Is there a simplification of this sum of quadratic sums? I have $a(1+2+3+...+N)^2+a(2+3+...+N)^2+a(3+..+N)^2+...+aN^2$, essentially a sum of quadratic terms holding an integer summation that get consecutively smaller  Within each term the summation up to N elements (which could also go to infinity if that makes life easier). Some experimentation lead me to believe that this converge to something when divided by $N^{-5}$, but don't take my word for it. Any simplification here possible?

Drop the $a$ for now.
Then, the $n$th term is, \begin{align}a_n&=\left(\displaystyle\sum_{i=n}^{N} i\right)^2
\\&=\left(\dfrac12N(N+1)-\dfrac12n(n-1)\right)^2
\\&=\dfrac14(N^2(N+1)^2+n^2(n-1)^2-2Nn(N+1)(n-1)).\end{align}
So, \begin{align}
\displaystyle\sum_{n=1}^N a_n
&=\frac14\sum_{n=1}^N N^2(N+1)^2+\frac14\sum_{n=1}^N n^2(n-1)^2-\frac14\sum_{n=1}^N 2Nn(N+1)(n-1)
\\&=\frac14N^3(N+1)^2+\frac14\sum_{n=1}^N n^2(n-1)^2-\frac12N(N+1)\sum_{n=1}^N n(n-1).\end{align}
Provided you know the closed-forms of $\sum_{n=1}^N n^k$ for $k=1,2,3,4$, you should be good!

If you can guess that the answer will be a degree $5$ polynomial (and it's not hard since each term is a degree $4$ polynomial), then another approach would be to first guess the closed-form to be $aN^5+bN^4+cN^3+dN^2+eN$.
Then for $N=1$, it should equal $1$.
So, we get $a+b+c+d+e=1$.
Similarly, for $N=2$, it should equal $13$ and thus, $32a+16b+8c+4d+2e=13$.
After getting $5$ (nasty) linear equations, you can solve them to obtain your answer.
Admittedly, this is yet another cumbersome approach but I thought I'd mention this since as you've already mentioned in your question, you already know that the closed-form will have degree $5$.
