Saudi Arabia Mathematical Competitions 2012

There are $mn$ squares of length side $1$, $(m-1)(n-1)$ squares of length side $2$, $(m-2)(n-2)$ squares of side length $3$, etc. Assume that $m\ge n$. The total number of squares in this configuration is $$\sum _{k=0}^{n-1}(m-k)(n-k).(1)$$ We have $$\begin{array}{rl}\sum _{k=0}^{n-1}(m-k)(n-k)& =\sum _{k=0}^{n-1}(mn-(m+n)k+k^2)\\ & =m{n}^2-(m+n)\frac{(n-1)n}{2}+\frac{(n-1)n(2n-1)}{6}\\ & =m{n}^2-m\frac{(n-1)n}{2}-\frac{(n-1)n^2}{2}+\frac{(n-1)n(2n-1)}{6}\\ & =\frac{n(n+1)(3m-n+1)}{6}.\end{array}$$