FP2 June 2013 Q1
1.
(a) Express \(\dfrac{2}{(2r + 1)(2r + 3)}\) in partial fractions. (2)
(b) Using your answer to (a), find, in terms of \(n\), \[\sum_{r=1}^{n}\frac{3}{(2r + 1)(2r + 3)}\] Give your answer as a single fraction in its simplest form. (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{2}{(2r + 1)(2r + 3)} = \dfrac{A}{2r + 1} + \dfrac{B}{2r + 3} =,\ \ \dfrac{1}{2r + 1} - \dfrac{1}{2r + 3}\) | M1,A1 |
| (2) |
Notes
M1 for any valid attempt to obtain the PFs
A1 for \(\dfrac{1}{2r + 1} - \dfrac{1}{2r + 3}\)
NB With no working shown award M1A1 if the correct PFs are written down, but M0A0 if either one is incorrect
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{7} + \ldots\dfrac{1}{2n + 1} - \dfrac{1}{2n + 3}\) | |
| \(= \dfrac{1}{3} - \dfrac{1}{2n + 3} = \dfrac{2n + 3 - 3}{3(2n + 3)}\) | M1 |
| \(\displaystyle\sum_{1}^{n}\dfrac{3}{(2r + 1)(2r + 3)} = \dfrac{3}{2} \times \dfrac{2n}{3(2n + 3)} = \dfrac{n}{2n + 3}\) | M1depA1 |
| (3) | |
| (5 marks) |
Notes
M1 for using their PFs to split each of the terms of the sum or of \(\displaystyle\sum\dfrac{2}{(2r + 1)(2r + 3)}\) into 2 PFs.
At least 2 terms at the start and 1 at the end needed to show the diagonal cancellation resulting in two remaining terms.
M1dep for simplifying to a single fraction and multiplying it by the appropriate constant
A1cao for \(\displaystyle\sum = \dfrac{n}{2n + 3}\)
NB: If \(r\) is used instead of \(n\) (including for the answer), only M marks are available.