FP1 January 2011 Q5

EdexcelOld spec7 marksSeries

(a) Use the results for \(\displaystyle\sum_{r=1}^{n} r\), \(\displaystyle\sum_{r=1}^{n} r^2\) and \(\displaystyle\sum_{r=1}^{n} r^3\), to prove that \[\sum_{r=1}^{n} r(r + 1)(r + 5) = \frac{1}{4}n(n + 1)(n + 2)(n + 7)\] for all positive integers \(n\). (5)

(b) Hence, or otherwise, find the value of \[\sum_{r=20}^{50} r(r + 1)(r + 5)\] (2)

Mark scheme (a)

Scheme	Marks
\(\displaystyle\sum_{r=1}^{n} r(r + 1)(r + 5)\)
\(\displaystyle= \sum_{r=1}^{n} r^3 + 6r^2 + 5r\) Multiplying out brackets and an attempt to use at least one of the standard formulae correctly.	M1
\(= \underline{\dfrac{1}{4}n^2(n + 1)^2 + 6 \cdot \dfrac{1}{6}n(n + 1)(2n + 1) + 5 \cdot \dfrac{1}{2}n(n + 1)}\) Correct expression.	A1
\(= \dfrac{1}{4}n^2(n + 1)^2 + n(n + 1)(2n + 1) + \dfrac{5}{2}n(n + 1)\)
\(= \dfrac{1}{4}n(n + 1)\big(n(n + 1) + 4(2n + 1) + 10\big)\) Factorising out at least \(n(n + 1)\)	dM1
\(= \dfrac{1}{4}n(n + 1)\big(n^2 + n + 8n + 4 + 10\big)\)
\(= \dfrac{1}{4}n(n + 1)\big(n^2 + 9n + 14\big)\) Correct 3 term quadratic factor	A1
\(= \dfrac{1}{4}n(n + 1)(n + 2)(n + 7)\) * Correct proof. No errors seen.	A1
	(5)

Mark scheme (b)

Scheme	Marks
\(\displaystyle S_n = \sum_{r=20}^{50} r(r + 1)(r + 5)\) \(= S_{50} - S_{19}\)
\(= \tfrac{1}{4}(50)(51)(52)(57) - \tfrac{1}{4}(19)(20)(21)(26)\) Use of \(S_{50} - S_{19}\)	M1
\(= 1889550 - 51870\)
\(= 1837680\) 1837680 Correct answer only 2/2	A1
	(2)
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