Question

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

$S={\sum }_{r=0}^{19}ta{n}^{-1}\left(\frac{1}{1+(n+r)(n+r+1)}\right)=ta{n}^{-1}\left(\frac{(n+r+1)-(n+r)}{1+(n+r)(n+r+1)}\right)$

$\Rightarrow S={\sum }_{r=0}^{19}\left(ta{n}^{-1}\right(n+r+1)-ta{n}^{-1}(n+r\left)\right)$

$=ta{n}^{-1}(n+1)-ta{n}^{-1}\left(n\right)=ta{n}^{-1}(n+20)-ta{n}^{-1}(n+19)$

$+ta{n}^{-1}(n+2)-ta{n}^{-1}(n+1)$

$ta{n}^{-1}(n+20)-ta{n}^{-1}(n+19)$

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

$\Rightarrow \left(C\right)$