A2 October 2020 Paper 2 Q3

OCR ACurrent spec6 marksSeries

3 In this question you must show detailed reasoning.

(a) Use partial fractions to show that \(\displaystyle\sum_{r=5}^{n} \frac{3}{r^2 + r - 2} = \frac{37}{60} - \frac{1}{n} - \frac{1}{n + 1} - \frac{1}{n + 2}\). [5]

(b) Write down the value of \(\displaystyle\lim_{n \to \infty}\left(\sum_{r=5}^{n} \frac{3}{r^2 + r - 2}\right)\). [1]

Mark scheme (a)

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]