Example 6. Given $a_{1}=1, a_{n+1}=2 a_{n}-3$, try to find the general term $a_{n}$ of the sequence.
Solution: Using method (1), from the recurrence relation $a_{n+1}=2 a_{n}-3$, we get
$$
\begin{array}{l}
a_{1}=1, \\
a_{2}=2 \times 1-3=2-3, \\
a_{3}=2^{2}-3 \times 2-3, \\
a_{4}=2^{3}-3 \times 2^{2}-3 \times 2-3, \\
a_{5}=2^{4}-3 \times 2^{3}-3 \times 2^{2}-3 \times 2-3,
\end{array}
$$

By analogy, we get
$$
\begin{array}{c}
a_{n}=2^{n-1}-3 \cdot 2^{n-2}-3 \cdot 2^{n-3}-\cdots-3 \cdot 2-3, \\
\therefore \quad a_{n}=2^{n-1}-3 \cdot 2^{n-1}+3 \\
=-2 \cdot 2^{n-1}+3=3-2^{n} .
\end{array}
$$
(This proves the necessary condition)
Conversely, when $a_{n}=3-2^{n}$, substituting into the right side of the original equation, by the recurrence relation we get the left side:
$$
\begin{array}{l}
2 a_{n}-3=2\left(3-2^{n}\right)-3=6-2^{n+1}-3 \\
\quad=3-2^{n+1}=a_{n+1} . \\
\therefore \quad a_{n}=3-2^{n} .
\end{array}
$$
(Thus proving the sufficient condition)
Using method (2), first find the difference sequence
$$
\begin{array}{l}
\dot{a}_{n+1}=2 a_{n}-3 . \\
a_{n}=2 a_{n-1}-3 .
\end{array}
$$
$$
\text { From }(A)-(B), a_{n+1}-a_{n}=2\left(a_{n}-a_{n-1}\right) \text {. }
$$

Let $\dot{b}_{n}=a_{n+1}-a_{n}$, then $\left\{b_{n}\right\}$ is the difference sequence of $\left\{a_{n}\right\}$. And $\left\{b_{11}\right\}$ is a geometric sequence with the first term $b_{1}=a_{2}-a_{1}$ $=(2 \cdot 1-3)-1=-2$, and common ratio 2. (The characteristic of this method is to derive the recurrence relation of the difference sequence from the original sequence's recurrence relation)
Therefore, when $n \geqslant 2$,
$$
\begin{aligned}
a_{n} & =a_{1}+\sum_{i=1}^{n-1} b_{i}=\frac{-2\left(2^{n-1}-1\right)}{2-1} \\
& =3-2^{n} .
\end{aligned}
$$

This formula also holds when $\pi=1$. Hence $a_{n}=3-2^{n}$.
Using method (3), let $a_{n}=c_{n}+k$, substituting into the recurrence relation, then
$$
\begin{array}{l}
a_{n+1}=c_{n+1}+k=2\left(c_{n}+k\right)-3 . \\
\therefore c_{n+1}=2 c_{n}+k-3 . \\
\text { Let } k-3=0, k=3, \text { then we have } \\
c_{n+1}=2 c_{n}, \quad c_{1}=a_{1}-k=1-3=-2 .
\end{array}
$$
$\therefore$ The sequence $\left\{c_{n}\right\}$ is a geometric sequence with the first term -2 and common ratio 2.
$$
\therefore c_{n}=-2 \cdot 2^{n-1} \text {. }
$$

Therefore, $a_{n}=-2 \cdot 2^{n-1}+3=-2^{n}+3$.
(In this method, the value of $k$ in the assumption $a_{n}-k=c_{n}$ is obtained by substituting $k$ for $a_{n+1}$ and $a_{n}$ in the original sequence's recurrence relation $a_{n+1}=2 a_{n}-3$, solving the equation, i.e., $k=2 k-3$, yielding $k=3$.)