Question

If $\sum _{k=1}^n{\tan }^{-1}\left(\frac{2k}{2+k^2+k^4}\right)={\tan }^{-1}\left(\frac{6}{7}\right)$, then the value of $n$ is equal to

Answer 1

We know, ${\tan }^{-1}\left(x\right)-{\tan }^{-1}\left(y\right)={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)$
Now, ${\tan }^{-1}\left(\frac{2k}{2+k^2+k^4}\right)$
$={\tan }^{-1}\left(\frac{2k}{1+(k^4+1+2k^2-k^2)}\right)$
$={\tan }^{-1}\left(\frac{2k}{1+{(k^2+1)}^2-k^2}\right)$
$={\tan }^{-1}\left(\frac{(k^2+1+k)-(k^2+1-k)}{1+(k^2+1+k)(k^2+1-k)}\right)$
Hence, $\sum _{k=1}^k{\tan }^{-1}\left(\frac{2k}{2+k^2+k^4}\right)$
$=\sum _{k=1}^k{\tan }^{-1}(k^2+1+k)-{\tan }^{-1}(k^2+1-k)$

Putting the values of $k$, we get
${\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(7\right)-{\tan }^{-1}\left(3\right)+...+{\tan }^{-1}(n^2+n+1)-{\tan }^{-1}(n^2-n+1)={\tan }^{-1}\left(\frac{6}{7}\right)$
$\Rightarrow {\tan }^{-1}(n^2+n+1)-{\tan }^{-1}\left(1\right)={\tan }^{-1}\left(\frac{6}{7}\right)$
$\Rightarrow {\tan }^{-1}(n^2+n+1)={\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(\frac{6}{7}\right)$
$\Rightarrow {\tan }^{-1}(n^2+n+1)={\tan }^{-1}\left(\frac{1+\frac{6}{7}}{1-\frac{6}{7}\times 1}\right)$
$\Rightarrow {\tan }^{-1}(n^2+n+1)={\tan }^{-1}\left(13\right)$
$\Rightarrow n^2+n+1=13$
$\Rightarrow n^2+n-12=0$
$\Rightarrow (n-3)(n+4)=0$
$\Rightarrow n=3$ or $n=-4$
As $n$ cannot be negative, so $n=3$