Question

If $\sum _{n=1}^k{\text{Tan}}^{-1}\left(\frac{1}{n^2+3n+3}\right)={\text{Tan}}^{-1}\alpha$

• A. $\frac{k}{k+2}$
• B. $\frac{2k}{2k+1}$
• C. $\frac{k}{2k+5}$
• D. $\frac{3k}{4k+5}$

Answer 1

Answer: (C)

We have, $\sum _{n=1}^k{\tan }^{-1}\left(\frac{1}{n^2+3n+3}\right)={\tan }^{-1}\alpha$
$\sum _{n=1}^k{\tan }^{-1}\left(\frac{(n+2)-(n+1)}{1+(n+2)(n+1)}\right)={\tan }^{-1}\alpha$
$\Rightarrow \sum _{n=1}^k\left({\tan }^{-1}(n+2)-{\tan }^{-1}(n+1)\right)={\tan }^{-1}\alpha$
$\Rightarrow \left({\tan }^{-1}3-{\tan }^{-1}2\right)+\left({\tan }^{-1}4-{\tan }^{-1}2\right)\dots$ ${\tan }^{-1}(k+2)-{\tan }^{-1}(k+1)={\tan }^{-1}\alpha$
${\tan }^{-1}(k+2)-{\tan }^{-1}2={\tan }^{-1}\alpha$
${\tan }^{-1}\left(\frac{k+2-2}{1+(k+2)(2)}\right)={\tan }^{-1}\alpha$
$\Rightarrow \frac{k}{1+2k+4}=\alpha$
$\Rightarrow \alpha =\frac{k}{2k+5}$