Summing a series by the method of differences

Series

A-level Further Maths (9FM0) · Core Pure

ORIGINAL

Partial fractions first, then write out enough terms to see exactly which ones survive the cancellation.

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Question

Use the method of differences to show that

\displaystyle\sum_{r=1}^{n} \dfrac{2}{(r+2)(r+4)} = \dfrac{n(7n+25)}{12(n+3)(n+4)}

Every step worked, with the reasoning.

1
$\dfrac{2}{(r+2)(r+4)} \equiv \dfrac{1}{r+2} - \dfrac{1}{r+4}$
Partial fractions: $2 \equiv A(r+4) + B(r+2)$ ; $r = -2$ gives $2 = 2A$ , so $A = 1$ , and $r = -4$ gives $2 = -2B$ , so $B = -1$ .
2
$\displaystyle\sum_{r=1}^{n} \dfrac{2}{(r+2)(r+4)} = \Big(\dfrac13 - \dfrac15\Big) + \Big(\dfrac14 - \dfrac16\Big) + \Big(\dfrac15 - \dfrac17\Big) + \dots + \Big(\dfrac{1}{n+1} - \dfrac{1}{n+3}\Big) + \Big(\dfrac{1}{n+2} - \dfrac{1}{n+4}\Big)$
Write out the first few and the last two terms so the cancellation pattern is visible.
3
$= \dfrac13 + \dfrac14 - \dfrac{1}{n+3} - \dfrac{1}{n+4}$
Each $-\tfrac{1}{r+4}$ cancels with the $+\tfrac{1}{r+4}$ two brackets later; only the first two positive and the last two negative terms survive.
4
$= \dfrac{7(n+3)(n+4) - 12(n+4) - 12(n+3)}{12(n+3)(n+4)}$
$\tfrac13 + \tfrac14 = \tfrac{7}{12}$ ; write everything over the common denominator $12(n+3)(n+4)$ .
5
$= \dfrac{7n^2 + 49n + 84 - 24n - 84}{12(n+3)(n+4)} = \dfrac{7n^2 + 25n}{12(n+3)(n+4)} = \dfrac{n(7n+25)}{12(n+3)(n+4)}$
Expand, collect like terms and factorise the numerator to reach the given form.

Answer: $\displaystyle\sum_{r=1}^{n} \dfrac{2}{(r+2)(r+4)} = \dfrac{n(7n+25)}{12(n+3)(n+4)}$ , as required.