Summation problem

Relevant wiki: Telescoping Series - Sum

$\begin{aligned} \alpha &= \displaystyle \sum_{k=1}^{50} \dfrac{k}{k^4+k^2+1} \\ &= \displaystyle \sum_{k=1}^{50} \dfrac{k}{(k^2+k+1)(k^2-k+1)} \\ &= \displaystyle \dfrac 12 \sum_{k=1}^{50} \dfrac{2k}{(k^2+k+1)(k^2-k+1)} \\ &= \displaystyle \dfrac 12 \sum_{k=1}^{50} \dfrac{(k^2+k+1) - (k^2-k+1)}{(k^2+k+1)(k^2-k+1)} \\ &= \displaystyle \dfrac 12 \sum_{k=1}^{50} \dfrac{1}{k^2-k+1} - \dfrac{1}{k^2+k+1} \\ &= \dfrac 12 \left( \dfrac{1}{1^2-1+1} - \dfrac{1}{{50}^2+50+1} \right) \\ &= \dfrac 12 \left( 1 - \dfrac{1}{2551} \right) \\ &= \dfrac{1275}{2551} \end{aligned} \\ \implies 2551 \alpha = \ \boxed{1275}$

Write denominator $k^4+k^2+1$ = $(k^2+1)^2-k^2$ = $(k^2+1-k)(k^2+1+k)$ .

Now numerator as { $(k^2+1+k)-(k^2+1-k)$ }/2

Now by using telescopic cancellations.

We get the value of sequence to be 1275/2551

So 2551(1275/2551) = 1275.