Question

If $\displaystyle\sum^n_{k - 1} \tan^{-1} \left( \frac{1}{k^2 + k + 1} \right) = \tan^{-1} (\theta) $ , then $\theta$ =

• A. $\frac{n}{n + 2}$
• B. $\frac{n}{n + 1}$
• C. $1$
• D. $\frac{n}{n -1 }$

Answer 1

The Correct Option is A

Given, $\displaystyle\sum_{k=1}^{n} \tan ^{-1}\left(\frac{1}{k^{2}+k+1}\right)=\tan ^{-1} \theta$
Now, $\displaystyle\sum_{k=1}^{n} \tan ^{-1}\left(\frac{1}{k^{2}+k+1}\right)=\displaystyle\sum_{k=1}^{n} \tan ^{-1} \frac{[(k+1)-k]}{1+k(k+1)}$
$=\displaystyle\sum_{k=1}^{n}\left[\tan ^{-1}(k+1)-\tan ^{-1} k\right]$
$=\tan ^{-1} 2-\tan ^{-1} 1+\tan ^{-1} 3-\tan ^{-1} 2+\tan ^{-1} 4$
$-\tan ^{-1} 3+\ldots+\tan ^{-1}(n+1)-\tan ^{-1} n$
$=\tan ^{-1}(n+1)-\tan ^{-1} l =\tan ^{-1}\left[\frac{n+1-1}{1+n+1}\right]$
$=\tan ^{-1}\left[\frac{n}{2+n}\right]$
$\therefore \tan ^{-1}\left(\frac{n}{2+n}\right)=\tan ^{-1} \theta$
$ \Rightarrow \theta=\frac{n}{n+2}$