What is the sum of the $81$ products in the $9 \times 9$ multiplication grid? What is an easy way to solve this problem? I believe that the value in each box is the product of $x$ and $y$.

Suppose the 9 × 9 multiplication grid, shown here, were filled in completely. What would be the sum of the 81 products?


 A: We want to find the following:
$$\sum_{i=1}^9 \sum_{j=1}^9 ij$$
Factor out the $i$ from the first summation:
$$\sum_{i=1}^9 i\left(\sum_{j=1}^9 j\right)$$
Note that $\sum_{j=1}^9 j=45$.
$$\sum_{i=1}^9 i\cdot 45$$
Factor out the $45$:
$$45\left(\sum_{i=1}^9 i\right)$$
Note that $\sum_{i=1}^9 i=45$.
$$45\cdot 45=2025$$
A: The sum of the products in the top row would just be $(1+2+3+4+5+6+7+8+9)=45$
Then the next row would be $(2+4+6+8+10+12+14+16+18) = 2\times45 = 90$
So the top two rows sum to $(1+2)\times 45 = 135$
Then it becomes obvious that the full sum of the products is the product of the sums, ie. $45\times45 = 2025$
A: Use the distributive principle.  The sum of the entries in the $2$ column is $2$ times the sum of the numbers $1$ through $9$, so the sum of all the entries is the sum of the numbers $1$ through $9$ times (what?)
A: The sum is essentially $$\sum_{a=1}^9 \sum_{b=1}^9 ab =\sum_{a=1}^9 a \sum_{b=1} ^9 b=\sum_{a=1}^9 a \frac{9\cdot 10}{2}=\left(\frac{9\cdot 10}{2}\right)^2$$
A: I will focus on an intuition for finding the result. Once you have the formula, other answers are perfect in providing the means for proving it technically.
Let us start with a small size version. If you take the top $3\times 3$ table only with numbers $1$, $2$, $3$ (to save some electrons), the sum of products is:
$$ s= 1\times 1 +1\times 2 + 1\times 3 + 2\times 1 +2\times 2 + 2\times 3 + 3\times 1 +3\times 2 + 3\times 3 \,. $$
You see that you can rewrite this sum of nine products as:
$$s = (1+2+3)\times (1+2+3)\,,$$
using the distribution of multiplication over additions.
This form is more interesting, because you can see a pattern. The sum of the first integers, multiplied by itself. 
It can be reassuring to verify it works: you get  $36$, which you can check by hand ($6+12+18$). You can test it with the  $2\times2$ or $4\times4$ matrix, to verify it is not a coincidence.
The hint is apparent in the squared shape of the table. The same works for bigger tables too. Each product of $i$ and $j$ in this order is contained once and only once in the product $(1,\ldots,i,\ldots,n)\times(1,\ldots,j,\ldots,n)$.
Now you have to find the sum of integers. If you forget the generic expression for the  sum of the $n$ first integer, have this   figure in mind:

You pack together two triangles which give you a $n\times (n+1)$ rectangle, of area $n(n+1)$, the double of the area of each triangle. 
So finally the answer is  $\left(n(n+1)/2\right)^2$. With $n=9$, you get $2025$.
Once you have the method, which is relatively simple, you can easily turn it into a more formal proof, via induction for instance, as given in other answers.
A: Hint $\quad \begin{eqnarray} &\color{#c00}{1+2+3}\\ +\ &\color{#0a0}{2+4+6}\\ +\ &\color{blue}{3+6+9}\end{eqnarray}$ $\quad  =\quad \begin{eqnarray} &\color{#c00}1\,(1+2+3)\\ +\ &\color{#0a0}2\,(1+2+3)\\ +\ &\color{blue}3\,(1+2+3)\end{eqnarray}$ $\quad =\quad  (\color{#c00}1 + \color{#0a0}2 + \color{blue} 3)(1+2+3)\  \ =\ \  6\times 6$

A: Let's prove by induction that the sum of an ${n}\times{n}$ grid is $\frac{n^4+2n^3+n^2}{4}$:
First, show that this is true for $n=1$:
$\sum\limits_{x=1}^{1}\sum\limits_{y=1}^{1}xy=\frac{1^4+2\cdot1^3+1^2}{4}$
Second, assume that this is true for $n$:
$\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy=\frac{n^4+2n^3+n^2}{4}$
Third, prove that this is true for $n+1$:
$\sum\limits_{x=1}^{n+1}\sum\limits_{y=1}^{n+1}xy=$
$\color\red{\sum\limits_{x=1}^{n}\sum\limits_{y=1}^{n}xy}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$
$\color\red{\frac{n^4+2n^3+n^2}{4}}+\left(\sum\limits_{x=1}^{n}x(n+1)\right)+\left(\sum\limits_{y=1}^{n}y(n+1)\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\sum\limits_{x=1}^{n}x\right)+(n+1)\left(\sum\limits_{y=1}^{n}y\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)\left(\frac{n^2+n}{2}\right)+(n+1)(n+1)=$
$\frac{n^4+2n^3+n^2}{4}+\frac{n^3+2n^2+n}{2}+\frac{n^3+2n^2+n}{2}+n^2+2n+1=$
$\frac{n^4+2n^3+n^2}{4}+n^3+2n^2+n+n^2+2n+1=$
$\frac{n^4+2n^3+n^2}{4}+n^3+3n^2+3n+1=$
$\frac{n^4+6n^3+13n^2+12n+4}{4}=$
$\frac{(n+1)^4+2(n+1)^3+(n+1)^2}{4}$
Please note that the assumption is used only in the part marked red.

Therefore, the sum of a ${9}\times{9}$ grid is $\frac{9^4+2\cdot9^3+9^2}{4}=2025$.
