Disclaimer: This is for homework, but I'm just stuck on this one small part of a larger problem.
I'm having trouble figuring out how to get the following summation in closed form.
$$\sum_{j=1}^i 4ij$$
Since the index is j, am I able to move the i out of the summation as a constant the way I am with the 4? If I do that, do I need to square the i because it is the stopping point of the summation?
Am I able to separate the summation into two, like this?
$$\left(\sum_{j=1}^i 4i\right)\left(\sum_{j=1}^i j\right)$$
The $i$ is indeed a constant, so you treat it exactly as like the $4$:
$$\sum_{j=1}^i4ij=4i\sum_{j=1}^ij=4i\left(\frac{i(i+1)}2\right)=2i^2(i+1)\;.$$
It might help you to write out some terms: the sum is
$$4i\cdot1+4i\cdot2+4i\cdot3+\ldots+4i\cdot i\;,$$
which clearly has a factor of $4i$ in each term that can be factored out to yield
$$4i(1+2+3+\ldots+i)\;.$$