AS June 2018 Paper 1 Q6

EdexcelCurrent spec10 marksSeries

(a) Use the standard results for \(\displaystyle\sum_{r=1}^{n} r^2\) and \(\displaystyle\sum_{r=1}^{n} r\) to show that\[\sum_{r=1}^{n}(3r - 2)^2 = \frac{1}{2}n\left[6n^2 - 3n - 1\right]\]for all positive integers \(n\). (5)

(b) Hence find any values of \(n\) for which\[\sum_{r=5}^{n}(3r - 2)^2 + 103\sum_{r=1}^{28} r\cos\left(\frac{r\pi}{2}\right) = 3n^3\] (5)

Mark scheme (a)

Scheme	Marks	AO
\((3r - 2)^2 = 9r^2 - 12r + 4\)	B1	1.1b
\(\displaystyle\sum_{r=1}^{n}\left(9r^2 - 12r + 4\right) = 9 \times \frac{1}{6}n(n + 1)(2n + 1) - 12 \times \frac{1}{2}n(n + 1) + \ldots\)	M1	2.1
\(= 9 \times \dfrac{1}{6}n(n + 1)(2n + 1) - 12 \times \dfrac{1}{2}n(n + 1) + 4n\)	A1	1.1b
\(= \dfrac{1}{2}n\left[3(n + 1)(2n + 1) - 12(n + 1) + 8\right]\)	dM1	1.1b
\(= \dfrac{1}{2}n\left[6n^2 - 3n - 1\right]\)*	A1*	1.1b
	(5)

Notes

(a) Do not allow proof by induction (but the B1 could score for \((3r - 2)^2 = 9r^2 - 12r + 4\) if seen)

B1: Correct expansion

M1: Substitutes at least one of the standard formulae into their expanded expression

A1: Fully correct expression

dM1: Attempts to factorise \(\dfrac{1}{2}n\) having used at least one standard formula correctly. Dependent on the first M mark and dependent on there being an \(n\) in all terms.

A1*: Obtains the printed result with no errors seen

Mark scheme (b)

Scheme	Marks	AO
\(\displaystyle\sum_{r=5}^{n}(3r - 2)^2 = \frac{1}{2}n\left(6n^2 - 3n - 1\right) - \frac{1}{2}(4)\left(6(4)^2 - 3 \times 4 - 1\right)\)	M1	3.1a
\(\displaystyle\sum_{r=1}^{28} r\cos\left(\frac{r\pi}{2}\right) = 0 - 2 + 0 + 4 + 0 - 6 + 0 + 8 + 0 - 10 + 0 + 12 + \ldots\)	M1	3.1a
\(3n^3 - \dfrac{3}{2}n^2 - \dfrac{1}{2}n - 166 + 103 \times 14 = 3n^3\) \(\Rightarrow 3n^2 + n - 2552 = 0\)	A1	1.1b
\(\Rightarrow 3n^2 + n - 2552 = 0 \Rightarrow n = \ldots\)	M1	1.1b
\(n = 29\)	A1	2.3
	(5)
	(10 marks)

Notes

M1: Uses the result from part (a) by substituting \(n = 4\) and subtracts from the result in (a) in order to find the first sum in terms of \(n\).

M1: Identifies the periodic nature of the second sum by calculating terms. This may be implied by a sum of 14.

A1: Uses their sum and the given result to form the correct 3 term quadratic

M1: Solves their three term quadratic to obtain at least one value for \(n\)

A1: Obtains \(n = 29\) only or obtains \(n = 29\) and \(n = -\dfrac{88}{3}\) and rejects the \(-\dfrac{88}{3}\)