${t_r} = \frac{1}{{4{r^2} – 1}}$

$ = \frac{1}{{(2r – 1)(2r + 1)}} = \frac{1}{2}\left( {\frac{1}{{2r – 1}} – \frac{1}{{2r + 1}}} \right)$

${S_n} = \sum\limits_{r = 1}^n {\frac{1}{2}\left( {\frac{1}{{2r – 1}} – \frac{1}{{2r + 1}}} \right)} $

${S_n} = \frac{1}{2}\left[ {\left( {\frac{1}{1} – \frac{1}{3}} \right) + \left( {\frac{1}{3} – \frac{1}{5}} \right) + \left( {\frac{1}{5} – \frac{1}{7}} \right) + ……. + \left( {\frac{1}{{2n – 1}} – \frac{1}{{2n + 1}}} \right)} \right]$

${S_n} = \frac{1}{2}\left[ {1 – \frac{1}{{2n + 1}}} \right] = \frac{n}{{2n + 1}}$