AS June 2020 Paper 1 Q5

AQACurrent spec4 marksSeries

(a) Show that\[r^2(r + 1)^2 - (r - 1)^2r^2 = pr^3\]

where \(p\) is an integer to be found. [1 mark]

(b) Hence use the method of differences to show that\[\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2\]

[3 marks]

Mark scheme (a)

Scheme	Marks	AO
Completes a rigorous argument to show that \(r^2(r + 1)^2 - (r - 1)^2r^2 = 4r^3\) Must show at least one intermediate step.	R1	2.1

Typical solution

\[\begin{aligned}r^2(r + 1)^2 - (r - 1)^2r^2 &= r^2(r^2 + 2r + 1) - r^2(r^2 - 2r + 1) \\ &= r^4 + 2r^3 + r^2 - r^4 + 2r^3 - r^2 = 4r^3\end{aligned}\]

Mark scheme (b)

Scheme	Marks	AO
Uses the result from (a) with their \(p\) to express \(\sum r^3\) in terms of \(\sum r^2(r + 1)^2 - (r - 1)^2r^2\) with one pair of terms of the sum written correctly.	M1	1.1a
Writes down at least three pairs of terms including the first and last pair of terms of the sum.	A1	1.1b
Completes a reasoned argument using the method of differences to show that \(\displaystyle\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2\)	R1	2.1
	(4 marks)

Typical solution

\[\begin{aligned}\sum_{r=1}^{n} 4r^3 &= \sum_{r=1}^{n} \left[r^2(r + 1)^2 - (r - 1)^2r^2\right] \\ &= 1^2 \times 2^2 - \cancel{0^2 \times 1^2} \\ &\quad + \cancel{2^2 \times 3^2} - \cancel{1^2 \times 2^2} \\ &\quad + \ldots\ldots\ldots \\ &\quad + \cancel{(n - 1)^2n^2} - \cancel{(n - 2)^2(n - 1)^2} \\ &\quad + n^2(n + 1)^2 - \cancel{(n - 1)^2n^2} \\ &= n^2(n + 1)^2 - 0^2 \times 1^2\end{aligned}\]\[\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2\]