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} $
