FP1 January 2010 Q8

EdexcelOld spec12 marksInduction Series

(a) Prove by induction that, for any positive integer \(n\), \[\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n + 1)^2\] (5)

(b) Using the formulae for \(\displaystyle\sum_{r=1}^{n} r\) and \(\displaystyle\sum_{r=1}^{n} r^3\), show that \[\sum_{r=1}^{n}(r^3 + 3r + 2) = \frac{1}{4}n(n + 2)(n^2 + 7)\] (5)

Mark scheme (a)

Scheme	Marks
\(\displaystyle\sum_{r=1}^{1} r^3 = 1^3 = 1\) and \(\dfrac{1}{4} \times 1^2 \times 2^2 = 1\)	B1
Assume true for \(n = k\): \(\displaystyle\sum_{r=1}^{k+1} r^3 = \frac{1}{4}k^2(k + 1)^2 + (k + 1)^3\)	B1
\(\dfrac{1}{4}(k + 1)^2\left[k^2 + 4(k + 1)\right] = \dfrac{1}{4}(k + 1)^2(k + 2)^2\)	M1 A1
\(\therefore\) True for \(n = k + 1\) if true for \(n = k\). True for \(n = 1\), \(\therefore\) true for all \(n\).	A1cso
	(5)

Notes

(a) Correct method to identify \((k + 1)^2\) as a factor award M1

\(\dfrac{1}{4}(k + 1)^2(k + 2)^2\) award first A1

All three elements stated somewhere in the solution award final A1

Mark scheme (b)

Scheme	Marks
\(\sum r^3 + 3\sum r + \sum 2 = \dfrac{1}{4}n^2(n + 1)^2 + 3\left(\dfrac{1}{2}n(n + 1)\right),\ + 2n\)	B1, B1
\(= \dfrac{1}{4}n\left[n(n + 1)^2 + 6(n + 1) + 8\right]\)	M1
\(= \dfrac{1}{4}n\left[n^3 + 2n^2 + 7n + 14\right] = \dfrac{1}{4}n(n + 2)(n^2 + 7)\) (*)	A1 A1cso
	(5)

Notes

(b) Attempt to factorise by \(n\) for M1

\(\dfrac{1}{4}\) and \(n^3 + 2n^2 + 7n + 14\) for first A1

Mark scheme (c)

Scheme	Marks
\(\displaystyle\sum_{15}^{25} = \sum_{1}^{25} - \sum_{1}^{14}\) with attempt to sub in answer to part (b)	M1
\(= \dfrac{1}{4}(25 \times 27 \times 632) - \dfrac{1}{4}(14 \times 16 \times 203) = 106650 - 11368 = 95282\)	A1
	(2)
	[12]