FP1 June 2011 Q7

EdexcelOld spec10 marksSeries

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

(b) Hence show that \[\sum_{r=n+1}^{3n} (2r - 1)^2 = \frac{2}{3}n\left(an^2 + b\right)\] where \(a\) and \(b\) are integers to be found. (4)

Mark scheme (a)

Scheme	Marks
\(\left\{\mathrm{S}_n =\right\} \displaystyle\sum_{r=1}^{n} (2r - 1)^2\)
\(= \displaystyle\sum_{r=1}^{n} 4r^2 - 4r + 1\) \(= \underline{4.\dfrac{1}{6}n(n + 1)(2n + 1) - 4.\dfrac{1}{2}n(n + 1)} + n\) Multiplying out brackets and an attempt to use at least one of the two standard formulae correctly. First two terms correct. \(+\,n\)	M1 A1 B1
\(= \tfrac{2}{3}n(n + 1)(2n + 1) - 2n(n + 1) + n\)
\(= \tfrac{1}{3}n\{2(n + 1)(2n + 1) - 6(n + 1) + 3\}\) Attempt to factorise out \(\tfrac{1}{3}n\) Correct expression with \(\tfrac{1}{3}n\) factorised out with no errors seen.	M1 A1
\(= \tfrac{1}{3}n\{2(2n^2 + 3n + 1) - 6(n + 1) + 3\}\) \(= \tfrac{1}{3}n\{4n^2 + 6n + 2 - 6n - 6 + 3\}\) \(= \tfrac{1}{3}n(4n^2 - 1)\)
\(= \tfrac{1}{3}n(2n + 1)(2n - 1)\) Correct proof. No errors seen.	A1 *
	(6)

Notes

Note that there are no marks for proof by induction.

Mark scheme (b)

Scheme	Marks
\(\displaystyle\sum_{r=n+1}^{3n} (2r - 1)^2 = \mathrm{S}_{3n} - \mathrm{S}_n\)
\(= \dfrac{1}{3}.3n(6n + 1)(6n - 1) - \dfrac{1}{3}n(2n + 1)(2n - 1)\) Use of \(\mathrm{S}_{3n} - \mathrm{S}_n\) or \(\mathrm{S}_{3n} - \mathrm{S}_{n+1}\) with the result from (a) used at least once. Correct unsimplified expression. E.g. Allow \(2(3n)\) for \(6n\).	M1 A1
\(= n(36n^2 - 1) - \tfrac{1}{3}n(4n^2 - 1)\)
\(= \tfrac{1}{3}n(108n^2 - 3 - 4n^2 + 1)\) Factorising out \(\tfrac{1}{3}n\) (or \(\tfrac{2}{3}n\))	dM1
\(= \tfrac{1}{3}n(104n^2 - 2)\)
\(= \tfrac{2}{3}n(52n^2 - 1)\) \(\tfrac{2}{3}n(52n^2 - 1)\)	A1
\(\{a = 52,\ b = -1\}\)
	(4)
	(10 marks)

Notes

Note that (b) says hence so they have to be using the result from (a)