3.2.1 * Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=\frac{1}{3} a_{n}+1, n=1,2, \cdots$. Find the general term of the sequence $\left\{a_{n}\right\}$.
Analysis 1: By substituting $n$ for $n+1$ in $a_{n+1}=\frac{1}{3} a_{n}+1$, we get $a_{n}=\frac{1}{3} a_{n-1}+1$. Therefore,
$$
a_{n+1}-a_{n}=\frac{1}{3}\left(a_{n}-a_{n-1}\right)(n \geqslant 2).
$$

By iterating the above equation, we have
$$
a_{n+1}-a_{n}=\frac{1}{3}\left(a_{n}-a_{n-1}\right)=\frac{1}{3^{2}}\left(a_{n-1}-a_{n-2}\right)=\cdots=\frac{1}{3^{n-1}}\left(a_{2}-a_{1}\right).
$$

Since $a_{2}=\frac{1}{3} a_{1}+1=\frac{4}{3}$, we have $a_{n+1}-a_{n}=\frac{1}{3^{n}}$. Combining this with $a_{n+1}=\frac{1}{3} a_{n}+1$, we get $a_{n}-\frac{3}{2}-\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}$, which also holds for $n=1$. Therefore, $a_{n}=\frac{3}{2}-\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}$.

Comment: The solution method used in this problem is the "iteration method". The basic idea is to appropriately transform the given recurrence relation so that it can continuously substitute terms with smaller subscripts for certain terms with larger subscripts, eventually establishing a certain connection between the general term and the initial term, thereby finding the general term.

Analysis 2: From $a_{n+1}=\frac{1}{3} a_{n}+1$, we can derive $a_{n+1}-\frac{3}{2}=\frac{1}{3}\left(a_{n}-\frac{3}{2}\right)$. Therefore, the sequence $\left\{a_{n}-\frac{3}{2}\right\}$ is a geometric sequence with the first term $a_{1}-\frac{3}{2}=-\frac{1}{2}$ and common ratio $\frac{1}{3}$. Hence, $a_{n} \cdots \frac{3}{2}=\left(\frac{1}{2}\right) \cdot\left(\frac{1}{3}\right)^{n-1}$, which means $a_{n}=\frac{3}{2}-\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}$.