$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{2 \over {4{r^2}}}} \right) = \sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(2r + 1) - (2r - 1)} \over {1 + (2r + 1)(2r - 1)}}} \right)} }$ = $\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}(2r + 1) - {{\tan }^{ - 1}}(2r - 1)}$ = ${\tan ^{ - 1}}(101) - {\tan ^{ - 1}}1 = {\tan ^{ - 1}}{{50} \over {51}}$