AS June 2018 Paper 1 Q7
7
(i) Express \(\dfrac{1}{2r - 1} - \dfrac{1}{2r + 1}\) as a single fraction. [2]
(ii) Find how many terms of the series\[\frac{2}{1 \times 3} + \frac{2}{3 \times 5} + \frac{2}{5 \times 7} + \ldots + \frac{2}{(2r - 1)(2r + 1)} + \ldots\]are needed for the sum to exceed \(0.999\,999\). [7]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{2r - 1} - \dfrac{1}{2r + 1} = \dfrac{2r + 1 - (2r - 1)}{(2r - 1)(2r + 1)} = \dfrac{2}{(2r - 1)(2r + 1)}\) | M1 A1 | 1.1a 1.1 |
| [2] |
Notes
M1: combining fractions
A1: or \(\dfrac{2}{4r^2 - 1}\)
mark final answer
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{2}{1 \times 3} + \dfrac{2}{3 \times 5} + \dfrac{2}{5 \times 7} + \ldots + \dfrac{2}{(2n - 1)(2n + 1)}\) \(= \displaystyle\sum_{r=1}^{n} \frac{2}{(2r - 1)(2r + 1)}\) | M1 | 3.1a |
| \(= \displaystyle\sum_{r=1}^{n}\left(\frac{1}{2r - 1} - \frac{1}{2r + 1}\right)\) | M1 | 3.1a |
| \(= 1 - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{7} + \ldots + \dfrac{1}{2n - 1} - \dfrac{1}{2n + 1}\) | M1 | 2.1 |
| \(= 1 - \dfrac{1}{2n + 1}\) | A1 | 2.1 |
| so \(1 - \dfrac{1}{2n + 1} \gt 0.999\,999\) | M1 | 1.1a |
| \(\Rightarrow 2n + 1 \gt 1\,000\,000 \Rightarrow n \gt 499\,999.5\) | M1 | 1.1 |
| \(\Rightarrow\) Number of terms is \(500\,000\) | A1cao | 3.2a |
| [7] |
Notes
M1: (1st) \(r\)th term is \(\dfrac{2}{(2r - 1)(2r + 1)}\) soi
M1: (2nd) splitting into partial fractions soi
M1: (3rd) must end with \(\ldots - \dfrac{1}{2n + 1}\)
not \(\ldots - \dfrac{1}{2r + 1}\)
A1: \(1 - \dfrac{1}{2r + 1}\) is A0
M1: (4th) or \(\dfrac{2n}{2n + 1} \gt 0.999\,999\) allow equality, condone \(r\)
dep use of difference method [e.g. not by inspection]
M1: (5th) re-arranging (correctly) for \(n\)
A1cao: trial and error: must show both \(n = 499999\) and \(n = 500000\) for final A1