Example 6 (2006 Hope Cup Training Question) Given the function $f(x)=\frac{1}{2^{-1}}$, the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is $S_{n}=6-$ $2 a_{n}+f(n-1)$, try to find the general term formula $a_{n}$ of the sequence $\left\{a_{n}\right\}$.
Solution: Since $f(x)=\frac{1}{2^{x-1}}$, then $f(n)=\frac{1}{2^{n-1}}, f(n-1)=\frac{1}{2^{n-2}}$.
Also, $S_{n}=6-2 a_{n}+f(n-1)=6-2 a_{n}+\frac{1}{2^{n-2}}$.
Therefore, $a_{1}=6-2 a_{1}+\frac{1}{2^{-1}}=8-2 a_{1}$, solving this gives $a_{1}=\frac{8}{3}$.
When $n \geqslant 2$, $a_{n}=S_{n}-S_{n-1}=\left(6-2 a_{n}+\frac{1}{2^{n-2}}\right)-\left(6-2 a_{n-1}+\frac{1}{2^{n-3}}\right)=-2 a_{n}+2 a$ $+\frac{4}{2^{n}}-\frac{8}{2^{n}}$.
Thus, $a_{n}=\frac{2}{3} a_{n-1}-\frac{2}{3} \times \frac{1}{2^{n-1}}$, which means $a_{n}-4 \times \frac{1}{2^{n}}=\frac{2}{3}\left(a_{n-1}-4 \times \frac{1}{2^{n-1}}\right)$.
Therefore, the sequence $\left\{a_{n}-4 \times \frac{1}{2^{n}}\right\}$ is a geometric sequence with the first term $\left(a_{1}-4 \times \frac{1}{2}\right)$ and common ratio $\frac{2}{3}$.
So, $a_{n}-4 \times \frac{1}{2^{n}}=\left(a_{1}-4 \times \frac{1}{2}\right) \cdot\left(\frac{2}{3}\right)^{n-1}$, which means $a_{n}=\left(\frac{2}{3}\right)^{n}+\frac{4}{2^{n}}(n \geqslant 2)(\cdot)$.
When $n=1$, $a_{1}=\frac{8}{3}=\frac{2}{3}+\frac{4}{2}$, which means when $n=1$, $a_{1}$ also satisfies (*). Hence, the general term formula for $\left\{a_{n}\right\}$ is $a_{n}=\left(\frac{2}{3}\right)^{n}+\frac{4}{2^{n}}\left(n \in \mathbf{N}^{*}\right)$.
Comment: When finding the general term of a sequence, it is important to verify whether $a_{1}$ satisfies the derived general term formula.