Question

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

Answer 1

$ \displaystyle\sum_{ k =1}^{10} \frac{ k }{ k ^4+ k ^2+1}=\frac{ m }{ n }$
$ \Rightarrow \frac{1}{2} \displaystyle\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( \displaystyle\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$