The value of cot ( ∑ _ n = 1 ^ 50 ^ - 1 ( 1 1 + n + n ^ 2 )) is

Now $\mathop{∑}_{n = 1}^{50}\tan^{- 1}\left(\dfrac{1}{1 + n + {n}^{2}}\right) = \mathop{∑}_{n = 1}^{50}\tan^{- 1}\left(\dfrac{\left(n + 1\right) - n}{1 + n\left(n + 1\right)}\right)$

$= \mathop{∑}_{n = 1}^{50}\left[\tan^{- 1}\left(n + 1\right) - \tan^{- 1}n\right]$

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

$= \tan^{- 1}\left(\dfrac{50}{52}\right) = \cot^{- 1}\left(\dfrac{26}{25}\right)$

So $\cot \left(\mathop{∑}_{n = 1}^{50}\tan^{- 1}\left(\dfrac{1}{1 + n + {n}^{2}}\right)\right) = \cot \left(\cot^{- 1}\dfrac{26}{25}\right) = \dfrac{26}{25}$