Example 4. How many $n$-digit numbers can be formed using $0,1,2, 3, 4$ such that exactly one digit is 1.
Let it be called “qualified number”, and the total number of all “qualified numbers” is denoted as $x_{\mathrm{n}}$.

The number of “qualified numbers” starting from 0, 1, 2, 3, 4 are denoted as $y_{\mathrm{n}}, z_{\mathrm{n}}, u_{\mathrm{n}}, v_{\mathrm{n}}, w_{\mathrm{n}}$ respectively.

By symmetry, 1 and 3 are both odd numbers, and they have adjacent numbers on both sides, so it must be that $z_{\mathrm{n}}=v_{\mathrm{n}}$.
0 and 4 are both even numbers, and they have adjacent numbers on only one side, so it must be that $y_{\mathrm{n}}=w_{\mathrm{n}}$.
Thus, it can be simplified to $y_{\mathrm{n}}, z_{\mathrm{n}}, u_{\mathrm{n}}, z_{\mathrm{n}}, y_{\mathrm{n}}$.
Therefore, we have $x_{\mathrm{n}}=2 y_{\mathrm{n}}+2 z_{\mathrm{n}}+u_{\mathrm{n}}$. In the form of $\times \times \cdots \times$, thus $y_{\mathrm{n}}=z_{\mathrm{n}-1}$.

Numbers starting with 1 must have the form $10 \times \times \cdots$ $\times$ and $12 \times \times \cdots \times$, thus
$$
z_{\mathrm{n}}=y_{\mathrm{n}-1}+u_{\mathrm{n}-1} \text {. }
$$

Numbers starting with 2 must have the form $21 \times \times \cdots \times$ and $23 \times \times \cdots \times$, thus
$$
u_{\mathrm{n}}=z_{\mathrm{n}-1}+z_{\mathrm{n}-1}=2 z_{\mathrm{n}-1} \text {. }
$$

From (2), (3), and (4), we get
$$
\begin{aligned}
z_{\mathrm{u}} & =y_{\mathrm{n}-1}+u_{\mathrm{n}-1}=z_{\mathrm{n}-2}+2 z_{\mathrm{n}-2} \\
& =3 z_{\mathrm{n}-2} .
\end{aligned}
$$

It is known that $z_{1}=1, z_{2}=2$ (i.e., $10, 12$).
Thus, from (5), we get
$$
z_{2 \mathrm{n}}=2 \cdot 3^{\mathrm{n}-1}, z_{2 \mathrm{n}+1}=3^{\mathrm{n}}
$$

From (2), $y_{2 n}=z_{20-1}=3^{n-1}$,
$$
y_{2 n}=z_{2 n-1}=3^{n-1} \text {. }
$$

From (4), $u_{\mathrm{in}}=2 z_{2 \mathrm{n}-1}=2 \cdot 3^{n-1}$,
$$
u_{2 \mathrm{n}+1}=2 z_{2 \mathrm{n}}=4 \cdot 3^{n-1} \text {. }
$$

Therefore, we have
$$
\begin{aligned}
x_{2 \mathrm{n}}= & 2 y_{2 \mathrm{n}}+2 z_{2 \mathrm{n}}+u_{2 \mathrm{n}} \\
& =2 \cdot 3^{n-1}+4 \cdot 3^{n-1}+2 \cdot 3^{n-1} \\
& =8 \cdot 3^{n-1} \cdot(n \geqslant 1)
\end{aligned}
$$

Similarly, $x_{2 n+1}=14 \cdot 3^{n-1}(n \geqslant 1)$
$$
H x_{1}=5 \text {. }
$$