特征方程
我们通常会需要求解形如\(f_{n + 2} = af_{n + 1} + bf_{n}\)的通项公式，其中\(f_0\)和\(f_1\)已知
我们不妨设\(f_n\)是一个等比数列，公比为\(q\)
\[\begin{aligned}
f_{n + 2} &= af_{n + 1} + bf_{n} \\
q^2f_n &= aqf_n + bf_n \\
q^2 - aq - b &= 0 \\
\end{aligned}
\]
解这个方程即可得到两根\(q = q_0，q = q_1\)
所以\(f_n = (q_0)^n = (q_1)^n\)
同时\(q_0\)和\(q_0\)的线性组合也满足通项公式，
所以可以写成\(f_n = \alpha q_0^n + \beta q_1^n\)
再利用\(f_0\)和\(f_1\)联立即可解出\(\alpha\)和\(\beta\)，代入即为\(f_n\)通项
以斐波那契数列为例：
\[\begin{aligned}
f_{n + 2} &= f_{n + 1} + f_{n} \\
q^2f_n &= qf_n + f_n \\
q^2 - q - 1 &= 0 \\
\end{aligned}
\]
解得
\[q_1 = \frac{1 + \sqrt{5}}{2},q_2 = \frac{1 - \sqrt{5}}{2}
\]
令
\[f_n = \alpha(\frac{1 + \sqrt{5}}{2})^{n} + \beta(\frac{1 - \sqrt{5}}{2})^{n}
\]
则
\[\begin{aligned}
\alpha + \beta = 0\\
\alpha\frac{1 + \sqrt{5}}{2} + \beta\frac{1 - \sqrt{5}}{2} = 1\\
\end{aligned}
\]
解得
\[\alpha = \frac{\sqrt{5}}{5},\beta = -\frac{\sqrt{5}}{5}
\]
综上
\[f_n = \frac{\sqrt{5}}{5}(\frac{1 + \sqrt{5}}{2})^{n} - \frac{\sqrt{5}}{5}(\frac{1 - \sqrt{5}}{2})^{n}
\]
