FP2 June 2010 Q1
1.
(a) Express \(\dfrac{3}{(3r-1)(3r+2)}\) in partial fractions. (2)
(b) Using your answer to part (a) and the method of differences, show that \[\sum_{r=1}^{n}\frac{3}{(3r-1)(3r+2)} = \frac{3n}{2(3n+2)}\] (3)
(c) Evaluate \(\displaystyle\sum_{r=100}^{1000}\frac{3}{(3r-1)(3r+2)}\), giving your answer to 3 significant figures. (2)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{3r-1} - \dfrac{1}{3r+2}\) | M1 A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\sum_{r=1}^{n}\frac{3}{(3r-1)(3r+2)} = \frac{1}{2} - \frac{1}{5} + \frac{1}{5} - \frac{1}{8} + \frac{1}{8} - \frac{1}{11} + \ldots\frac{1}{3n-1} - \frac{1}{3n+2}\) | M1 A1ft |
| \(= \dfrac{1}{2} - \dfrac{1}{3n+2} = \dfrac{3n}{2(3n+2)}\) * | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Sum \(= \mathrm{f}(1000) - \mathrm{f}(99)\) | |
| \(\dfrac{3000}{6004} - \dfrac{297}{598} = 0.00301\) or \(3.01 \times 10^{-3}\) | M1 A1 |
| (2) | |
| (7 marks) |