AS June 2018 Paper 1 Q15

AQACurrent spec4 marksSeries

(a) Show that\[\frac{1}{r + 2} - \frac{1}{r + 3} = \frac{1}{(r + 2)(r + 3)}\]

[1 mark]

(b) Use the method of differences to show that\[\sum_{r=1}^{n} \frac{1}{(r + 2)(r + 3)} = \frac{n}{3(n + 3)}\]

[3 marks]

Mark scheme (a)

Scheme	Marks	AO
Shows the result is true with at least one intermediate step.	B1	1.1b

Typical solution

\[\begin{aligned}\frac{1}{r + 2} - \frac{1}{r + 3} &= \frac{r + 3 - (r + 2)}{(r + 2)(r + 3)} \\ &= \frac{1}{(r + 2)(r + 3)}\end{aligned}\]

Mark scheme (b)

Scheme	Marks	AO
Writes at least three corresponding terms of \(\frac{1}{r + 2}\) and \(\frac{1}{r + 3}\) Must include the 1^st and n^th terms and at least the 2^nd term or the \((n - 1)\)^th term.	M1	1.1a
Correctly uses the method of differences to reduce the sum to two terms.	A1	1.1b
Completes fully correct proof to reach the required result. This mark is only available if all previous marks have been awarded.	R1	2.1
	(4 marks)

Typical solution

\[\begin{aligned}\sum_{r=1}^{n}\left(\frac{1}{(r + 2)(r + 3)}\right) &= \sum_{r=1}^{n}\left(\frac{1}{r + 2} - \frac{1}{r + 3}\right) \\ &= \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{6}\right) + \ldots\ldots \\ &\quad + \left(\frac{1}{n + 1} - \frac{1}{n + 2}\right) + \left(\frac{1}{n + 2} - \frac{1}{n + 3}\right) \\ &= \frac{1}{3} - \frac{1}{n + 3} \\ &= \frac{n + 3 - 3}{3(n + 3)} \\ &= \frac{n}{3(n + 3)}\end{aligned}\]