Mathematics · 2013
JEE Main · 23 April 2013, Shift 1 · Q69
If S = tan^-1(1/(n^2+ n +1))+ tan^-1(1/(n^2+3 n+3))+… + tan^-1(1/(1+( n +19)( n +20))), then tan S is equal to:
If $\displaystyle \mathrm{S}=\tan ^{-1}\left(\frac{1}{\mathrm{n}^2+\mathrm{n}+1}\right)+$
$\displaystyle \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)+\ldots$
$\displaystyle +\tan ^{-1}\left(\frac{1}{1+(\mathrm{n}+19)(\mathrm{n}+20)}\right)$, then
$\displaystyle \tan S$ is equal to :
Official answer
(3)
$\displaystyle \frac{20}{n^2+20 n+1}$
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JEE Main 2013 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.