Example 2 Write the number 1 at both ends of a circle's diameter, and let this diameter divide the circle into two semicircles. Write the sum of the adjacent two numbers (i.e., $1+1=2$) at each division point. Then, divide each of the resulting four $\frac{1}{4}$ circles, and write the sum of the adjacent two numbers (i.e., write $1+2=3$ or $2+1=3$) at each division point, and so on. If this procedure is carried out $n$ times, try to write out the sum of all the numbers at the division points on the circumference.
Solution: It is easy to find that $S_{1}=2, S_{2}=6, S_{3}=18, \cdots$. Let the sum after the $k$-th operation be $S_{k}$, then when performing the $(k+1)$-th operation, each number filled in is the sum of the two adjacent numbers that appeared after the $k$-th operation, so each of these numbers is added exactly twice, i.e., the sum of the newly filled numbers is $2 S_{k}$. Therefore, we have the recurrence relation:
$$
\begin{array}{l}
\quad S_{k+1}=S_{k}+2 S_{k}=3 S_{k}, S_{1}=2 . \\
\quad \text { Therefore, } S_{n}=3 S_{n-1}=3^{2} S_{n-2}=\cdots=3^{n-1} S_{1} \\
=2 \times 3^{n-1} .
\end{array}
$$