Mathematics · 2022
JEE Main · 27 June 2022, Shift 2 · Q18
The value of cot (Σ_n=1^50 tan^-1(1/(1+n+n^2))) is
The value of $\displaystyle \cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^2}\right)\right)$ is
Official answer
(1)
$\displaystyle \frac{26}{25}$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.