Question

Find the sum upto $n$ terms
$\frac{1}{\mathrm{1.2.3}}+\frac{1}{\mathrm{2.3.4}}+\frac{1}{\mathrm{3.4.5}}+.....$

Answer 1

$\sum _{x=1}^n\frac{1}{x(x+1)(x+2)}=\sum _{x=1}^n\frac{(x+2)-2(x+1)+x}{x(x+1)(x+2)}$

$\sum _{x=1}^n\frac{1}{x(x+1)}-\sum _{x=1}^n\frac{2}{2(x+2)}+\sum _{x=1}^n\frac{1}{(x+1)(x+2)}$

$=\sum _{x=1}^n\frac{(x+1)-x}{x(x+1)}-\sum _{x=1}^n\frac{(x+2)-x}{x(x+2)}+\sum _{x=1}^n\frac{(x+2)-(x+1)}{(x+1)(x+2)}$

$=\sum _{x=1}^n\frac{1}{x}-\frac{1}{x+1}-\sum _{x=1}^n\frac{1}{x}-\frac{1}{x+2}+\sum _{x=1}^n\frac{1}{x+1}-\frac{1}{x+2}$

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

$(1-\frac{1}{n+1})-(1-\frac{1}{n+1}+\frac{1}{2}-\frac{1}{n+2})+(\frac{1}{2}-\frac{1}{n+2})$

$0$