Question

$\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{(5n-4)(5n+1)}$

Answer 1

We have, $\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+\frac{1}{16.21}+...+\frac{1}{(5n-4)(5n+1)}$

Let $T_r$be the rth term of the given series. Then,

$T_r=\frac{1}{(5r-4)(5r+1)},r=1,2,....,n\Rightarrow T_r=\frac{1}{5}[\frac{1}{5r-2}-\frac{1}{5r+1}]\therefore requiredsum=\frac{1}{5}{\sum }_{r-1}^n{T}_r=\frac{1}{5}{\sum }_{r-1}^n[\frac{1}{5r-4}-\frac{1}{5r+1}]=\frac{1}{5}[(1-\frac{1}{6})+(\frac{1}{6}-\frac{1}{11})+(\frac{1}{11}-\frac{1}{14})...(\frac{1}{5n-4}-\frac{1}{5n+1})]=\frac{1}{5}[1-\frac{1}{5n+1}]=\frac{1}{5}\left[\frac{5n+1-1}{5n+1}\right]=\frac{1}{5}\times \frac{5n}{5n+1}=\frac{n}{5n+1}Hence,requiredsum=\frac{n}{5n+1}$