Question

If $S$ denotes the sum of first $24$ terms of series $\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+.\dots \dots$ then $S=$

• A. $\frac{300}{49}$
• B. $\frac{300}{51}$
• C. $\frac{295}{49}$
• D. None

Answer 1

Answer: (A)

$\sum _{n=1}^{24}\frac{n^2}{\left(2n-1\right)\left(2n+1\right)}=\frac{1}{4}\sum _{n=1}^{24}\frac{4n^2-1+1}{\left(2n-1\right)\left(2n+1\right)}$
$\frac{1}{4}\left[\sum _{n=1}^{24}1+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\right]$
$=\frac{1}{4}\left[24+\frac{1}{2}\left[\frac{1}{1}-\frac{1}{49}\right]\right]$
$=6+\frac{6}{49}=\frac{300}{49}$