FP2 June 2014 (R) Q1
1.
(a) Express \(\dfrac{2}{4r^2 - 1}\) in partial fractions. (2)
(b) Hence use the method of differences to show that \[\sum_{r=1}^{n} \frac{1}{4r^2 - 1} = \frac{n}{2n + 1}\] (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{2}{4r^2 - 1} = \dfrac{A}{2r + 1} + \dfrac{B}{2r - 1}\) | |
| \(2 = A(2r - 1) + B(2r + 1)\ \Rightarrow A = -1,\ B = 1\) | |
| \(\dfrac{2}{4r^2 - 1} = \dfrac{1}{2r - 1} - \dfrac{1}{2r + 1}\) | M1A1 |
| (2) |
Notes
M1 complete method for finding PFs
A1 both PFs correct
Award M1A1 for both PFs seen correct w/o working. M0A0 otherwise
| Scheme | Marks |
|---|---|
| \((2)\displaystyle\sum_{r=1}^{n} \frac{1}{4r^2 - 1} = \sum_{r=1}^{n}\left(\frac{1}{2r - 1} - \frac{1}{2r + 1}\right)\) | |
| \(= 1 - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{7} + \ldots + \dfrac{1}{2n - 1} - \dfrac{1}{2n + 1} = 1 - \dfrac{1}{2n + 1}\) | M1A1ft |
| \(= \dfrac{2n + 1 - 1}{2n + 1}\) | |
| \(\displaystyle\sum_{r=1}^{n} \frac{1}{4r^2 - 1} = \frac{n}{2n + 1}\) * | A1 |
| (3) | |
| (5 marks) |
Notes
M1 showing fractions with their PFs. Min 2 at start and 1 at end. Must start at 1 and end at \(n\). Required sum may be used or 2 x sum
A1ft Identify 2 non-cancelling fractions, follow through their PFs - sum or 2 x sum
A1cso correct final answer