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}+.\ldots \ldots $ then $S=$

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

Answer 1

Answer: (A)

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