AS June 2024 Paper 1 Q3

EdexcelCurrent spec10 marksSeries

(a) Use the standard results for summations to show that, for all positive integers \(n\),\[\sum_{r=1}^{n} r^2(r + 1) = \frac{1}{12}n(n + 1)(n + 2)(an + b)\]where \(a\) and \(b\) are integers to be determined. (4)

(b) Hence show that, for all positive integers \(k\),\[\sum_{r=k+1}^{3k} r^2(r + 1) = \frac{1}{3}k(3k + 1)\left(Ak^2 + Bk + C\right)\]where \(A\), \(B\) and \(C\) are integers to be determined. (3)

(c) Hence, using algebra and making your method clear, determine the value of \(k\) for which\[25\sum_{r=k+1}^{3k} r^2(r + 1) = 192k^3(3k + 1)\] (3)

Mark scheme (a)

Scheme	Marks	AO
\(\displaystyle\sum_{r=1}^{n} r^2(r + 1) = \sum_{r=1}^{n} r^3 + r^2 = \frac{1}{4}n^2(n + 1)^2 + \frac{1}{6}n(n + 1)(2n + 1)\)	M1 A1	1.1b 1.1b
\(= \dfrac{1}{12}n(n + 1)\left[3n(n + 1) + 2(2n + 1)\right]\)	dM1	1.1b
\(= \dfrac{1}{12}n(n + 1)\left[3n^2 + 7n + 2\right] = \dfrac{1}{12}n(n + 1)(n + 2)(3n + 1)\) cso	A1	2.1
	(4)

Notes

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

A1: Fully correct expression

dM1: Attempts to factorise \(\dfrac{1}{12}n(n + 1)\) or \(\dfrac{1}{3}n(n + 1)\) having used at least one standard formula correctly at any stage. Dependent on the first M mark.
If they show no method for factorising (use a calculator) they can go from \(3n^3 + 10n^2 + 9n + 2 = (n + 1)(n + 2)(3n + 1)\)

A1: Obtains the correct expression or the correct values of \(a\) and \(b\), with no errors seen

Mark scheme (b)

Scheme	Marks	AO
\(\displaystyle\sum_{r=k+1}^{3k} r^2(r + 1) = \frac{1}{12}(3k)(3k + 1)(3k + 2)(9k + 1) - \frac{1}{12}(k)(k + 1)(k + 2)(3k + 1)\)	M1	3.1a
\(= \dfrac{1}{12}k(3k + 1)\left[3(3k + 2)(9k + 1) - (k + 1)(k + 2)\right]\) or \(= \dfrac{1}{3}k(3k + 1)\left[\dfrac{3}{4}(3k + 2)(9k + 1) - \dfrac{1}{4}(k + 1)(k + 2)\right]\)	M1	1.1b
\(= \dfrac{1}{12}k(3k + 1)\left[80k^2 + 60k + 4\right]\) \(= \dfrac{1}{3}k(3k + 1)\left(20k^2 + 15k + 1\right)\) cso	A1	1.1b
	(3)

Notes

M1: Uses the result from part (a) and adopts a correct strategy by attempting \(\displaystyle\sum_{r=1}^{3k} r^2(r + 1) - \sum_{r=1}^{k} r^2(r + 1)\)

M1: Factorises out at least \(k(3k + 1)\) at any stage, could be done by inspection

A1: Obtains the correct expression with no errors seen.

Note If a candidate does not use part (a) but restarts they can still score marks for the same reasons. If unsure please send to review

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{25}{3}k(3k + 1)\left(20k^2 + 15k + 1\right) = 192k^3(3k + 1)\) either \(\Rightarrow 25\left(20k^2 + 15k + 1\right) = 576k^2 \Rightarrow 76k^2 - 375k - 25 = 0\) Or \(\Rightarrow 25k\left(20k^2 + 15k + 1\right) = 576k^3 \Rightarrow 76k^3 - 375k^2 - 25k = 0\) Or \(\Rightarrow 25(3k + 1)\left(20k^2 + 15k + 1\right) = 576k^2(3k + 1) \Rightarrow 228k^3 - 1049k^2 - 450k - 25 = 0\) Or \(-76k^4 + \dfrac{1049}{3}k^3 + 150k^2 + \dfrac{25}{3}k = 0\)	M1	1.1b
\(76k^2 - 375k - 25 = 0 \Rightarrow k = \ldots\) \(76k^3 - 375k^2 - 25k = 0 \Rightarrow k = \ldots\) \(-76k^4 + \dfrac{1049}{3}k^3 + 150k^2 + \dfrac{25}{3}k = 0 \Rightarrow k = \ldots\)	M1	1.1b
\(k = 5\) (only)	A1	2.3
	(3)
	(10 marks)

Notes

M1: Uses the given equation, substitutes their answer from part (b) and simplifies to either reach

\(Ak^4 + Bk^3 + Ck^2 + Dk\{= 0\}\)
\(k\left(Ak^3 + Bk^2 + Ck + D\right)\{= 0\}\) or \((3k + 1)\left(Ak^3 + Bk^2 + Ck\right)\{= 0\}\)
\(Ak^3 + Bk^2 + Ck\{= 0\}\)
a 3TQ or \(k(3k + 1)\left(Ak^2 + Bk + C\right)\{= 0\}\)

this can be implied by a correct value for \(k\) if left unsimplified

M1: Solves their equation as long as solving their \((b) = 192k^3(3k + 1)\) to find a non zero value for \(k\), including by calculator. You may need to check this.

A1: Selects the appropriate correct answer of \(k = 5\). Any other solutions must be clearly rejected.
Note: Correct answer with no working is M0M0A0