思路

设$f_1=A,f_2=B,f_n(n>2)=f_{n-1}+f_{n-2},$则$f_3$就是ANSWER $A+B$。

推导

设$q^n=q^{n-1}+q^{n-2},$解得$(q≠0)q_1=\frac{1+\sqrt{5}}{2},q_2=\frac{1-\sqrt{5}}{2}$。
再设$\begin{cases} f_n=aq_1^{n-1}+bq_2^{n-1} \\ f_1=A,f_2=B \end{cases}$
代入，即得$\begin{cases} a+b=A \\ a(\frac{1+\sqrt{5}}{2})+b(\frac{1-\sqrt{5}}{2})=B \end{cases}$
解二元一次方程组得$\begin{cases} a=\frac{B-A(\frac{1-\sqrt{5}}{2})}{\sqrt{5}} \\ b=\frac{A(\frac{1+\sqrt{5}}{2})-B}{\sqrt{5}} \end{cases}$
再代入得$f_n=\frac{1}{\sqrt{5}}((B-A(\frac{1-\sqrt{5}}{2}))(\frac{1+\sqrt{5}}{2})^{n-1}+(A(\frac{1+\sqrt{5}}{2})-B)(\frac{1-\sqrt{5}}{2})^{n-1})$

解答

将$n=3$代入$f_n=\frac{1}{\sqrt{5}}((B-A(\frac{1-\sqrt{5}}{2}))(\frac{1+\sqrt{5}}{2})^{n-1}+(A(\frac{1+\sqrt{5}}{2})-B)(\frac{1-\sqrt{5}}{2})^{n-1})$
得 $\begin{aligned} A+B &= f_3\\ &= \frac{1}{\sqrt{5}}((B-A(\frac{1-\sqrt{5}}{2}))(\frac{1+\sqrt{5}}{2})^{2}+(A(\frac{1+\sqrt{5}}{2})-B)(\frac{1-\sqrt{5}}{2})^{2})\\ &= \frac{1}{\sqrt{5}}((B-A(\frac{1-\sqrt{5}}{2}))(\frac{3+\sqrt{5}}{2})+(A(\frac{1+\sqrt{5}}{2})-B)(\frac{3-\sqrt{5}}{2})) \end{aligned}$

实现

只要计算$\frac{1}{\sqrt{5}}((B-A(\frac{1-\sqrt{5}}{2}))(\frac{3+\sqrt{5}}{2})+(A(\frac{1+\sqrt{5}}{2})-B)(\frac{3-\sqrt{5}}{2}))$,但怎么写？