The sum

Relevant wiki: Partial Fractions , Telescoping Series

$\begin{aligned} S &=\sum_{n=1}^{50} \dfrac{n}{n^4+n^2+1}\\ &=\sum_{n=1}^{50} \dfrac{n}{(n^2+1)^2-n^2}\\ &=\sum_{n=1}^{50} \dfrac{n}{(n^2+n+1)(n^2-n+1)}\\ &=\dfrac{1}{2}\sum_{n=1}^{50} \dfrac{1}{n^2-n+1}-\dfrac{1}{n^2+n+1}\\ &=\dfrac{1}{2}\sum_{n=1}^{50} \dfrac{1}{n^2-n+1}-\dfrac{1}{(n+1)^2-(n+1)+1}\\ &=\dfrac{1}{2}\left(\dfrac{1}{1^2-1+1}-\dfrac{1}{(50+1)^2-(50+1)+1}\right)\\ &=\dfrac{1275}{2551}≈0.499 \end{aligned}$

$\displaystyle \begin{aligned} S & = \sum_{n \ = \ 1}^{50} \frac{n}{1 + n^2 + n^4} \\ & = \sum_{n \ = \ 1}^{50} \left( \frac{1}{2\left(n^2 - n + 1\right)} - \frac{1}{2\left(n^2 + n + 1\right)} \right) \\ & = \underbrace{\left( \frac 12 - \frac 16\right)}_{a_1} + \underbrace{\left(\frac 16 - \frac{1}{14}\right)}_{a_2} + \underbrace{\left(\frac{1}{14} - \frac{1}{26}\right)}_{a_3} + \cdots + \underbrace{\left(\frac{1}{4706} - \frac{1}{4902}\right)}_{a_{49}} + \underbrace{\left(\frac{1}{4902} - \frac{1}{5102}\right)}_{a_{50}} \\ & = \frac 12 - \frac{1}{5102} \\ & \approx \boxed{0.499} \end{aligned}$