Summation of $\dfrac{1}{1+1^2+1^4}+ \dfrac{2}{1+2^2+2^4}+\dfrac{3}{1+3^2+3^4}...$ 
Find the sum of $\dfrac{1}{1+1^2+1^4}+ \dfrac{2}{1+2^2+2^4}+\dfrac{3}{1+3^2+3^4}...$ till n terms. 

The $i^{th}$ term is given by $\dfrac{i}{1+i^2+i^4}$. I did a similar problem previously but there I was able to decompose the fraction into partial fractions. Here, I am unable to do so. What would be an efficeinent way to solve it then? (Just a hint would suffice) 
 A: HINT: $1+i^2+i^4=(1+i^2)^2-i^2$
A: You can get the partial fraction decomposition by writing $x^4+x^2+1$ as the product of two quadratics $(Ax^2+Bx+C)(Dx^2+Ex+F)$, and doing some casework to get the variables. Since the coefficient of $x^4$ is $1$, and $1$'s only factor is $1$, then both $A$ and $D$ equal $1$. We then have:
$$(x^2+Bx+C)(x^2+Ex+F)$$
The coefficient of the constant term is $1$, so following the same logic, we get:
$$(x^2+Bx+1)(x^2+Ex+1)$$
Since there is no $x^3$ term, and we know that the only ways of getting a $x^3$ is from $Bx*x^2$ and $x^2*Ex$, then these terms must cancel each other. We can conclude $B = -E$, and we obtain:
$$(x^2+Bx+1)(x^2-Bx+1)$$
$$\Rightarrow (x^4-Bx^3+x^2)+(Bx^3-B^2x+Bx)+(x^2-Bx+1)$$
$$\Rightarrow x^4+(2-B^2)x^2+1$$
Then since $(2-B^2)x^2=x^2$, then $2-B^2=1$, or $B=1$. Then collecting all the variables into the same expression, we get that the factorisation of $x^4+x^2+1$ equals $(x^2+x+1)(x^2-x+1)$. You can multiply this out to verify my answer if you wish to.
Done!
