The value of ( _ n=1 ^ 50 ^ -1 ( 1 1+n+n^ 2 ) ) is
Mathematics · JEE Main · NTA Exams — Sets, Relations and Functions
The value of \(\cot \left(\sum_{n=1}^{50} \tan ^{-1}\left(\frac{1}{1+n+n^{2}}\right)\right)\) is
Answer
(A)
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