<s>[INST] 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?


 [/INST] 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 $\frac{9^4+2\cdot9^3+9^2}{4}=2025$.
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