1. As shown in the figure, there is a sequence of curves $P_{0}, P_{1}, P_{2}, \cdots$. It is known that the area of the equilateral triangle enclosed by $P_{0}$ is 1, and $P_{k+1}$ is obtained from $P_{k}$ by the following operation: divide each side of $P_{k}$ into three equal parts, construct an equilateral triangle outward using the middle segment of each side, and then remove the middle segment $(k=0,1,2, \cdots)$. Let $S_{n}$ be the area of the figure enclosed by the curve $P_{n}$.
Find the general term formula for the sequence $\left\{S_{n}\right\}$.
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1. For $P_{0}$, it is easy to see that each edge of $P_{0}$ becomes 4 edges of $P_{1}$, so the number of edges of $P_{1}$ is $3 \cdot 4$; similarly, for $P_{1}$, each edge of $P_{1}$ becomes 4 edges of $P_{2}$, so the number of edges of $P_{2}$ is $3 \cdot 4^{2}$, thus it is not difficult to get that the number of edges of $P_{n}$ is $3 \cdot 4^{n}$.

Given that the area of $P_{0}$ is $S_{0}=1$, comparing $P_{1}$ with $P_{0}$. It is easy to see that $P_{1}$ adds a small equilateral triangle on each edge of $P_{0}$, with an area of $\frac{1}{3^{2}}$, and $P_{0}$ has 3 edges, so $S_{1}=S_{0}+3 \cdot \frac{1}{3^{2}}=1+\frac{1}{3}$, then comparing $P_{2}$ with $P_{1}$, it is known that $P_{2}$ adds a small equilateral triangle on each edge of $P_{1}$, with an area of $\frac{1}{3^{2}} \cdot \frac{1}{3^{2}}$, and $P_{1}$ has $3 \cdot 4$ edges, so $S_{2}=S_{1}+3 \cdot 4 \cdot \frac{1}{3^{4}}=1+\frac{1}{3}+\frac{4}{3^{3}}$
Similarly, we have: $S_{3}=S_{2}+3 \cdot 4^{2} \cdot \frac{1}{3^{6}}=1+\frac{1}{3}+\frac{4}{3^{3}}+\frac{4^{2}}{3^{5}}$
Thus, $S_{n}=1+\frac{1}{3}+\frac{4}{3^{3}}+\frac{4^{2}}{3^{5}}+\cdots+\frac{4^{n-1}}{3^{2 n-1}}=1+\sum_{k=1}^{n} \frac{4^{k-1}}{3^{2 k-1}}=1+\frac{3}{4} \sum_{k=1}^{n}\left(\frac{4}{9}\right)^{k}$
$$
\begin{array}{l}
=1+\frac{3}{4} \cdot \frac{\frac{4}{9}\left[1-\left(\frac{4}{9}\right)^{n}\right]}{1-\frac{4}{9}} \\
=1+\frac{3}{5}\left[1-\left(\frac{4}{9}\right)^{n}\right]=\frac{8}{5}-\frac{3}{5} \cdot\left(\frac{4}{9}\right)^{n}
\end{array}
$$

Now, we use mathematical induction to prove (1). For $n=1$, it is known from the above that (1) holds.
Assume that for $n=k$, we have $S_{k}=\frac{8}{5}-\frac{3}{5} \cdot\left(\frac{4}{9}\right)^{k}$.
For $n=k+1$, it is easy to see that after the $(k+1)$-th operation, comparing $P_{k+1}$ with $P_{k}$, $P_{k+1}$ adds a small equilateral triangle on each edge of $P_{k}$, with an area of $\frac{1}{3^{2(k+1)}}$, and $P_{k}$ has $3 \cdot 4^{k}$ edges, so
$S_{k+1}=S_{k}+3 \cdot 4^{k} \cdot \frac{1}{3^{2(k+1)}}=S_{k}+\frac{4^{k}}{3^{2 k+1}}=\frac{8}{5}-\frac{3}{5} \cdot\left(\frac{4}{9}\right)^{k+1}$. In summary, by mathematical induction, (1) is proved.