AS June 2024 Paper 1 Q9

AQACurrent spec7 marksSeries

(a) Show that, for all positive integers \(r\),\[\frac{r + 1}{r + 2} - \frac{r}{r + 1} = \frac{1}{(r + 1)(r + 2)}\] [1 mark]

(b) Hence, using the method of differences, show that\[\sum_{r=1}^{n} \frac{1}{(r + 1)(r + 2)} = \frac{n}{an + b}\]

where \(a\) and \(b\) are integers to be determined. [3 marks]

Mark scheme (a)

Scheme	Marks	AO
Completes a rigorous argument to show that \(\dfrac{r + 1}{r + 2} - \dfrac{r}{r + 1} = \dfrac{1}{(r + 1)(r + 2)}\) Must include the LHS, at least one intermediate step, and the RHS.	R1	2.1
	(1)

Typical solution

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

Mark scheme (b)

Scheme	Marks	AO
Writes the first two pairs (or last two pairs) of corresponding terms of \(\dfrac{r + 1}{r + 2}\) and \(\dfrac{r}{r + 1}\)	M1	1.1a
Writes at least the 1^st pair and the \(n\)^th pair of corresponding terms of \(\dfrac{r + 1}{r + 2}\) and \(\dfrac{r}{r + 1}\) and shows the pattern of cancelling. The pattern of cancelling is possibly implied by a correct expression.	M1	1.1a
Completes a reasoned argument using the method of differences to obtain \(\dfrac{n}{2n + 4}\) Must include the 1^st and \(n\)^th terms and at least one pair of cancelling terms when completing the method of differences process.	R1	2.1
	(3)

Typical solution

\[\sum_{r=1}^{n} \frac{1}{(r + 1)(r + 2)} = \sum_{r=1}^{n}\left(\frac{r + 1}{r + 2} - \frac{r}{r + 1}\right)\]\[\begin{aligned} = {} & \cancel{\dfrac{2}{3}} - \dfrac{1}{2} \\[6pt] + {} & \cancel{\dfrac{3}{4}} - \cancel{\dfrac{2}{3}} \\[6pt] + {} & \ldots \\[6pt] + {} & \cancel{\dfrac{n}{n + 1}} - \cancel{\dfrac{n - 1}{n}} \\[6pt] + {} & \dfrac{n + 1}{n + 2} - \cancel{\dfrac{n}{n + 1}} \\[6pt] = {} & \dfrac{n + 1}{n + 2} - \dfrac{1}{2} \\[6pt] = {} & \dfrac{2(n + 1) - 1(n + 2)}{2(n + 2)} \\[6pt] = {} & \dfrac{n}{2n + 4}\end{aligned}\]

Mark scheme (c)

Scheme	Marks	AO
Substitutes \(n = 2000\) or \(n = 1000\) into their \(\dfrac{n}{an + b}\)	M1	1.1a
Substitutes \(n = 2000\) and \(n = 1000\) into their \(\dfrac{n}{an + b}\) and subtracts.	M1	3.1a
Obtains the correct result. Ignore an approximated answer.	A1	1.1b
	(3)
	(7 marks)

Typical solution

\[\begin{aligned}\sum_{r=1}^{2000} \frac{1}{(r + 1)(r + 2)} &= \frac{2000}{2 \times 2000 + 4} \\ &= \frac{500}{1001}\end{aligned}\]\[\begin{aligned}\sum_{r=1}^{1000} \frac{1}{(r + 1)(r + 2)} &= \frac{1000}{2 \times 1000 + 4} \\ &= \frac{250}{501}\end{aligned}\]\[\begin{aligned}\sum_{r=1001}^{2000} \frac{1}{(r + 1)(r + 2)} &= \frac{500}{1001} - \frac{250}{501} \\ &= \frac{250}{501501}\end{aligned}\]