FP2 June 2017 Q1
1.
(a) Show that, for \(r \gt 0\) \[\frac{1}{r^2} - \frac{1}{(r + 1)^2} \equiv \frac{2r + 1}{r^2(r + 1)^2}\] (1)
(b) Hence prove that, for \(n \in \mathbb{N}\) \[\sum_{r=1}^{n} \frac{2r + 1}{r^2(r + 1)^2} = \frac{n(n + 2)}{(n + 1)^2}\] (3)
(c) Show that, for \(n \in \mathbb{N},\ n \gt 1\) \[\sum_{r=n}^{3n} \frac{6r + 3}{r^2(r + 1)^2} = \frac{an^2 + bn + c}{n^2(3n + 1)^2}\] where \(a\), \(b\) and \(c\) are constants to be found. (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{r^2} - \dfrac{1}{(r + 1)^2} = \dfrac{(r + 1)^2 - r^2}{r^2(r + 1)^2} = \dfrac{2r + 1}{r^2(r + 1)^2}\) Correct proof (minimum as shown) (\((r + 1)^2\) or \(r^2 + 2r + 1\) Can be worked in either direction. | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{n}\left(\frac{1}{r^2} - \frac{1}{(r + 1)^2}\right) = 1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{9} \ldots\ldots + \left(\frac{1}{n^2}\right) - \frac{1}{(n + 1)^2}\) Terms of the series with \(r = 1,\ r = n\) and one of \(r = 2,\ r = n - 1\) should be shown. | M1 |
| \(1 - \dfrac{1}{(n + 1)^2}\) Extracts correct terms that do not cancel | A1 |
| \(\dfrac{(n + 1)^2 - 1}{(n + 1)^2} = \dfrac{n(n + 2)}{(n + 1)^2}\) * Correct completion with no errors | A1*cso |
| (3) |
Notes
Alternative for (b) - by induction. NB: No marks available if result in (a) is not used.
| Scheme | Marks |
|---|---|
| Assume true for \(n = k\) | |
| \(\displaystyle\sum_{r=1}^{k+1}\frac{2r + 1}{r^2(r + 1)^2} = \frac{k(k + 2)}{(k + 1)^2} + \frac{1}{(k + 1)^2} - \frac{1}{(k + 2)^2}\) Uses \(\displaystyle\sum_{r=1}^{k}\) together with the \((k + 1)\)th term as 2 fractions (see (a)) | M1 |
| \(= \dfrac{k^2 + 2k + 1}{(k + 1)^2} - \dfrac{1}{(k + 2)^2}\) | |
| \(1 - \dfrac{1}{(k + 2)^2} = \dfrac{k^2 + 4k + 3}{(k + 2)^2} = \dfrac{(k + 1)(k + 3)}{(k + 2)^2}\) Combines the 3 fractions to obtain a single fraction. Must be correct but numerator need not be factorised. | A1 |
| Show true for \(n = 1\) This must be seen somewhere | |
| Hence proved by induction Complete proof with no errors and a concluding statement. | A1 |
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=n}^{3n}\frac{6r + 3}{r^2(r + 1)^2} = 3\left(\frac{3n(3n + 2)}{(3n + 1)^2} - \frac{(n - 1)(n + 1)}{n^2}\right)\) Attempts to use \(\mathrm{f}(3n) - \left(\mathrm{f}(n - 1)\text{ or }\mathrm{f}(n)\right)\) 3 may be missing | M1 |
| \(= 3\left(\dfrac{3n^3(3n + 2) - (3n + 1)^2\left(n^2 - 1\right)}{n^2(3n + 1)^2}\right)\) Attempt at common denominator, Denom to be \(n^2(3n + 1)^2\) or \((n + 1)^2(3n + 1)^2\) Numerator to be difference of 2 quartics. 3 may be missing | dM1 |
| \(= \dfrac{24n^2 + 18n + 3}{n^2(3n + 1)^2}\) cao | A1cao |
| (3) | |
| (7 marks) |
Notes
Alternative for part (c)
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=n}^{3n}\frac{6r + 3}{r^2(r + 1)^2} = 3\left(\frac{1}{n^2} - \frac{1}{(3n + 1)^2}\right)\) OR: \(3\left(\dfrac{1}{(n + 1)^2} - \dfrac{1}{(3n + 1)^2}\right)\) Attempts the difference of 2 terms (either difference accepted) 3 may be missing | M1 |
| \(= 3\left(\dfrac{(3n + 1)^2 - n^2}{n^2(3n + 1)^2}\right)\) Valid attempt at common denominator for their fractions 3 may be missing | dM1 |
| \(= \dfrac{24n^2 + 18n + 3}{n^2(3n + 1)^2}\) cao | A1 |
If (b) and/or (c) are worked with \(r\) instead of \(n\) do NOT award the final A mark for the parts affected.
This applies even if \(r\) is changed to \(n\) at the end.