Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

We know, ${tan}^{-1}\left(x\right)-{tan}^{-1}\left(y\right)={tan}^{-1}\left(\frac{x-y}{1+xy}\right)$

Now, ${tan}^{-1}\left(\frac{2k}{2+k^2{+k}^4}\right)$ = ${tan}^{-1}\left(\frac{2k}{1+k^2+1{+k}^4}\right)$

$={tan}^{-1}\left(\frac{2k}{2+k^2{+k}^4}\right)={tan}^{-1}\left(\frac{2k}{1+(k^4+1+2k^2-k^2)}\right)={tan}^{-1}\left(\frac{2k}{1+{(k^2+1)}^2-k^2}\right)={tan}^{-1}\left(\frac{(k^2+1+k)-(k^2+1-k)}{1+(k^2+1+k)(k^2+1-k)}\right)$

Hence, ${\sum }_{k=1}^{k=n}{tan}^{-1}\left(\frac{2k}{2+k^2{+k}^4}\right)$

=${\sum }_{k=1}^{k=n}{tan}^{-1}(k^2+1+k)-{tan}^{-1}(k^2+1-k)$

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 $n=3$