Consider all points $(m,n)$ where $m$ and $n$ are natural numbers, $m\leq10$, and $n\leq5$. What is the average distances from each point to the origin $(0,0)$?
I know how to find the distance $d$ between the points $(x_1,y_1)$ and $(x_2,y_2)$; $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.
In this case, the first point is $(m,n)$ and the second point is the origin, therefore the distance, $d=\sqrt{m^2+n^2}$.
It is annoying to find all distances, summing them, and dividing them by the number of distances, is not it annoying?
What if the given conditions, $m\leq10$ and $n\leq5$, were changed to $m\leq49$ and $n\leq36$?!
There is a quick solution for a related problem, even though it's not quite what you are asking. If you are willing to compute squares of distances of all points, then you get $$ \begin{split} f(m,n) &= \sum_{i=1}^m \sum_{k=1}^n \left(i^2 + k^2\right) \\ &= \sum_{i=1}^m \sum_{k=1}^n i^2 + \sum_{i=1}^m \sum_{k=1}^n k^2 \\ &= n \sum_{i=1}^m i^2 + m \sum_{k=1}^n k^2 \\ &= n \frac{m(m+1)(2m+1)}{6} + m \frac{n(n+1)(2n+1)}{6} \\ &= \frac{mn}{6} \left[(m+1)(2m+1) + (n+1)(2n+1)\right] \end{split} $$
As expected, $f(m,n)$ is symmetric in $m,n$, in other words, $f(m,n) = f(n,m)$...