A2 June 2025 Paper 1 Q6
6 In this question you must show detailed reasoning.
The series \(S_n\) is given by \(S_n = \left(\dfrac{1}{5} \times \dfrac{1}{15}\right) + \left(\dfrac{1}{15} \times \dfrac{1}{25}\right) + \cdots + \left(\dfrac{1}{10n - 5} \times \dfrac{1}{10n + 5}\right)\) for \(n \in \mathbb{Z}^+\).
Let \(S_\infty = \displaystyle\lim_{n \to \infty} S_n\).
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{1}{(10r - 5)(10r + 5)} \equiv \dfrac{A}{10r - 5} + \dfrac{B}{10r + 5}\) \(\Rightarrow A(10r + 5) + B(10r - 5) \equiv 1\) | M1 | 3.1a |
| \(\dfrac{1}{(10r - 5)(10r + 5)} \equiv \dfrac{1}{10}\left(\dfrac{1}{10r - 5} - \dfrac{1}{10r + 5}\right)\) | A1 | 1.1 |
| \(\displaystyle\sum_{r=1}^{n} \frac{1}{10r - 5} \times \frac{1}{10r + 5} = \left(\frac{1}{50} - \frac{1}{150}\right) + \left(\frac{1}{150} - \frac{1}{250}\right) + \ldots\) \(\displaystyle + \left(\frac{1}{100n - 150} - \frac{1}{100n - 50}\right) + \left(\frac{1}{100n - 50} - \frac{1}{100n + 50}\right)\) | M1 | 2.1 |
| \(= \dfrac{1}{50}\left(1 - \dfrac{1}{2n + 1}\right)\) | B1 | 1.1 |
| \(\lt \dfrac{1}{50}\) (for all \(n\)) as \(\dfrac{1}{2n + 1} \gt 0\) (for all \(n\)) | A1 | 2.2a |
| [5] |
Notes
M1: For correct identity oe e.g. \(P(2r + 1) + Q(2r - 1) \equiv 1\) following \(\dfrac{1}{(2r - 1)(2r + 1)} \equiv \dfrac{P}{2r - 1} + \dfrac{Q}{2r + 1}\) - condone if in terms of \(n\) – can be implied from correct partial fractions
A1: oe, allow for \(\dfrac{1}{(10r - 5)(10r + 5)} \equiv \dfrac{A}{10r - 5} + \dfrac{B}{10r + 5}\) followed by \(A = \dfrac{1}{10}, B = -\dfrac{1}{10}\) - condone if in terms of \(n\)
M1: Uses method of differences for their two term partial fractions of the correct form (where their \(A\) and \(B\) must have different signs). Must show at least the cases for \(r = 1, 2\) and \(n\) or \(r = 1, n - 1\) and \(n\) (so either first two terms and last term or first term and last two terms) with subtraction between terms
Condone missing or incorrect factor of \(\dfrac{1}{50}\)
B1: oe e.g. \(\dfrac{1}{50} - \dfrac{1}{100n + 50}\)
A1: www - must explaining why the sum is less than \(\dfrac{1}{50}\) - must consider the sign of \(\pm\dfrac{1}{2n + 1}\) correctly (oe) - dependent on all previous marks
| Scheme | Marks | AO |
|---|---|---|
| DR \(S_\infty = \dfrac{1}{50}\) \(\left(S_\infty = \dfrac{1}{2450} + S_k \Rightarrow\right)\) | B1* | 2.2a |
| \(\dfrac{1}{50} = \dfrac{1}{2450} + \dfrac{1}{50}\left(1 - \dfrac{1}{2k + 1}\right)\) | M1dep* | 3.1a |
| \(k = 24\) | A1 | 1.1 |
| [3] |
Notes
B1*: soi
M1dep*: Using their \(S_k\) from part (a) which must be of the form \(a - \dfrac{b}{2n + 1}\) where \(a\) is non-zero and \(b \gt 0\), in the given equation, to get an equation in \(k\) e.g. \(\dfrac{24}{1225} = \dfrac{1}{50}\left(1 - \dfrac{1}{2k + 1}\right)\)
A1: cao - SCB1 for the correct answer with no working