What is the sum of the $81$ products in the $9 \times 9$ multiplication grid?

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