AS June 2019 Paper 1 Q7

AQACurrent spec5 marksSeries

(a) Show that\[\frac{1}{r - 1} - \frac{1}{r + 1} \equiv \frac{A}{r^2 - 1}\]

where \(A\) is a constant to be found. [1 mark]

(b) Hence use the method of differences to show that\[\sum_{r=2}^{n} \frac{1}{r^2 - 1} \equiv \frac{an^2 + bn + c}{4n(n + 1)}\]

where \(a\), \(b\) and \(c\) are integers to be found. [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Obtains the required result with \(A = 2\). Must show at least one intermediate step with no incorrect steps. Condone the LHS not appearing in their working.	B1	1.1b

Typical solution

\[\begin{aligned}\frac{1}{r - 1} - \frac{1}{r + 1} &\equiv \frac{r + 1}{(r - 1)(r + 1)} - \frac{r - 1}{(r - 1)(r + 1)} \\ &\equiv \frac{2}{r^2 - 1}\end{aligned}\]

Mark scheme (b)

Scheme	Marks	AO
Writes at least five corresponding terms of \(\dfrac{k}{r - 1}\) and \(\dfrac{k}{r + 1}\) Must include the terms for \(r = 2\), \(r = 3\), \(r = n - 1\), \(r = n\) and for either \(r = 4\) or \(r = n - 2\) Condone \(\frac{1}{0} - \frac{1}{2}\) also included.	M1	1.1a
Correctly uses the method of differences to reduce the expression to four terms, or equivalent.	A1	1.1b
Multiplies by \(\frac{1}{2}\) (may be seen at any stage).	M1	1.1a
Completes fully correct working to reach the required result. This mark is only available if all previous marks have been awarded.	R1	2.1
	(5 marks)

Typical solution

\[\begin{aligned}\sum_{r=2}^{n}\left(\frac{2}{r^2 - 1}\right) &= \sum_{r=2}^{n}\left(\frac{1}{r - 1} - \frac{1}{r + 1}\right) \\ &= \left(\frac{1}{1} - \cancel{\frac{1}{3}}\right) + \left(\frac{1}{2} - \cancel{\frac{1}{4}}\right) + \left(\cancel{\frac{1}{3}} - \cancel{\frac{1}{5}}\right) + \ldots\ldots \\ &\quad + \left(\cancel{\frac{1}{n - 3}} - \cancel{\frac{1}{n - 1}}\right) + \left(\cancel{\frac{1}{n - 2}} - \frac{1}{n}\right) + \left(\cancel{\frac{1}{n - 1}} - \frac{1}{n + 1}\right) \\ &= \frac{1}{1} + \frac{1}{2} - \frac{1}{n} - \frac{1}{n + 1}\end{aligned}\]\[\begin{aligned}\sum_{r=2}^{n}\left(\frac{1}{r^2 - 1}\right) &= \frac{1}{2}\left(\frac{3}{2} - \frac{(n + 1) + n}{n(n + 1)}\right) \\ &= \frac{3}{4} - \frac{2n + 1}{2n(n + 1)} \\ &= \frac{3(n^2 + n) - 2(2n + 1)}{4n(n + 1)} \\ &= \frac{3n^2 - n - 2}{4n(n + 1)}\end{aligned}\]