Example 3. Given the sequence $\left\{a_{\mathbf{n}}\right\}$, where $a_{1}=1$, and it satisfies $a_{n+1}=1+2 a_{n}$. Find the general term $a_{n}$.
$$
\begin{array}{c}
\text { Sol } \because a_{n+1}=1+2 a_{n}, \\
\therefore a_{n}=1+2 a_{n-1} \text {. Subtracting the two equations gives } \\
a_{n+1}-a_{n}=2\left(a_{n}-a_{n-1}\right), \text { and } a_{1}=1, \\
a_{2}=1+2 \times 1=3.
\end{array}
$$

From the relation $a_{n+1}-a_{n}=2\left(a_{n}-a_{n-1}\right)$, we can deduce:
$$
\begin{array}{l}
a_{2}-a_{1}=2, \\
a_{3}-a_{2}=2\left(a_{2}-a_{1}\right)=2 \times 2=2^{2}, \\
a_{4}-a_{3}=2\left(a_{3}-a_{2}\right)=2 \times 2^{2}=2^{3}, \\
\cdots \cdots \\
a_{n}-a_{n-1}=2^{n-1}.
\end{array}
$$

Adding these $n-1$ equations, we get $a_{n}-a_{1}$.
$$
\begin{array}{l}
=2+2^{2}+2^{3}+\cdots+2^{n-1}=2 \times \frac{2^{n-1}-1}{2-1} \\
=2^{n}-2. \text { Therefore, } a_{n}=a_{1}+2^{n}-2=2^{n}-1
\end{array}
$$