Edexcel IAL Mathematics FP2.2.1 method of differences

Practise evaluating finite series by decomposing terms, cancelling successive differences, and giving exact formulae or values.

Exam points

rewrite rational terms with partial fractions so adjacent terms cancel
find a sum from r = 2 to n and express it in the requested form with constants
use a formula for S_n to calculate an exact sum over a restricted range

[Maximum number: 4]

Hence, using the method of differences, show that for all integer values of n,

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

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

$\frac18\sum_{r=1}^{n}\left(\frac{1}{2r-1}-\frac{2}{2r+1}+\frac{1}{2r+3}\right)$ $=\frac18\left(1-\frac13-\frac{1}{2n+1}+\frac{1}{2n+3}\right)$ $=\frac18\left(\frac{2(2n+1)(2n+3)-3(2n+3)+3(2n+1)}{3(2n+1)(2n+3)}\right)$ $=\frac{n(n+2)}{3(2n+1)(2n+3)}$
M1
A1
dM1
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FP2.2.1 - Summation of simple finite series n 1 ∑ question 1