A2 October 2020 Paper 2 Q3
3 In this question you must show detailed reasoning.
| Scheme | Marks | AO |
|---|---|---|
| DR \((r + 2)(r - 1)\) | B1 | 1.1 |
| \(\dfrac{A}{r - 1} + \dfrac{B}{r + 2}\) | M1 | 1.1 |
| \(A = 1,\ B = -1\) | A1 | 1.1 |
| \(\therefore \displaystyle\sum_{r=5}^{n} \frac{3}{r^2 + r - 2} = \sum_{r=5}^{n} \frac{1}{r - 1} - \sum_{r=5}^{n} \frac{1}{r + 2}\) | M1 | 1.1 |
| \(= \displaystyle\sum_{r=5}^{n} \frac{1}{r - 1} - \sum_{r=8}^{n+3} \frac{1}{r - 1}\) | ||
| \(= \displaystyle\sum_{r=5}^{7} \frac{1}{r - 1} - \sum_{r=n+1}^{n+3} \frac{1}{r - 1}\) | ||
| \(\therefore \displaystyle\sum_{r=5}^{n} \frac{3}{r^2 + r - 2} =\) | A1 | 1.1 |
| \(= \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} - \dfrac{1}{n} - \dfrac{1}{n + 1} - \dfrac{1}{n + 2}\) | ||
| \(= \dfrac{37}{60} - \dfrac{1}{n} - \dfrac{1}{n + 1} - \dfrac{1}{n + 2}\) | ||
| [5] |
Notes
B1: Correct factorisation of denominator soi
M1: Correct form for partial fractions
M1: Using partial fractions, separating into two sums, re-indexing so that the summands have identical form and cancelling central terms.
Might see start and end terms explicitly. eg
\(\displaystyle\sum_{r=5}^{n} \frac{1}{r - 1} - \sum_{r=8}^{n+3} \frac{1}{r - 1} = \frac{1}{5 - 1} + \frac{1}{6 - 1} + \frac{1}{7 - 1} + \sum_{r=8}^{n} \frac{1}{r - 1} - \left(\sum_{r=8}^{n} \frac{1}{r - 1} + \frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2}\right)\)
Might see formal substitution of index. eg
Let \(R = r + 3 \Rightarrow r + 2 = R - 1\)
\(\therefore \displaystyle\sum_{r=5}^{n} \frac{1}{r - 1} - \sum_{r=5}^{n} \frac{1}{r + 2} = \sum_{r=5}^{n} \frac{1}{r - 1} - \sum_{R=8}^{n+3} \frac{1}{R - 1}\)
A1: AG.
Alternative method for last 2 marks
| Scheme | Marks |
|---|---|
| \(=\) \(\begin{array}{cccccc} \dfrac{1}{4} & - & \dfrac{1}{7} & \ldots & - & \dfrac{1}{n - 1} \\[10pt] \dfrac{1}{5} & - & \dfrac{1}{8} & \dfrac{1}{n - 3} & - & \dfrac{1}{n} \\[10pt] \dfrac{1}{6} & - & \dfrac{1}{9} & \dfrac{1}{n - 2} & - & \dfrac{1}{n + 1} \\[10pt] \dfrac{1}{7} & - & \ldots & \dfrac{1}{n - 1} & - & \dfrac{1}{n + 2} \end{array}\) | M1 |
| \(= \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} - \dfrac{1}{n} - \dfrac{1}{n + 1} - \dfrac{1}{n + 2}\) | A1 |
| \(= \dfrac{37}{60} - \dfrac{1}{n} - \dfrac{1}{n + 1} - \dfrac{1}{n + 2}\) | |
| [5] |
M1: At least these terms. M1 can be ft from any \(A\), \(B\) having opposite signs. For M1, condone omission of \(\dfrac{1}{7}\) or \(-\dfrac{1}{n - 1}\)
A1: AG. Correct cancellation to AG. Requires joined argument. Must have either clear diagonal cancellations with one explicit cancellation or less clear cancellation with numerical and algebraic cancellation shown or described
| Scheme | Marks | AO |
|---|---|---|
| \(= \dfrac{37}{60}\) or awrt 0.617 | B1 | 2.2a |
| [1] |