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?

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. $45\times45 = 2025$

The sum is essentially $$\sum_{a=1}^9 \sum_{b=1}^9 ab =\sum_{a=1}^9 a \sum_{b=1} ^9 b=\sum_{a=1}^9 a \frac{9\cdot 10}{2}=\left(\frac{9\cdot 10}{2}\right)^2$$

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 $\sum_{i=1}^9 i=45$. $$45\cdot 45=2025$$

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?)

Hint $\quad \begin{eqnarray} &\color{#c00}{1+2+3}\\ +\ &\color{#0a0}{2+4+6}\\ +\ &\color{blue}{3+6+9}\end{eqnarray}$ $\quad =\quad \begin{eqnarray} &\color{#c00}1\,(1+2+3)\\ +\ &\color{#0a0}2\,(1+2+3)\\ +\ &\color{blue}3\,(1+2+3)\end{eqnarray}$ $\quad =\quad (\color{#c00}1 + \color{#0a0}2 + \color{blue} 3)(1+2+3)\ \ =\ \ 6\times 6$