A2 June 2023 Paper 2 Q7

AQACurrent spec3 marksSeries

7 Show that

\[\sum_{r=11}^{n+1} r^3 = \frac{1}{4}\left(n^2 + an + b\right)\left(n^2 + an + c\right)\]

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

Mark scheme

Scheme	Marks	AO
Expresses the sum as the difference of two series.	M1	3.1a
Obtains a correct (unsimplified) expression in terms of \(n\) for the sum.	A1	1.1b
Obtains the required result.	R1	2.1
	(3 marks)

Typical solution

\[\begin{aligned} \sum_{r=11}^{n+1} r^3 &= \sum_{r=1}^{n+1} r^3 - \sum_{r=1}^{10} r^3 \\ &= \frac{1}{4}(n + 1)^2(n + 2)^2 - \frac{1}{4}(10)^2(11)^2 \\ &= \frac{1}{4}\Big\{\big((n + 1)(n + 2) + 110\big)\big((n + 1)(n + 2) - 110\big)\Big\} \\ &= \frac{1}{4}\Big\{\left(n^2 + 3n + 112\right)\left(n^2 + 3n - 108\right)\Big\} \end{aligned}\]