If S is the sum of the first 10 terms of the series

$${\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....$$

then tan(S) is equal to :

S = $${\tan ^{ - 1}}\left( {{1 \over 3}} \right) + {\tan ^{ - 1}}\left( {{1 \over 7}} \right) + {\tan ^{ - 1}}\left( {{1 \over {13}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {21}}} \right) + ....$$ <br><br>= $${\tan ^{ - 1}}\left( {{1 \over {1 + 1 \times 2}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {1 + 2 \times 3}}} \right) + ...$$ <br><br>$\therefore$ T_r = ${\tan ^{ - 1}}\left( {{1 \over {1 + r \times \left( {r + 1} \right)}}} \right)$ <br><br>= tan^–1(r + 1) – tan^–1r <br><br>$\therefore$ T₁ = tan^–12 – tan^–11 <br><br>T₂ = tan^–13 – tan^–12 <br><br>T₃ = tan^–14 – tan^–13 <br>. <br>. <br>. <br><br>T₁₀ = tan^-111 – tan^–110 <br><br>$\therefore$ S = tan^–111 – tan^–11 = ${\tan ^{ - 1}}\left( {{{11 - 1} \over {1 + 11}}} \right)$ <br><br>$\therefore$ tan(S) = $\tan \left( {{{\tan }^{ - 1}}\left( {{{11 - 1} \over {1 + 11}}} \right)} \right)$ <br><br>= ${{{11 - 1} \over {1 + 11}}}$ = ${{10} \over {12}} = {5 \over 6}$