Example 2-37 Find the value of the following $n$-order determinant:
$$
d_{n}=\left|\begin{array}{llllll}
2 & 1 & 0 & 0 & \cdots & 0 \\
1 & 2 & 1 & 0 & \cdots & 0 \\
0 & 1 & 2 & 1 & \cdots & 0 \\
& & & \vdots & & \\
0 & 0 & 0 & 0 & \cdots & 2
\end{array}\right|
$$

According to the properties of determinants
$$
d_{n}-2 d_{n-1}+d_{n-2}=0, d_{1}=2, d_{2}=3
$$

The corresponding characteristic equation is
$$
\begin{array}{l}
x^{2}-2 x+1=(x-1)^{2}=0 \\
x_{1}=x_{2}=1
\end{array}
$$

Thus, $x=1$ is a double root.
Therefore,
$$
d_{n}=(A+B n)(1)^{n}=A+B n
$$

When $n=1$, we have $d_{1}=A+B=2$,
When $n=2$, we have $d_{2}=A+2 B=3$.

Hence
$$
\left\{\begin{array}{l}
A+B=2 \\
A+2 B=3
\end{array}, \quad A=B=1\right.
$$
which means
$$
d_{n}=n+1
$$