I am wondering how you would compute a double sum such as $$\sum_{n=0}^{p}\sum_{m=0}^n (m+n)^2$$
I understand methods for finding closed forms of sums using Discrete Calculus from reading Concrete Mathematics where for example one could view $\sum_{m=1}^p m^2 = \sum_{m=1}^p \Delta(\frac{1}{3}m^{\underline{3}}+\frac{1}{2}m^{\underline{2}})$ and then using the fact the the sum telescopes. I run into trouble applying that approach here because I get terms such as $(2n+1)^{\underline{2}}$. Is it possible to realize this as the forward difference of something? If it's not then what is a way to evaluate the sum?
I'm offhand unaware how to determine the summations using forward differences. Instead, with the Faulhaber's formula examples, and eti902's comment suggestion, we get for the inner summation
$$\begin{equation}\begin{aligned} \sum_{m=0}^{n}(m^2+2mn+n^2) & = \sum_{m=0}^{n}m^2 + 2n\sum_{m=0}^{n}m + n^2\sum_{m=0}^{n}1 \\ & = \left(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\right)+2n\left(\frac{n^2+n} {2}\right) + n^2(n+1) \\ & = \left(\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\right)+ (n^3+n^2) + (n^3+n^2) \\ & = \frac{7n^3}{3} + \frac{5n^2}{2} + \frac{n}{6} \end{aligned}\end{equation}$$
Alternatively, based on ShyamalSayak's comment suggestion, using $i = m+n$, we instead get
$$\begin{equation}\begin{aligned} \sum_{m=0}^{n}(m+n)^2 & = \sum_{i=n}^{2n}i^2 \\ & = \sum_{i=0}^{2n}i^2 - \sum_{i=0}^{n-1}i^2 \\ & = \left(\frac{(2n)^3}{3} + \frac{(2n)^2}{2} + \frac{2n}{6}\right) - \left(\frac{(n-1)^3}{3} + \frac{(n-1)^2}{2} + \frac{n-1}{6}\right) \\ & = \left(\frac{8n^3}{3} + 2n^2 + \frac{n}{3}\right) - \left(\frac{n^3-3n^2+3n-1}{3} + \frac{n^2-2n+1}{2} + \frac{n-1}{6}\right) \\ & = \frac{7n^3}{3} + \frac{5n^2}{2} + \frac{n}{6} \end{aligned}\end{equation}$$
In either case, the outer summation is then
$$\begin{equation}\begin{aligned} \sum_{n=0}^{p}\left(\frac{7n^3}{3} + \frac{5n^2}{2} + \frac{n}{6}\right) & = \frac{7}{3}\sum_{n=0}^{p}n^3 + \frac{5}{2}\sum_{n=0}^{p}n^2 + \frac{1}{6}\sum_{n=0}^{p}n \\ & = \frac{7}{3}\left(\frac{p^4}{4}+\frac{p^3}{2}+\frac{p^2}{4}\right) + \frac{5}{2}\left(\frac{p^3}{3}+\frac{p^2}{2}+\frac{p}{6}\right) + \frac{1}{6}\left(\frac{p^2+p}{2}\right) \\ & = \frac{7p^4}{12} + 2p^3 + \frac{23}{12}p^2 + \frac{p}{2} \\ & = \frac{p(p+1)(7p^2+17p+6)}{12} \\ & = \frac{p(p+1)(p+2)(7p+3)}{12} \end{aligned}\end{equation}$$