Question

$\frac{1}{\mathrm{2.3.4.5}}+\frac{1}{\mathrm{3.4.5.6}}+\frac{1}{\mathrm{4.5.6.7}}+..$ to n terms

Answer 1

$\frac{1}{\mathrm{2.3.4.5}}+\frac{1}{\mathrm{3.4.5.6}}+...t_{n}terms$

$n^{th}$ term of the series is

$\frac{1}{n(1+1)(n+2)(n+3)},n⩾2$

$=\frac{1}{3}(\frac{(n+3)}{(n+2)n(n+1)(n+3)}-\frac{n}{n(n+1)(n+2)(n+3)})$

$=\frac{1}{3}(\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)})$

on solving, we get,

$\frac{1}{3}[\sum _{n=2}^{k-1}\frac{1}{n(n+1)(n+2)}-\sum _{n=3}^k\frac{1}{(n+1)(n+2)(n+3)}]$

$=\frac{1}{3}[\frac{1}{\mathrm{2.3.4}}-\frac{1}{(n+1)(n+2)(n+3)}]$