Question

$S=ta{n}^{-1}\left(\frac{1}{n^2+n+1}\right)+ta{n}^{-1}\left(\frac{1}{n^2+3n+3}\right)+....$$+ta{n}^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right),$ then tan S is equal to

• A. $\frac{20}{401+20n}$
• B. $\frac{n}{n^2+20n+1}$
• C. $\frac{20}{n^2+20n+1}$
• D. $\frac{n}{401+20n}$

Answer 1

The correct option is C $\frac{20}{n^2+20n+1}$

$S={\tan }^{-1}\left(\frac{1}{n^2+n+1}\right)+{\tan }^{-1}\left(\frac{1}{n^2+3n+3}\right)+...+{\tan }^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right)$

$\Rightarrow S={\tan }^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right)+{\tan }^{-1}\left(\frac{(n+2)-(n+1)}{1+(n+2)(n+1)}\right)+...+{\tan }^{-1}\left(\frac{(n+20)-(n+19)}{1+(n+19)(n+20)}\right)$

$\Rightarrow S={\tan }^{-1}(n+1)-{\tan }^{-1}n+{\tan }^{-1}(n+2)-{\tan }^{-1}(n+1)+...+{\tan }^{-1}(n+20)-{\tan }^{-1}(n+19)$

$\Rightarrow S=-{\tan }^{-1}n+{\tan }^{-1}(n+20)$

$\Rightarrow S={\tan }^{-1}(n+20)-{\tan }^{-1}n$

$\Rightarrow S={\tan }^{-1}\left(\frac{n+20-n}{1+n(n+20)}\right)$

$\Rightarrow S={\tan }^{-1}\left(\frac{20}{1+n^2+20n}\right)$

$\Rightarrow \tan S=\frac{20}{n^2+20n+1}$