China Mathematical Competition

(1) We perform the operation on $P_0$. It is easy to see that each side of $P_0$ becomes $4$ sides of $P_1$. So the number of sides of $P_1$ is $3\cdot 4$. In the same way, we operate on $P_1$. Each side of $P_1$ becomes $4$ sides of $P_2$. So the number of sides of $P_2$ is $3\cdot 4^2$. Consequently, it is not difficult to get that the number of sides of $P_n$ is $3\cdot 4^n$.

It is known that the area of $P_0$ is $S_0=1$. Comparing $P_1$ with $P_0$, it is easy to see that we add to $P_1$ a smaller equilateral triangle with area $\frac{1}{3^2}$ on each side of $P_0$. Since $P_0$ has $3$ sides, so $$S_1=S_0+3\cdot \frac{1}{3^2}=1+\frac{1}{3}.$$ Again, comparing $P_2$ with $P_1$, we see that $P_2$ has an additional smaller equilateral triangle with area $\frac{1}{3^2}\cdot \frac{1}{3^2}$ on each side of $P_1$, and $P_1$ has $3\cdot 4$ sides. So that $$S_2=S_1+3\cdot 4\cdot \frac{1}{3^4}=1+\frac{1}{3}+\frac{4}{3^3}.$$ Similarly, we have $$S_3=S_2+3\cdot 4\cdot \frac{1}{3^6}=1+\frac{1}{3}+\frac{4}{3^3}+\frac{4^2}{3^5}.$$ Hence, we have $$\begin{array}{rl}{S}_n& =1+\frac{1}{3}+\frac{4}{3^3}+\frac{4^2}{3^5}+\cdots +\frac{4^{n-1}}{3^{2n-1}}\\ & =1+\sum _{k=1}^n\frac{4^{k-1}}{3^{2k-1}}=1+\frac{3}{4}\sum _{k=1}^n{\left(\frac{4}{9}\right)}^k\\ & =1+\frac{3}{4}\cdot \frac{4}{9}\cdot \left[1-{\left(\frac{4}{9}\right)}^n\right]\\ & =1+\frac{3}{5}\left[1-{\left(\frac{4}{9}\right)}^n\right]\\ & =\frac{8}{5}-\frac{3}{5}\cdot {\left(\frac{4}{9}\right)}^n.\end{array}\text{\hspace{1em}(*)}$$

We will prove $(\ast )$ by mathematical induction as follows: When $n=1$, it is known that $(\ast )$ holds from above. Suppose, when $n=k$, we have $S_k=\frac{8}{5}-\frac{3}{5}\cdot {\left(\frac{4}{9}\right)}^k$. When $n=k+1$, it is easy to see that, after $k+1$ times of operations, by comparing $P_{k+1}$ with $P_k$, we have added to $P_{k+1}$ a smaller equilateral triangle with area $\frac{1}{3^{2(k+1)}}$ on each side of $P_k$ and $P_k$ has $3\cdot 4^k$ sides. So we get $$\begin{array}{rl}{S}_{k+1}& =S_k+3\cdot 4^k\cdot \frac{1}{3^{2(k+1)}}\\ & =S_k+\frac{4^k}{3^{2k+1}}=\frac{8}{5}-\frac{3}{5}\cdot {\left(\frac{4}{9}\right)}^{k+1}.\end{array}$$ By mathematical induction, $(\ast )$ is proved.

(2) From (1), we have $S_n=\frac{8}{5}-\frac{3}{5}\cdot {\left(\frac{4}{9}\right)}^n$. Therefore, $$\underset{n\to \infty }{\lim }{S}_n=\underset{n\to \infty }{\lim }\left[\frac{8}{5}-\frac{3}{5}\cdot {\left(\frac{4}{9}\right)}^n\right]=\frac{8}{5}.$$