Edexcel A-Level Mathematics AS Fp1 7 Series Questions

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

[Maximum number: 8]

In this question use the standard results for summations.

(a)

Show that for all positive integers n

\sum_{r=1}^{n}\left(12 r^{2}+2 r-3\right)=A n^{3}+B n^{2}

where A and B are integers to be determined.

[ 4 ]

$\sum_{r=1}^n(12r^2+2r-3)=12\sum r^2+2\sum r-3\sum1$ $=12\cdot\frac{n(n+1)(2n+1)}6+2\cdot\frac{n(n+1)}2-3n$ $=4n^3+7n^2$ $A=4,\quad B=7$

(b)

Hence determine the value of n for which

\sum_{r=1}^{2 n} r^{3}-\sum_{r=1}^{n}\left(12 r^{2}+2 r-3\right)=270

[ 4 ]

$\sum_{r=1}^{2n}r^3=\left(\frac{2n(2n+1)}2\right)^2=n^2(2n+1)^2$ $n^2(2n+1)^2-(4n^3+7n^2)=270$ $4n^4-6n^2=270$ $2n^4-3n^2-135=0=(2n^2+15)(n^2-9)$ n=3
M1 A1ft M1 A1

[Maximum number: 8]

(a)

Use the standard results for $\sum_{r=1}^{n} r^{2}$ and $\sum_{r=1}^{n} r$ to show that for all positive integers n

\sum_{r=0}^{n}(r+1)(r+2)=\frac{1}{3}(n+1)(n+2)(n+3)

[ 5 ]

$\sum_{r=0}^{n}(r+1)(r+2)=\sum_{r=0}^{n}(r^2+3r+2)$ $=\frac16n(n+1)(2n+1)+\frac32n(n+1)+2n+2$ $=(n+1)\left[\frac16n(2n+1)+\frac32n+2\right]$ $=\frac13(n+1)(n^2+5n+6)$ $=\frac13(n+1)(n+2)(n+3)$

(b)

Hence determine the value of

10 \times 11+11 \times 12+12 \times 13+\ldots+100 \times 101

[ 3 ]

Upper limit =99.
$10\times11+11\times12+\cdots+100\times101 =\sum_{r=0}^{99}(r+1)(r+2)-\sum_{r=0}^{8}(r+1)(r+2)$ $=\frac13(100)(101)(102)-\frac13(9)(10)(11)$ =343070

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (b)

Question 3

Question (a)

Question (b)