Let Sn=122−1+142−1+⋯1(2n)2−1, then value of S25 is

$S_{n} = \frac{1}{2^{2} - 1} + \frac{1}{4^{2} - 1} + \dots \frac{1}{{(2 n)}^{2} - 1}$

$S_{n} = \frac{1}{2^{2} - 1} + \frac{1}{4^{2} - 1} + \frac{1}{6^{2} - 1} + \dots u p t o 2 n t e r m s$

$t_{k} = \frac{1}{{(2 k)}^{2} - 1} = \frac{1}{(2 k + 1) (2 k - 1)}$

$= \frac{1}{2} [\frac{1}{2 k - 1} - \frac{1}{2 k + 1}]$

$S_{n} = \sum_{k = 1}^{n} t_{k} = \frac{1}{2} \sum_{k = 1}^{n} (\frac{1}{2 k - 1} - \frac{1}{2 k + 1})$

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

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

$∴ S_{n} = \frac{n}{2 n + 1}$

$S_{25} = \frac{25}{51}$

Let $S_{n} = \frac{1}{2^{2} - 1} + \frac{1}{4^{2} - 1} + \dots \frac{1}{{(2 n)}^{2} - 1}$ , then value of $S_{25}$ is