13^2-1+15^2-1+17^2-1++1201^2-1 is equal to:

Question

$\dfrac{1}{3^2-1}+\dfrac{1}{5^2-1}+\dfrac{1}{7^2-1}+\cdots+\dfrac{1}{201^2-1}$ is equal to:

Accepted Answer

$$S = \sum\limits_{r = 1}^{100} {{1 \over {{{(2n + 1)}^2} - 1}}} $$

$$ = \sum\limits_{r = 1}^{100} {{1 \over {(2n + 1 + 1)(2n + 1 - 1)}}} $$

$$ = \sum\limits_{r = 1}^{100} {{1 \over {2n(2n + 2)}}} $$

$$ = {1 \over 4}\sum\limits_{r = 1}^{100} {{1 \over {n(n + 1)}}} $$

$$ = {1 \over 4}\sum\limits_{r = 1}^{100} {{{(n + 1) - n} \over {n(n + 1)}}} $$

$$ = {1 \over 4}\sum\limits_{r = 1}^{100} {\left( {{1 \over n} - {1 \over {n + 1}}} \right)} $$

$$S = {1 \over 4}\left( {\left( {1 - {1 \over 2}} \right) + \left( {{1 \over 2} - {1 \over 3}} \right) + \left( {{1 \over 3} - {1 \over 4}} \right) + ...... + \left( {{1 \over {100}} - {1 \over {101}}} \right)} \right)$$

$$ \therefore $$ $$S = {1 \over 4}\left[ {{{100} \over {101}}} \right] = {{25} \over {101}}$$

Solutions · 1