Question

Evaluate sum of $n$ terms of the series $\frac{8}{5}+\frac{16}{65}+\frac{24}{325}+.....$

• A. $\left(\frac{4n^2+2n}{2n^2+2n+1}\right)$
• B. $\left(\frac{4n^2+2n}{2n^2+2n}\right)$
• C. $\left(\frac{4n^2+4n}{2n^2+2n+1}\right)$
• D. $\left(\frac{4n^2+4n}{2n^2+2n}\right)$

Answer 1

The correct option is C $\left(\frac{4n^2+4n}{2n^2+2n+1}\right)$

Let $S_n=\frac{8}{5}+\frac{16}{65}+\frac{24}{325}+.....+nterms$
$={\sum }_{r=1}^n\left(\frac{8r}{4r^4+1}\right)$
$={\sum }_{r=1}^n\left(\frac{8r}{(2r^2-2r+1)(2r^2+2r+1)}\right)$
$=2{\sum }_{r=1}^n\left\{\frac{1}{(2r^2-2r+1)}-\frac{1}{(2r^2+2r+1)}\right\}$
$=2\left\{\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{13}+\frac{1}{13}-......+\frac{1}{(2n^2-2n+1)}-\frac{1}{(2n^2+2n+1)}\right\}$
$=2\{1-\frac{1}{2n^2+2n+1}\}$
$=\left(\frac{4n^2+4n}{2n^2+2n+1}\right)$