Question

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

Answer 1

$\begin{array}{c}Sn=\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\\ =\frac{1}{2}\left(\frac{2}{\mathrm{1.2.3}}+\frac{2}{\mathrm{2.3.4}}+\frac{2}{\mathrm{3.4.5}}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\right)\\ =\frac{1}{2}\left(\frac{3-1}{\mathrm{1.2.3}}+\frac{4-2}{\mathrm{2.3.4}}+\frac{5-3}{\mathrm{3.4.5}}+...+\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\right)\\ =\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\\ =\frac{1}{2}\left(\frac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\right)\\ =\frac{1}{4}\left(\frac{n^2+3n}{\left(n+1\right)\left(n+2\right)}\right)\\ =\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\end{array}$