A2 June 2022 Paper 1 Q1

OCR MEICurrent spec7 marksSeries

(a) By considering \((r + 1)^3 - r^3\), find \(\displaystyle\sum_{r=1}^{n}\left(3r^2 + 3r + 1\right)\). [3]

(b) Use this result to find \(\displaystyle\sum_{r=1}^{n} r(r + 1)\), expressing your answer in fully factorised form. [4]

Mark scheme (a)

Scheme	Marks	AO
\((r + 1)^3 - r^3 = r^3 + 3r^2 + 3r + 1 - r^3\) \(= 3r^2 + 3r + 1\)	B1	1.1
\(\displaystyle\sum_{r=1}^{n}\left(3r^2 + 3r + 1\right) = \sum_{r=1}^{n}\left[(r + 1)^3 - r^3\right]\) \(= 2^3 - 1^3 + 3^3 - 2^3 + \cdots + n^3 - (n - 1)^3 + (n + 1)^3 - n^3\)	M1	2.5
\(= (n + 1)^3 - 1\)	A1	2.2a
	[3]

B1: Soi by M1A1

M1: Cannot use standard summation formulae for final 2 marks

A1: isw

Mark scheme (b)

Scheme	Marks	AO
\(\displaystyle 3\sum_{r=1}^{n} r(r + 1) + n = (n + 1)^3 - 1\)	M1*	2.5
\(\displaystyle\Rightarrow \sum_{r=1}^{n} r(r + 1) = \frac{1}{3}\left[(n + 1)^3 - 1 - n\right]\)	A1	3.1a
\(= \dfrac{1}{3}(n + 1)\left[(n + 1)^2 - 1\right]\)	M1dep*	1.1
\(= \dfrac{1}{3}n(n + 1)(n + 2)\)	A1	2.2a
	[4]

M1*: \(\sum_{r=1}^{n} 1 = n\) used and starting to rearrange

M1dep*: Factorising with \(n\) or \(n + 1\) correctly

A1: Allow SC2 for correct solution using standard summation formulae.