Pearson Edexcel IAL Mathematics FP1.7.1 Summation of simple finite series

Practise summing finite polynomial series by expanding terms, applying standard formulae and using resulting expressions to find constants or n.

Exam points

use standard sums to prove a printed formula and identify constants in factorised forms
express partial ranges as differences of sums, then solve the resulting equation for n

[Maximum number: 9]

(a)

Using the standard summation formulae, show that

\sum_{r=1}^{n} r(r+1)(r-1)=\frac{1}{4} n(n+A)(n+B)(n+C)

where A, B and C are constants to be determined.

[ 4 ]

$\sum_{r=1}^{n} r(r+1)(r-1)=\sum_{r=1}^{n} r^{3}-r$

B1

$=\frac{1}{4} n^{2}(n+1)^{2}-\frac{1}{2} n(n+1)$

M1

(Please note the mark above is incorrectly labelled as A1 on e-PEN)

$=\frac{1}{4} n(n+1)\left[n^{2}+n-2\right]=\frac{1}{4} n(n+1)(n+\ldots)(n+\ldots)$

М1

$=\frac{1}{4} n(n+1)(n-1)(n+2)$

A1

(4)

(b)

Determine the value of n for which

3 \sum_{r=1}^{n} r(r+1)(r-1)=17 \sum_{r=n}^{2 n} r^{2}

[ 5 ]

$\sum_{r=n}^{2n}r^2 =\frac16(2n)(2n+1)(4n+1)-\frac16(n-1)n(2n-1)$

M1 A1

$3\sum_{r=1}^{n}r(r+1)(r-1)=17\sum_{r=n}^{2n}r^2$ $\frac34n(n+1)(n-1)(n+2) =\frac{17}{6}n\{2(8n^2+6n+1)-(2n^2-3n+1)\}$ $18(n+1)(n-1)(n+2)=68(14n^2+15n+1)$ 18(n+1)(n-1)(n+2)=68(14n+1)(n+1)

dM1

18(n-1)(n+2)=68(14n+1) $18n^2-934n-104=0$

ddM1

n=52

A1

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FP1.7.1 - Summation of simple finite series question 1

Question (a)

Question (b)