$\sum_{n=1}^{50}\arctan\left(\frac{2n}{n^4-n^2+1}\right)$ Find the value of 
$$\sum_{n=1}^{50}\arctan\left(\frac{2n}{n^4-n^2+1}\right)$$

$$\frac{2n}{n^4-n^2+1}=\frac{2n}{1-n^2(1-n^2)}$$
I am not able to split it into sum or difference of two $\arctan$s.Please help me.

Notice that we can write
$$ \frac{2n}{n^4 - n^2 + 1} = \frac{(n^2+n) - (n^2-n)}{1 + (n^2+n)(n^2-n)}. $$
In view of the addition formula for the tangent, we find that
$$ \arctan \bigg( \frac{x-y}{1+xy} \bigg) = \arctan x - \arctan y $$
for any $x > y > 0$. Thus we have
$$ \arctan\bigg( \frac{2n}{n^4 - n^2 + 1} \bigg) = \arctan(n^2+n) - \arctan(n^2-n). $$
Then you can easily compute the sum by telescoping.
