Pearson Edexcel IAL Mathematics FP1.7 Series Question Bank

Practise finite series manipulations using standard summation results, changed limits and algebraic factorisation to determine constants or n.

Exam points

expand polynomial terms before applying standard sums for r, r^2 or r^3

Covered syllabus points

FP1.7.1 - Summation of simple finite series

[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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Covered syllabus points

FP1.7 - Series question 1

Question (a)

Question (b)