If $\sum _{k=1}^{10}\frac{k}{k^4+k^2+1}=\frac{m}{n}$, where $m$ and $n$ are co-prime, then $m+n$ is equal to

Given,

$\sum _{k=1}^{10}\frac{k}{k^4+k^2+1}=\frac{m}{n}$

$\Rightarrow \frac{1}{2}\sum _{k=1}^{10}\frac{\left(k^2+k+1\right)-\left(k^2-k+1\right)}{\left(k^2+k+1\right)\left(k^2-k+1\right)}=\frac{m}{n}$

$\left\{\text{by using the formula }{x}^4+x^2+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\right\}$

$\Rightarrow \frac{1}{2}\left(\sum _{k=1}^{10}\left(\frac{1}{\left(k^2-k+1\right)}-\frac{1}{k^2+k+1}\right)\right)=\frac{m}{n}$

$\Rightarrow \frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-.........-\frac{1}{111}\right)=\frac{m}{n}$

$\Rightarrow \frac{1}{2}\left(1-\frac{1}{111}\right)=\frac{m}{n}$

$\Rightarrow \frac{55}{111}=\frac{m}{n}$

So, $m+n=166$