A2 October 2021 Paper 1 Q10
10 Using an algebraic method, determine the least value of \(n\) for which \(\displaystyle\sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)} \geqslant 0.49\). [8]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{(2r - 1)(2r + 1)} = \dfrac{A}{2r - 1} + \dfrac{B}{2r + 1}\) | M1 | 3.1a |
| \(\Rightarrow A(2r + 1) + B(2r - 1) = 1\) | M1 | 1.1 |
| \(\Rightarrow A - B = 1,\ A + B = 0\) \(\Rightarrow A = \dfrac{1}{2}, B = -\dfrac{1}{2}\) | A1 | 1.1 |
| \(\displaystyle\sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)} = \frac{1}{2}\left(\begin{array}{l}\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \ldots \\ + \left(\frac{1}{2n - 3} - \frac{1}{2n - 1}\right) + \left(\frac{1}{2n - 1} - \frac{1}{2n + 1}\right)\end{array}\right)\) | M1 M1 | 3.1a 2.1 |
| \(= \dfrac{1}{2}\left(1 - \dfrac{1}{2n + 1}\right)\) oe | A1 | 1.1 |
| \(\dfrac{1}{2}\left(1 - \dfrac{1}{2n + 1}\right) \geqslant 0.49\) \(\Rightarrow n \geqslant 0.98n + 0.49\) \(\Rightarrow n \geqslant \dfrac{0.49}{0.02} = 24.5\) | M1 | 3.1a |
| \(\Rightarrow n = 25\) | A1 | 3.2a |
| [8] |
Notes
M1: partial fractions
M1: Allow any method to determine \(A\) and \(B\)
A1: Both values
M1: Use of differences
M1: Deal with subtraction
M1: Use of inequality on their formula
No marks for a purely numerical solution.