Find the sum of:
1/(1*2*3) + 1(2*3*4) + 1/(3*4*5) + ... + 1/(n(n+1)(n+2))
Telescopic series: $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\Longrightarrow \frac{1}{n(n+1)(n+2)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\frac{1}{n+2}=$ $\displaystyle \frac{1}{n(n+2)}$ $\displaystyle -\frac{1}{(n+1)(n+2)}$ $\displaystyle =\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$ .
I think you can take it from here
Tonio
note that in tonio's solution the first term telescopes not so fast than the second term which does faster.
another way on doing this is to see that $\displaystyle \frac{1}{n(n+1)(n+2)}=\frac{1}{2}\cdot \frac{n+2-n}{n(n+1)(n+2)}=\frac{1}{2}\left( \frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)} \right),$ and this one telescopes easily.