Question

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

Answer 1

Answer: 166

$\sum _{k=1}^{10}\frac{k}{k^4+k^2+1}=\frac{m}{n}$
$\Rightarrow \frac{1}{2}\sum _{k=1}^{10}\frac{\left(k^2+k+1\right)-\left(k^2-k+1\right)}{\left(k^2+k+1\right)\left(k^2-k+1\right)}$
$\Rightarrow \frac{1}{2}\left(\sum _{k=1}^{10}\left(\frac{1}{\left(k^2-k+1\right)}-\frac{1}{k^2+k+1}\right)\right)$
$\Rightarrow \frac{55}{111}=\frac{m}{n}$
$m+n=166$