Question

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

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

Answer 1

Answer: (C)

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