 2.1 |
| (1) |
Typical solution
\[\begin{aligned}\frac{1}{r^2} - \frac{1}{(r + 1)^2} &= \frac{(r + 1)^2 - r^2}{r^2(r + 1)^2} \\ &= \frac{r^2 + 2r + 1 - r^2}{r^2(r + 1)^2} \\ &= \frac{2r + 1}{r^2(r + 1)^2}\end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Writes at least two pairs of subtracting fractions. | M1 | 1.1a |
| Writes the \(n\)th pair of fractions and at least one pair of cancelling fractions. | M1 | 1.1a |
| Completes a reasoned argument using the method of differences to reach the required result. This mark is only available if at least the first two pairs of fractions and the \(n\)th pair are shown. Accept \(\dfrac{1n^2 + 2n}{(n + 1)^2}\) | R1 | 2.1 |
| (3) |
Typical solution
\[\sum_{r=1}^{n} \frac{2r + 1}{r^2(r + 1)^2} = \sum_{r=1}^{n}\left(\frac{1}{r^2} - \frac{1}{(r + 1)^2}\right)\]\[\begin{aligned} = {} & \dfrac{1}{1^2} - \cancel{\dfrac{1}{2^2}} \\[6pt] + {} & \cancel{\dfrac{1}{2^2}} - \cancel{\dfrac{1}{3^2}} \\[6pt] + {} & \ldots\ldots\ldots\ldots \\[6pt] + {} & \cancel{\dfrac{1}{(n - 1)^2}} - \cancel{\dfrac{1}{n^2}} \\[6pt] + {} & \cancel{\dfrac{1}{n^2}} - \dfrac{1}{(n + 1)^2} \\[6pt] = {} & 1 - \dfrac{1}{(n + 1)^2} \\[6pt] = {} & \dfrac{(n + 1)^2 - 1}{(n + 1)^2} \\[6pt] = {} & \dfrac{n^2 + 2n}{(n + 1)^2}\end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(n = 2c\) or \(n = c - 1\) into their \(\dfrac{an^2 + bn}{(n + 1)^2}\) | M1 | 1.1a |
| Substitutes \(n = 2c\) and \(n = c - 1\) into their \(\dfrac{an^2 + bn}{(n + 1)^2}\) and subtracts. | M1 | 3.1a |
| Obtains \(\dfrac{(3c + 1)(c + 1)}{c^2(2c + 1)^2}\) | R1 | 2.1 |
| (3) | ||
| (7 marks) |
Typical solution
\[\begin{aligned}\sum_{r=c}^{2c} \frac{2r + 1}{r^2(r + 1)^2} &= \sum_{r=1}^{2c} \frac{2r + 1}{r^2(r + 1)^2} - \sum_{r=1}^{c-1} \frac{2r + 1}{r^2(r + 1)^2} \\ &= \frac{(2c)^2 + 2(2c)}{(2c + 1)^2} - \frac{(c - 1)^2 + 2(c - 1)}{c^2} \\ &= \frac{4c^3(c + 1) - (2c + 1)^2(c^2 - 2c + 1 + 2c - 2)}{c^2(2c + 1)^2} \\ &= \frac{4c^3(c + 1) - (2c + 1)^2(c^2 - 1)}{c^2(2c + 1)^2} \\ &= \frac{(c + 1)\left(4c^3 - (4c^2 + 4c + 1)(c - 1)\right)}{c^2(2c + 1)^2} \\ &= \frac{(c + 1)\left(4c^3 - (4c^3 - 3c - 1)\right)}{c^2(2c + 1)^2} \\ &= \frac{(3c + 1)(c + 1)}{c^2(2c + 1)^2}\end{aligned}\]Notes
(corrected from the printed mark scheme: the M1 rows print “Substitutes \(r = 2c\) … \(r = c - 1\)”; the substitution is into the \(n\) of the part (b) result.)