Given big rectangle of size $x, y$, count sum of areas of smaller rectangles. Let's say we have two integers $x$ and $y$ that describe one rectangle, if this rectangle is splitten in exactly $x\cdot y$ squares, each of size $1\cdot 1$, count the sum of areas of all rectangles that can be formed from those squares.
For example if $x = 2, y = 2$, we have $4$ rectangles of size $(1,1)$, $2$ rectangles of size $(1, 2)$, $2$ rectangles of size $(2, 1)$ and $1$ rectangle of size $(4,4)$. So the total answer is $4\cdot 1+ 2\cdot2 + 2\cdot2+1\cdot4 = 16$
Is there easy way to solve this for different $x$ and $y$?
 A: I assume, in your example,  you mean there is one rectangle of size $(2, 2)$, because there are none $(4, 4)$.
In general, we have $(x - a + 1) \times (y - b + 1)$ rectangles of size $(a, b)$: we need to choose how many empty columns we leave before the rectangle $(0, 1, 2, \dots \text{ or } x-a)$, and how many rows we leave above it $(0, 1, 2, \dots \text{ or } y-b)$
So, in total, we have this area
$$
\sum_{a=1}^x \sum_{b=1}^y (x - a + 1) \times (y - b + 1) \times ab = \\
\sum_{a=1}^x \Big( a (x - a + 1) \sum_{b=1}^y b (y - b + 1)\Big) = \\
\big(\sum_{a=1}^x a (x - a + 1)\big) \big( \sum_{b=1}^y b (y - b + 1)\big)
$$
We can simplify this further by examining the sequence $A_n = \sum_{a=1}^n a (n - a + 1)$.
If we examine the first few elements of this sequence $(1, 4, 10, 20, 35, 56)$, we can notice that the sequence of differences $(1, 3, 6, 10, 15, 21)$ is exactly the sequence of triangular numbers. If that holds true in general, $A_n$ is the sum of the first $n$ triangular numbers, i.e. $A_n = \frac{n(n+1)(n+2)}{6}$ as seen here.
Here's a proof that this indeed holds in general:
$$
\begin{align}
A_n - A_{n-1} = 1 \times n &+ 2 \times (n-1) + 3 \times (n-2) + \dots + n \times 1 \\
&- 1 \times (n-1) - 2 \times (n-2) - 3 \times (n-3) - \dots = (n-1) \times 1 = \\
&= n + (2 - 1) \times (n - 1) + (3 - 2) \times (n - 2) + \dots + (n - 2 - n + 1) \times 1 \\
&=\sum_{i=1}^n i = \frac{n(n+1)}{2},
\end{align}
$$
which is indeed the $n^{\text{th}}$ triangle number.
Thus, the answer to the original question is
$$\frac{x(x+1)(x+2)}{6} \times \frac{y(y+1)(y+2)}{6}$$
and, in the case $x=y=2$ this is indeed 4.
A: It is useful to move from simpler cases to more complex ones!
Consider $1\times y$ rectangle. The sum of areas of all rectangles is:
$$\begin{align}A_{1\times y}&=1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1=\\
&=\sum_{i=1}^yi\cdot (y-i+1)=\\
&=(y+1)\sum_{i=1}^y i-\sum_{i=1}^yi^2=\\
&=(y+1)\cdot \frac{y(y+1)}{2}-\frac{y(y+1)(2y+1)}{6}=\\
&=\frac{y(y+1)[3y+3-2y-1]}{6}=\\
&=\frac{y(y+1)(y+2)}{6}.\end{align}$$
Consider $2\times y$ rectangle. The sum of areas of all rectangles is:
$$\begin{align}A_{2\times y}&=\color{red}{2}\times [1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1]+\\
&+\color{red}{1}\times [2\times y+4\times (y-1)+6\times (y-3)+\cdots +2y\times 1]=\\
&=\color{red}{2}\times [A_{1\times y}]+\color{red}{1}\times 2\times [A_{1\times y}]\\
&=(\color{red}2+\color{red}1\times 2)\times [A_{1\times y}]\\
&=4\cdot \frac{y(y+1)(y+2)}{6}.\end{align}$$
Consider $3\times y$ rectangle. The sum of areas of all rectangles is:
$$\begin{align}A_{3\times y}&=\color{red}{3}\times [1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1]+\\
&+\color{red}{2}\times [2\times y+4\times (y-1)+6\times (y-3)+\cdots +2y\times 1]+\\
&+\color{red}{1}\times [3\times y+6\times (y-1)+9\times (y-3)+\cdots +3y\times 1]=\\
&=(\color{red}{3}+\color{red}2\times 2+\color{red}1\times 3)\times [A_{1\times y}]=\\
&=10\cdot \frac{y(y+1)(y+2)}{6}.\end{align}$$
And now consider $x\times y$ rectangle. The sum of areas of all rectangles is:
$$\begin{align}A_{x\times y}&=\ \ \ \ \ \ \ \ \ \ \color{red}{x}\times [1\times y+2\times (y-1)+3\times (y-3)+\cdots +y\times 1]+\\
&+\color{red}{(x-1)}\times [2\times y+4\times (y-1)+6\times (y-3)+\cdots +2y\times 1]+\\
&+\color{red}{(x-2)}\times [3\times y+6\times (y-1)+9\times (y-3)+\cdots +3y\times 1]=\\
& \ \ \vdots\\
&+\ \ \ \ \ \ \ \ \ \ \ \color{red}{1}\times [x\times y+2x\times (y-1)+3x\times (y-3)+\cdots +xy\times 1]=\\
&=(\color{red}{x}+\color{red}{(x-1)}\times 2+\color{red}{(x-2)}\times 3+\cdots +\color{red}1\times x)\times [A_{1\times y}]=\\
&=[A_{x\times 1}]\cdot [A_{1\times y}]=\\
&=\frac{x(x+1)(x+2)}{6}\cdot \frac{y(y+1)(y+2)}{6},\end{align}$$
which complies with the brilliant answer given by Todor Markov (+1).
