FP2 June 2009 Q1
1.
(a) Express \(\dfrac{1}{r(r+2)}\) in partial fractions. (1)
(b) Hence show that \[\sum_{r=1}^{n}\frac{4}{r(r+2)} = \frac{n(3n+5)}{(n+1)(n+2)}.\] (5)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{r(r+2)} = \underline{\dfrac{1}{2r} - \dfrac{1}{2(r+2)}}\) | B1 aef |
| (1) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{n}\frac{4}{r(r+2)} = \sum_{r=1}^{n}\left(\frac{2}{r} - \frac{2}{r+2}\right)\) | |
| \(= \left(\underline{\dfrac{2}{1}} - \dfrac{2}{3}\right) + \left(\underline{\dfrac{2}{2}} - \dfrac{2}{4}\right) + \ \ldots\ldots\ \ldots + \left(\dfrac{2}{n-1} - \underline{\dfrac{2}{n+1}}\right) + \left(\dfrac{2}{n} - \underline{\dfrac{2}{n+2}}\right)\) List the first two terms and the last two terms | M1 |
| \(= \dfrac{2}{1} + \dfrac{2}{2};\ - \dfrac{2}{n+1} - \dfrac{2}{n+2}\) Includes the first two underlined terms and includes the final two underlined terms. \(\frac{2}{1} + \frac{2}{2} - \frac{2}{n+1} - \frac{2}{n+2}\) | M1 A1 |
| \(= 3 - \dfrac{2}{n+1} - \dfrac{2}{n+2}\) | |
| \(= \dfrac{3(n+1)(n+2) - 2(n+2) - 2(n+1)}{(n+1)(n+2)}\) \(= \dfrac{3n^2 + 9n + 6 - 2n - 4 - 2n - 2}{(n+1)(n+2)}\) Attempt to combine to an at least 3 term fraction to a single fraction and an attempt to take out the brackets from their numerator. | M1 |
| \(= \dfrac{3n^2 + 5n}{(n+1)(n+2)}\) | |
| \(= \dfrac{n(3n+5)}{(n+1)(n+2)}\) Correct Result | A1 cso AG |
| (5) | |
| (6 marks) |