If [_r = 1^50 ^ - 11 2r^2 = p ], then the value of tan p is :

Question

If $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}{1 \over {2{r^2}}} = p} $$, then the value of tan p is :

Accepted Answer

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

= $$\sum\limits_{r = 1}^{50} {{{\tan }^{ - 1}}(2r + 1) - {{\tan }^{ - 1}}(2r - 1)} $$

= $${\tan ^{ - 1}}(101) - {\tan ^{ - 1}}1 = {\tan ^{ - 1}}{{50} \over {51}}$$

Solutions · 1