Question

Sum to $n$ terms of the series $\frac{1}{(1+x)(1+2x)}+\frac{1}{(1+2x)(1+3x)}+\frac{1}{(1+3x)(1+4x)}+...$

Answer 1

Let $T_r$ be the general term of the series
$T_r=\frac{1}{(1+rx)(1+(r+1)x)}$
So $T_r=\frac{1}{x}\left[\frac{(1+(r+1)x)-(1+rx)}{(1+rx)(1+(r+1)x)}\right]=\frac{1}{x}[\frac{1}{rx}-\frac{1}{1+(r+1)x}]$
$T_r=f\left(r\right)-f(r+1)$
$\therefore$ $S=\sum T_r=T_1+T_2+T_3+...+T_n$
$=\frac{1}{x}[\frac{1}{1+x}-\frac{1}{1+(n+1)x}]=\frac{n}{(1+x)[1+(n+1)x]}$