Figure 0 consists of a square with side length 18 . For each integer $n \geq 0$, Figure $n+1$ consists of Figure $n$ with the addition of two new squares constructed on each of the squares that were added in Figure $n$. The side length of the squares added in Figure $n+1$ is $\frac{2}{3}$ of the side length of the smallest square(s) in Figure $n$. Define $A_{n}$ to be the area of Figure $n$ for each integer $n \geq 0$. What is the smallest positive integer $M$ with the property that $A_{n}<M$ for all integers $n \geq 0$ ?

![](https://cdn.mathpix.com/cropped/2024_04_30_d1e45f8438f865d01c92g-3.jpg?height=201&width=184&top_left_y=1179&top_left_x=407)

Figure 0

![](https://cdn.mathpix.com/cropped/2024_04_30_d1e45f8438f865d01c92g-3.jpg?height=388&width=385&top_left_y=993&top_left_x=732)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_04_30_d1e45f8438f865d01c92g-3.jpg?height=488&width=610&top_left_y=894&top_left_x=1232)

Figure 2
Figure 0 consists of 1 square with side length 18. Thus, $A_{0}=18^{2}$.

Figure 1 consists of Figure 0 plus the addition of 2 squares with side length $\frac{2}{3} \cdot 18$.

Thus, $A_{1}=18^{2}+2\left(\frac{2}{3} \cdot 18\right)^{2}$.

Figure 2 consists of Figure 1 plus the addition of 4 squares with side length $\frac{2}{3} \cdot \frac{2}{3} \cdot 18$.

Thus, $A_{2}=18^{2}+2\left(\frac{2}{3} \cdot 18\right)^{2}+4\left(\frac{2}{3} \cdot \frac{2}{3} \cdot 18\right)^{2}$.

In general, since twice as many squares are added at each step as at the step before, then $2^{n}$ squares are added to Figure $n-1$ to make Figure $n$.

Also, since the side length of the squares added at each step is $\frac{2}{3}$ of the side length of the squares added at the previous step, then the squares added for Figure $n$ have side length $18\left(\frac{2}{3}\right)^{n}$.

Therefore,

$$
\begin{aligned}
A_{n} & =18^{2}+2\left(\frac{2}{3} \cdot 18\right)^{2}+4\left(\frac{2}{3} \cdot \frac{2}{3} \cdot 18\right)^{2}+\cdots+2^{n}\left(18\left(\frac{2}{3}\right)^{n}\right)^{2} \\
& =18^{2}+2^{1} \cdot 18^{2}\left(\frac{2}{3}\right)^{2}+2^{2} \cdot 18^{2}\left(\frac{2}{3}\right)^{4}+\cdots+2^{n} \cdot 18^{2}\left(\frac{2}{3}\right)^{2 n}
\end{aligned}
$$

The right side is a geometric series with $n+1$ terms, first term $a=18^{2}$, and common ratio $r=2\left(\frac{2}{3}\right)^{2}=\frac{8}{9}$.

Therefore,

$$
A_{n}=\frac{18^{2}\left(1-\left(\frac{8}{9}\right)^{n+1}\right)}{1-\frac{8}{9}}=9 \cdot 18^{2}\left(1-\left(\frac{8}{9}\right)^{n+1}\right)=2916\left(1-\left(\frac{8}{9}\right)^{n+1}\right)
$$

Since $\left(\frac{8}{9}\right)^{n+1}>0$ for all integers $n \geq 0$, then $1-\left(\frac{8}{9}\right)^{n+1}2915$.

Therefore, the smallest positive integers $M$ with the property that $A_{n}<M$ for all integers $n \geq 0$ is $M=2916$.

ANSWER: 2916