Saudi Arabian 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
$$\sum_{k=0}^{n-1} (m-k)(n-k). \quad (1)$$
We have
$$\begin{aligned} \sum_{k=0}^{n-1} (m-k)(n-k) &= \sum_{k=0}^{n-1} (mn - (m+n)k + k^2) \\ &= mn^2 - (m+n) \frac{(n-1)n}{2} + \frac{(n-1)n(2n-1)}{6} \\ &= mn^2 - m \frac{(n-1)n}{2} - \frac{(n-1)n^2}{2} + \frac{(n-1)n(2n-1)}{6} \\ &= \frac{n(n+1)(3m-n+1)}{6}. \end{aligned}$$