$n \cdot n$ board, each square holds the number of rectangles that contain it

So the problem goes: we have a $n\cdot n$ board, where there is a number on each square, representing the number of rectangles of the board, which contain that square. What is the sum of all such numbers?

I'm really just a beginner at discrete mathemathics, and we didn't finish this problem at uni, and it's been bothering me since I'm unsure what to do here.

An idea was to put the board in a coordinate system and view each square as $(i,j)$ which are $\in$ $P_{a,b,c,d}$ (rectangles). And from there proceeding onto some strange summations, but that proved useless. An approach using matrices also didn't get too far.

I'm really lost here so any hints whatsoever would be really useful, I wish I had more ideas to offer!

Thank you in advance!

Imagine a formula for the number of rectangles containing the square at $(i, j)$. Such a rectangle goes horizontally from $a$ to $b$ for some $a,b$ where $a\le i \le b$, and goes vertically from $c$ to $d$ for some $c,d$ where $c\le j\le d$.
For a square at $(i,j)$, there are $i$ possibilities for $a$, $n-i+1$ possibilities for $b$, $j$ possibilities for $c$, and $n-j+1$ possibilities for $d$. Since the choices for $a,b,c,d$ are independent, there are $ij(n-i+1)(n-j+1)$ possible rectangles containing the square at $(i,j)$. So your final answer will be $$\sum_{i=1}^n\sum_{j=1}^n ij(n-i+1)(n-j+1)$$ Do you think you can evaluate this sum?
Now, to further simplify things, this sum is also equal to the number of ways to choose a rectangle and then an individual square within this rectangle. This is equivalent to choosing a pair of left and right edge coordinates $x_1, x_2$ satisfying $0 \le x_1 < x_2 \le n$; a pair of bottom and top edge coordinates $y_1, y_2$ satisfying $0 \le y_1 < y_2 \le n$; and then the bottom left corner of the square $(x_0, y_0)$ satisfying $x_1 \le x_0 < x_2$ and $y_1 \le y_0 < y_2$. By replacing $x_0$ by $x_0 + 1$, $x_2$ by $x_2 + 1$, and similarly for $y$, we get a bijection with the set of 6-tuples $(x_1, x_2', x_0', y_1, y_2', y_0')$ with $0 \le x_1 < x_0' < x_2' \le n+1$ and $0 \le y_1 < y_0' < y_2' \le n+1$. It is now easy to see this is equal to $\binom{n+2}{3}^2$.