斐波那契数列
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等比数列
有斐波那契数列
\begin{aligned}
\left \{
\begin{matrix}
F_1=1\\
F_2=1\\
F_i=F_{i-1}+F_{i-2}
\end{matrix}
\right .
\end{aligned}
设
\begin{aligned}
F_i+AF_{i-1}=B(F_{i-1}+AF_{i-2})
\end{aligned}
其中
\begin{aligned}
&\left \{
\begin{matrix}
AB=1\\
B-A=1
\end{matrix}
\right .\\
&\therefore A=\frac{\pm \sqrt{5}-1}{2},B=\frac{\pm \sqrt{5}+1}{2}
\end{aligned}
设
\begin{aligned}
&F'_i=F_{i+1}+AF_{i}\\
&\therefore F'_i=BF'_{i-1}\\
&\therefore F'_i=B^{i-1}F'_1=(\frac{\pm \sqrt{5}+1}{2})^i\\
&\therefore
\begin{aligned}
F_{i+1}&=F'_i-AF_i\\
&=F'_i-A(F'_{i-1}-AF_{i-1})\\
&=F'_i-A(F'_{i-1}-A(F'_{i-2}-AF_{i-2}))\\
&=\sum_{j=0}^{i-1}(-A)^jF'_{i-j}+(-A)^{i}\\
&=\sum_{j=0}^{i}(-A)^jB^{i-j}\\
&=(-A)^i\sum_{j=0}^{i}(-\frac{B}{A})^{j}\\
&=(-A)^i\frac{(-\frac{B}{A})^{i+1}-1}{-\frac{B}{A}-1}\\
&=\frac{B^{i+1}-(-A)^{i+1}}{A+B}
\end{aligned}
\end{aligned}
此时可以将两个解任意一个代入，求出即为通项公式