The value of (_n=1^50^-111+n+n^2) is

Question

The value of $\cot\left(\displaystyle\sum_{n=1}^{50}	an^{-1}\dfrac{1}{1+n+n^2}ight)$ is

Accepted Answer

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

$$ = \cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{{(n + 1) - n} \over {1 + (n + 1)n}}} \right)} } \right)$$

$$ = \cot \left( {\sum\limits_{n = 1}^{50} {({{\tan }^{ - 1}}(n + 1) - {{\tan }^{ - 1}}n} } \right)$$

$$ = \cot ({\tan ^{ - 1}}51 - {\tan ^{ - 1}}1)$$

$$ = \cot \left( {{{\tan }^{ - 1}}\left( {{{51 - 1} \over {1 + 51}}} \right)} \right)$$

$$ = \cot \left( {{{\cot }^{ - 1}}\left( {{{52} \over {50}}} \right)} \right)$$

$$ = {{26} \over {25}}$$

Solutions · 1