$${1 \over {{3^2} - 1}} + {1 \over {{5^2} - 1}} + {1 \over {{7^2} - 1}} + .... + {1 \over {{{(201)}^2} - 1}}$$ is equal to

$S = \sum\limits_{r = 1}^{100} {{1 \over {{{(2n + 1)}^2} - 1}}}$<br><br>$= \sum\limits_{r = 1}^{100} {{1 \over {(2n + 1 + 1)(2n + 1 - 1)}}}$<br><br>$= \sum\limits_{r = 1}^{100} {{1 \over {2n(2n + 2)}}}$<br><br>$= {1 \over 4}\sum\limits_{r = 1}^{100} {{1 \over {n(n + 1)}}}$<br><br>$= {1 \over 4}\sum\limits_{r = 1}^{100} {{{(n + 1) - n} \over {n(n + 1)}}}$<br><br>$$ = {1 \over 4}\sum\limits_{r = 1}^{100} {\left( {{1 \over n} - {1 \over {n + 1}}} \right)} $$<br><br>$$S = {1 \over 4}\left( {\left( {1 - {1 \over 2}} \right) + \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right) + ...... + \left( {{1 \over {100}} - {1 \over {101}}} \right)} \right)$$<br><br>$\therefore$ $S = {1 \over 4}\left[ {{{100} \over {101}}} \right] = {{25} \over {101}}$