Question

$\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+.....+\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}$

Answer 1

$S_n=\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}$

$=\frac{1}{2}\left[\frac{3-1}{\mathrm{1.2.3}}+\frac{4-2}{\mathrm{2.3.4}}+...+\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}\right]$

$=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]$4

$=\frac{1}{2}\left[\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]$

$=\frac{1}{2}\left[\frac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\right]$

$=\frac{1}{2}\left[\frac{n^2+3n}{2\left(n+1\right)\left(n+2\right)}\right]$

$=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}$