Edexcel A-Level Mathematics AS Fp1 7 1 Summation of Simple Finite Series Questions

Practise Edexcel IAL FP1.7.1 by expanding polynomial summands, applying standard sums for r, r² and r³, factoring results and evaluating finite ranges.

Exam points

expand polynomial summands and apply standard sums for r, r² and r³ to prove a closed form
factor the resulting polynomial and compare coefficients to identify constants in the required form
evaluate a finite-range sum by subtracting prefix sums and solve the resulting equation for n or a value

[Maximum number: 8]

(a)

Use the standard results for summations to show that, for all positive integers n,

\sum_{r=1}^{n} r\left(2 r^{2}-3 r-1\right)=\frac{1}{2} n(n+1)^{2}(n-2)

[ 4 ]

$\sum_{r=1}^{n}r(2r^2-3r-1) =\sum_{r=1}^{n}(2r^3-3r^2-r).$ $=2\cdot\frac14n^2(n+1)^2 -3\cdot\frac16n(n+1)(2n+1) -\frac12n(n+1).$ $=\frac12n(n+1)[n(n+1)-(2n+1)-1].$ $=\frac12n(n+1)(n^2-n-2) =\frac12n(n+1)^2(n-2).$
A1*

(b)

Hence show that, for all positive integers n,

\sum_{r=n}^{2 n} r\left(2 r^{2}-3 r-1\right)=\frac{1}{2} n(n-1)(a n+b)(c n+d)

where a, b, c and d are integers to be determined.

[ 4 ]

Using part (a),
$\sum_{r=n}^{2n}r(2r^2-3r-1) =\sum_{r=1}^{2n}r(2r^2-3r-1) -\sum_{r=1}^{n-1}r(2r^2-3r-1).$ $=\frac12(2n)(2n+1)^2(2n-2) -\frac12(n-1)n^2(n-3).$ $=\frac12n(n-1)[4(2n+1)^2-n(n-3)].$ $=\frac12n(n-1)(15n^2+19n+4).$ $=\frac12n(n-1)(15n+4)(n+1).$

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Edexcel A-Level Mathematics AS Fp1 7 1 Summation of Simple Finite Series Questions question 1

Question (a)

Question (b)