Example $1^{\bullet}$. Consider the sequence $\left\{a_{n}\right\}$ defined inductively by $a_{1}=1, a_{n+1}=\frac{1}{2} a_{n}+1$ $(n=1,2,3, \cdots)$. Try to find the general term of the sequence $\left\{a_{n}\right\}$.
Solution: Given $a_{n+1}=\frac{1}{2} a_{4}+1$, we have
$$
a_{n}=\frac{1}{2} a_{n-1}+1,
$$

Thus, $a_{n+1}-a_{n}=\left(\frac{1}{2} a_{n}+1\right)$
$$
\begin{array}{l}
-\left(\frac{1}{2} a_{n-1}+1\right) \\
=\frac{1}{2}\left(a_{n}-a_{n-1}\right)
\end{array}
$$

Let $a_{n+1}-a_{n}=b_{n} ， \quad\left(b_{n}\right.$ be the $n$-th term of the difference sequence)
$$
\begin{array}{c}
\text { Then } b_{n}=\frac{1}{2} b_{n-1}, \\
b_{1}=a_{2}-a_{1}=\left(\frac{1}{2} \cdot 1+1\right)-1=\frac{1}{2} .
\end{array}
$$

When $n \geqslant 2$, we have
$$
\begin{aligned}
a_{n} & =1+\sum_{k^{-1}}^{n-1}\left(\frac{1}{2}\right)^{k} \\
& =2-\left(\frac{1}{2}\right)^{n-1} .
\end{aligned}
$$

The above formula also holds when $n \geqslant 1$.
Therefore, $a_{n}=2-\left(\frac{1}{2}\right)^{n-1}$.
Alternative solution: Let $k=\frac{1}{2} k i 1$, we get $k=2$.
Assume $a_{n}-2=c_{n}$, substitute into $a_{n+1}=\frac{1}{2} a_{n}+1$, then we have $c_{n+1}+2 \frac{1}{2}\left(c_{n}+2\right)+1$, hence $c_{n+1}=\frac{1}{2} c_{n}$.
$$
\begin{aligned}
\because c_{1} & =a_{1}-2=1-2=-1, \\
\therefore c_{n} & =-1 \cdot\left(\frac{1}{2}\right)^{n-1} .
\end{aligned}
$$

Thus, $\quad a_{n}=-\left(\frac{1}{2}\right)^{n-1}+2$.