Example 1. Let the sequence $\left\{\mathrm{a}_{\mathrm{n}}\right\}$ satisfy $\mathrm{a}_{1}=\mathrm{a}_{2}=1$, $a_{k+2}=a_{k+1}+a_{k} \cdot(k=1,2,3, \cdots)$ Find $a_{n}$.
Solving the equation $\mathrm{x}^{2}-\mathrm{x}-1=0$, we get $x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=\frac{1-\sqrt{5}}{2}$. According to Theorem 2, the general term is $\mathrm{a}_{\mathrm{n}}=\mathrm{A}_{0}\left(\frac{1+\sqrt{5}}{2}\right)^{\mathrm{n}-1}$
$$
\begin{array}{l}
+\mathrm{B}_{0}\left(\frac{1-\sqrt{5}}{2}\right)^{\mathrm{n}-1} \cdot \\
\text { By } \mathrm{a}_{1}=\mathrm{a}_{2}=1 \text {, we get the system of equations } \\
A_{0}+\mathrm{B}_{0}=1 ; \\
\frac{1+\sqrt{5}}{2} \mathrm{~A}_{0}+\frac{1-\sqrt{5}}{2} \mathrm{~B}_{0}=1 .
\end{array}
$$

Therefore, $\mathrm{A}_{0}=\frac{1}{2 \sqrt{5}}, \mathrm{~B}_{0}=-\frac{1-\sqrt{5}}{2 \sqrt{5}} \cdot$ Thus, the general term is
$$
\begin{array}{l}
a_{n}=-\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}\right. \\
\left.-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right] .
\end{array}
$$

The sequence in Example 1 is the Fibonacci sequence.