How to work out a double summation without manually finding each answer? I have the double summation
$$
\sum_{p = 1}^{2} \sum_{q = p}^{3} {(p-1)q}
$$
I know how to work this out if I was to go through and add up every term. My question is, how would I do this in a shorter way, so if the 2 and the 3 were 200 and 300, I could still do it just as easily. Examples I have found don't have the "q = p" on the 2nd summation and this is what I think is throwing me off.  Could anyone explain how to do this?
Thanks

The easiest way is to note that the $p-1$ factor does not depend on $q,$ so that $$\sum_{p=1}^2\sum_{q=p}^3(p-1)q=\sum_{p=1}^2(p-1)\sum_{q=p}^3q.$$ The second sum is the same as adding up the first $3$ positive integers, then subtracting the first $p-1$ of them. In general, the formula for the sum of the first $n$ positive integers is $$\sum_{q=1}^nq=\frac{n(n+1)}2,$$ so the sum of the first $3$ is $6$, and the sum of the first $p-1$ is $\frac{(p-1)p}2.$ Thus, $$\sum_{q=p}^3q=6-\frac{(p-1)p}2=\frac{12}2+\frac{-p^2+p}2=\frac{-p^2+p+12}2,$$ and so our nested sum becomes $$-\frac12\sum_{p=1}^2(p-1)(p^2-p-12).$$ Finally, noting that the summand is $0$ when $p=1$ due to the factor $p-1,$ our nested sum becomes $$-\frac12\sum_{p=2}^2(p-1)(p^2-p-12)=-\frac12(2-1)(2^2-2-12)=-\frac12\cdot1\cdot-10=5.$$ Had we been dealing with a larger number than $2,$ we'd have wanted to expand $(p-1)(p^2-p-12)=p^3-2p^2-11p+12,$ then use formulas for sums of the first $n$ cubes, squares, positive numbers, and sums of constants to get the final answer.
