Find the sum of:

1/(1*2*3) + 1(2*3*4) + 1/(3*4*5) + ... + 1/(n(n+1)(n+2))

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- Feb 15th 2010, 08:19 AMbearej50Series...
Find the sum of:

1/(1*2*3) + 1(2*3*4) + 1/(3*4*5) + ... + 1/(*n*(*n*+1)(*n*+2)) - Feb 15th 2010, 09:17 AMtonio

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 (Wink)

Tonio - Feb 15th 2010, 10:44 AMKrizalid
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. - Feb 15th 2010, 06:45 PMtonio