FP1 June 2014 (R) Q5

EdexcelOld spec8 marksSeries

(a) Use the standard results for \(\displaystyle\sum_{r=1}^{n} r\) and \(\displaystyle\sum_{r=1}^{n} r^3\) to show that \[\sum_{r=1}^{n} r(r^2 - 3) = \frac{1}{4}n(n + 1)(n + 3)(n - 2)\] (5)

(b) Calculate the value of \(\displaystyle\sum_{r=10}^{50} r(r^2 - 3)\) (3)

Mark scheme (a)

Scheme	Marks
\(r(r^2 - 3) = r^3 - 3r\) \(r^3 - 3r\)	B1
\(\displaystyle\sum_{r=1}^{n} r(r^2 - 3) = \sum_{r=1}^{n} r^3 - 3\sum_{r=1}^{n} r\)
\(= \dfrac{1}{4}n^2(n + 1)^2 - \dfrac{3}{2}n(n + 1)\) M1: An attempt to use at least one of the standard formulae correctly. A1: Correct expression	M1A1
\(= \dfrac{1}{4}n(n + 1)\big(n(n + 1) - 6\big)\) Attempt factor of \(\dfrac{1}{4}n(n + 1)\) before given answer	M1
\(= \dfrac{1}{4}n(n + 1)(n^2 + n - 6)\)
\(= \dfrac{1}{4}n(n + 1)(n + 3)(n - 2)\) cso	A1
	(5)

Mark scheme (b)

Scheme	Marks
\(\displaystyle\sum_{r=10}^{50} r(r^2 - 3) = \mathrm{f}(50) - \mathrm{f}(9 \text{ or } 10)\) Require some use of the result in part (a) for method.	M1
\(= \dfrac{1}{4}(50)(51)(53)(48) - \dfrac{1}{4}(9)(10)(12)(7)\) Correct expression	A1
\(= 1621800 - 1890\)
\(= 1619910\) cao	A1
	(3)
	(8 marks)