Mathematical Reflections 2017, Issue 3 - Problem O410

Problem:
On each cell of a chess board it is written a number equal to the amount of the rectangles that contain this cell. Find the sum of all the numbers.

Proposed by Robert Bosch, USA 

Solution:
Consider an $n \times n$ chessboard and let $S$ be the required sum. Observe that writing on each cell the number equal to the amount of the rectangles containing the cell is equivalent to perform the following operation on the chessboard: at the beginning write zero on each cell of the chessboard, then for each $i \times j$ rectangle ($1 \leq i,j \leq n$) add $1$ to all its cells. So, we only have to find the total number of rectangles, each counted with the number its of cells. The number of $i \times j$ rectangles is $(n+1-i)(n+1-j)$, where $1 \leq i,j \leq n$. Since each $i \times j$ rectangle contains $ij$ cells, then we have
\begin{eqnarray*} S&=& \sum_{i=1}^n \sum_{j=1}^n ij(n+1-i)(n+1-j) \\ &=& \sum_{i=1}^n i(n+1-i) \sum_{j=1}^n j(n+1-j) \\ &=& \left(\sum_{k=1}^n k(n+1-k)\right)^2 \\ &=& {n+2 \choose 3}^2.
\end{eqnarray*}