TITLE: Double Sum with a Neat Result
QUESTION [6 upvotes]: Based on an interesting question here (second question), I have devised a similar one. 
Evaluate the following double sum without expansion and substitution of standard sum-of-integers formula. 
$$\sum_{x=1}^n\sum_{y=1}^n (n-x+y)$$

REPLY [3 votes]: Here's a slightly different way to look at it. First, we rewrite the sum as:
$$\sum_{x=1}^{n} \sum_{y=1}^{n} (n-x+y) = \sum_{(x,y)\in\{1,\dots,n\}^2} (n-x+y) \enspace,$$
with the usual meaning of $A^2$ as the set of all ordered pairs of elements of the set $A$. Then, we observe that for $x\neq y$, the sum of the two terms corresponding to the pairs $(x,y)$ and $(y,x)$ is:
$$(n-x+y) + (n-y+x) = 2n$$
Therefore, we can think of each of the two pairs as contributing $n$ to the sum. For $x=y$, the term corresponding to $(x,y)$ is just $n$.
Conclusion: every pair contributes $n$ to the sum, so the sum is $n^2\cdot n= n^3$.