Question

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

Answer 1

If we denote the series by $u_1+u_2+u_3+.....+u_n,$
we have $u_1=\frac{1}{x}\left(\frac{1}{1+x}-\frac{1}{1+2x}\right),$
$u_2=\frac{1}{x}\left(\frac{1}{1+2x}-\frac{1}{1+3x}\right),$
$u_3=\frac{1}{x}\left(\frac{1}{1+3x}-\frac{1}{1+4x}\right),$
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$u_n=\frac{1}{x}\left(\frac{1}{1+nx}-\frac{1}{1+\overline{n+1}\cdot x}\right),$
Thus by adding, $S_n=\frac{1}{x}\left(\frac{1}{1+x}-\frac{1}{1+\overline{n+1}\cdot x}\right)$
$=\frac{n}{(1+x)(1+\overline{n+1}\cdot x)}$.