FP1 June 2018 Q4

EdexcelOld spec9 marksSeries

(a) Use the standard results for \(\displaystyle\sum_{r=1}^{n} r\) and \(\displaystyle\sum_{r=1}^{n} r^2\) to show that, for all positive integers \(n\), \[\sum_{r=1}^{n}\left(r^2 - r - 8\right) = \frac{1}{3}n(n - a)(n + a)\] where \(a\) is a positive integer to be determined. (4)

(b) Hence, or otherwise, state the positive value of \(n\) that satisfies \[\sum_{r=1}^{n}\left(r^2 - r - 8\right) = 0\] (1)

Given that \[\sum_{r=3}^{17}\left(kr^3 + r^2 - r - 8\right) = 6710 \qquad \text{where } k \text{ is a constant}\]

Mark scheme (a)

Scheme	Marks
\(\displaystyle\sum_{r=1}^{n}\left(r^2 - r - 8\right)\)
\(= \dfrac{1}{6}n(n + 1)(2n + 1) - \dfrac{1}{2}n(n + 1) - 8n\)	M1 A1
\(= \dfrac{1}{6}n\left((2n + 1)(n + 1) - 3(n + 1) - 48\right)\)	M1
\(= \dfrac{1}{6}n\left(2n^2 + 3n + 1 - 3n - 3 - 48\right)\)
\(= \dfrac{1}{6}n\left(2n^2 - 50\right)\)
\(= \dfrac{2}{6}n\left(n^2 - 25\right)\)
\(= \dfrac{1}{3}n(n - 5)(n + 5)\)	A1
	(4)

Notes

M1: At least one of the first two terms is correct.

A1: Correct expression

M1: An attempt to factorise out at least \(n\).

A1: Achieves the correct answer.

Scheme	Marks
\(n = 5\)	B1 cao
	(1)

Mark scheme (c)

Scheme	Marks
\(\left(\dfrac{k}{4}(17^2)(18^2) - \dfrac{k}{4}(3^2)(2^2)\right) + \left(\dfrac{1}{3}(17)(22)(12) - \dfrac{1}{3}(2)(-3)(7)\right)\)	M1 M1
\(\left\{\textstyle\sum = 6710 \Rightarrow\right\}\ 23409k - 9k + 1496 + 14 = 6710 \Rightarrow k = \dfrac{2}{9}\)	ddM1 A1 cso
	(4)
	(9 marks)

Notes

M1: Applying at least one of \(n = 17\) or \(n = 2\) to both \(\dfrac{1}{4}n^2(n + 1)^2\) and their \(\dfrac{1}{3}n(n - 5)(n + 5)\)

M1: Applying \(n = 17\) and \(n = 2\) only to both \(\dfrac{1}{4}n^2(n + 1)^2\) and their \(\dfrac{1}{3}n(n - 5)(n + 5)\). Require differences only for both brackets.

ddM1: Sets their sum to 6710 and solves to give \(k = \ldots\)

A1 cso: \(k = \dfrac{2}{9}\) or \(0.\dot{2}\)