AS June 2021 Paper 1 Q11

AQACurrent spec4 marksSeries

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

[1 mark]

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

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

Mark scheme (a)

Scheme	Marks	AO
Obtains the correct result including at least one intermediate step. All lines must be correct. Condone original expression omitted.	B1	2.1
	(1)

Typical solution

\[\frac{1}{(r - 1)!} - \frac{1}{r!} = \frac{r}{r!} - \frac{1}{r!} = \frac{r - 1}{r!}\]

Mark scheme (b)

Scheme	Marks	AO
Writes the first two pairs (or last two pairs) of corresponding terms of \(\dfrac{1}{(r - 1)!}\) and \(\dfrac{1}{r!}\)	M1	1.1a
Writes at least the first pair and the last pair of corresponding terms of \(\dfrac{1}{(r - 1)!}\) and \(\dfrac{1}{r!}\) and shows the pattern of cancelling.	M1	1.1a
Completes a fully correct proof to reach the required result with \(a = 1\) and \(b = -1\) 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)
	(4 marks)

Typical solution

\[\begin{aligned}\sum_{r=1}^{n} \frac{r - 1}{r!} &= \sum_{r=1}^{n} \left(\frac{1}{(r - 1)!} - \frac{1}{r!}\right) \\ &= \frac{1}{0!} - \cancel{\frac{1}{1!}} \\ &\quad + \cancel{\frac{1}{1!}} - \cancel{\frac{1}{2!}} \\ &\quad + \cancel{\ldots\ldots} \\ &\quad + \cancel{\frac{1}{(n - 2)!}} - \cancel{\frac{1}{(n - 1)!}} \\ &\quad + \cancel{\frac{1}{(n - 1)!}} - \frac{1}{n!} \\ &= \frac{1}{0!} - \frac{1}{n!} \\ &= 1 - \frac{1}{n!}\end{aligned}\]