# 💯An Ingelious Way to Solve Sequences and Series Method of Differences Questions

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## Sequences and Series Method of Differences

Here’s an “ingelious” way to solve Method of Differences questions related to Sequences and Series.
Let’s look at a typical JC A Level H2 Mathematics question:
To find the sum to $n^{th}$ terms:

$\sum_{r=1}^{\infty} \frac{3r+4}{(r)(r+1)(r+2)}$
$=\sum_{r=1}^{\infty} \frac{2}{r} – \frac{1}{r+1} – \frac{1}{r+2}$
instead of evaluating all the terms where most will be cancelled eventually:
$= [ 2 – \frac{1}{2} – \frac{1}{3}$
$[ + 1 – \frac{1}{3} – \frac{1}{4}$
$[ +\frac{2}{3} – \frac{1}{4} – \frac{1}{5}$
$+ …$
$[ + \frac{1}{n-1} – \frac{1}{n} – \frac{1}{n+1}$
$[ + \frac{1}{n} – \frac{1}{n+1} – \frac{1}{n+2}$
$= \frac{5}{2} – \frac{1}{n+2} – \frac{1}{n+1}$

Let $f(r) = \frac{1}{r}$ so that the terms can be expressed out easily:

$\sum_{r=1}^{\infty} 2f(r) – f(r+1) – f(r+2)$
$2f(1) – f(2) – f(3)$
$+ 2f(2) – f(3) – f(4)$
$+ 2f(3) – f(4) – f(5)$
$+ …$
$+ 2f(n-1) – f(n) – f(n+1)$
$+ 2f(n) – f(n+1) – f(n+2)$
$+ 2f(1) +f(2) – f(n+1) – f(n+2)$
$=\frac{5}{2} – \frac{1}{n+2} – \frac{1}{n+1}$