How to simplify a Summation within a (nested) summation: $\sum_{a=0}^{\ T/2 -1}$ $\sum_{b=2a}^{\ T-1} b*b $ How Would you Simplify a summation with-in a summation,
like so
$\sum_{a=0}^{\ T/2 -1}$ $\sum_{b=2a}^{\ T-1} b*b $
I honestly have tried numerous approaches to simply the inner part first, but I get stuck, plus I am confused on exactly what would be considered a legitimate mathematical operation when working on the inner part first.
Online tools can help simplify it, but they don't provide any explanation
 A: Hint: It is helpful to take a look at Faulhaber's formula.
We obtain

\begin{align*}
\color{blue}{\sum_{a=0}^{\frac{T}{2}-1}}\color{blue}{\sum_{b=2a}^{T-1}b^2}
&=\sum_{a=0}^{\frac{T}{2}-1}\left(\sum_{b=1}^{T-1}b^2-\sum_{b=1}^{2a-1}b^2\right)\tag{1}\\
&=\sum_{a=0}^{\frac{T}{2}-1}\frac{1}{6}(T-1)T(2T-1)
-\sum_{a=0}^{\frac{T}{2}-1}\frac{1}{6}(2a-1)2a(4a-1)\tag{2}\\
&\color{blue}{=\frac{1}{12}(T-1)T^2(2T-1)-\frac{1}{3}\sum_{a=0}^{\frac{T}{2}-1}\left(4a^3-3a^2+a\right)}\tag{3}
\end{align*}
and the right hand sum can be calculated using Faulhaber's formulas.

Comment:

*

*In (1) we represent the inner sum as difference of two sums which both start with index value $1$.


*In (2) we use Faulhaber's formula for the sum of squares: $\sum_{q=0}^n q^2=\frac{1}{6}n(n+1)(2n+1)$.


*In (3) we use $\sum_{a=0}^{\frac{T}{2}-1}1=\frac{T}{2}$.
Hint: You might also find chapter 2: Sums in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik helpful. It provides a thorough introduction in the usage of sums.
