Example 7. Given: $a_{1}=1, a_{n+1}=3+\frac{1}{2} a_{n}$ $(n \geqslant 1)$, find the general term of the sequence $a_{n}$.
Solution 1 $\because a_{1}=1, a_{n+1}=3+\frac{1}{2} a_{n}$, $(n \geqslant 1)$ then
$$
\begin{array}{l} 
a_{1}=1, a_{2}=\frac{1}{2}+3, \\
a_{3}=\frac{1}{2^{\frac{2}{2}}}+3 \times \frac{1}{2}+3, \\
a_{4}=\frac{1}{2^{3}}+3 \times \frac{1}{2^{2}}+3 \times \frac{1}{2}+3, \\
\cdots \cdots \\
a_{n}=\frac{1}{2^{n-1}}+3 \times \frac{1}{2^{n-2}}+3 \times \frac{1}{2^{n-3}}+\cdots \\
+3 \times \frac{1}{2}+3 \\
=\frac{1}{2^{n-1}}+3 \times\left(\frac{1}{2^{n-2}}+\frac{1}{2^{n-3}}+\cdots+\right. \\
\frac{1}{2}+1\left.=\frac{1}{2^{n-1}}+\frac{1}{3\left(1-\frac{1}{2}\right.}\right) \\
=6-\frac{5}{2^{n-1}} \cdot
\end{array}
$$
(Proof omitted)

Solution 2 $\because a_{n+1}=3+\frac{1}{2} a_{n}$
$$
\begin{array}{l}
-) \quad a_{n}=3+\frac{1}{2} a_{n-1} \\
a_{n+1}-a_{n}=\frac{1}{2}\left(a_{n}-a_{n-1}\right) . \\
(n \geqslant 2)
\end{array}
$$

Let $a_{n+1}-a_{n}=b_{n}$, then we get a new sequence $\{b_{n}\}$, where $b_{1}=a_{2}-a_{1}=\left(3+\frac{1}{2} a_{1}\right)-a_{1}$
$$
=\left(3+\frac{1}{2}\right)-1=\frac{5}{2} \text {, }
$$

and $b_{n}=\frac{1}{2} b_{n-1}$, i.e., $\frac{b_{n}}{b_{n-1}}=\frac{1}{2}$.
$\therefore$ The sequence $\{b_{n}\}$ is a geometric sequence with the first term $b_{1}=\frac{5}{2}$.
$$
\begin{aligned}
& \text { Also } b_{n-1}+b_{n-2}+\cdots+b_{2}+b_{1} \\
& =a_{n}-a_{n-1}+a_{n-1}-a_{n-2}+\cdots+a_{3} \\
-a_{2} & +a_{2}-a_{1} \\
& =a_{n}-a_{1} \text {. }
\end{aligned}
$$

That is, $a_{n}-a_{1}=b_{1}+b_{2}+\cdots+b_{n-}+{ }_{2} b_{n-1}$
$$
\begin{aligned}
& =\frac{\frac{5}{2}\left(1-\frac{1}{2^{n-1}}\right)}{1-\frac{1}{2}}=5-\frac{5}{2^{n^{-1}}} . \\
\therefore \quad a_{n} & =6-\frac{5}{2^{n-1}} .
\end{aligned}
$$