标签：span 斐波那契 斐波那契数 sqrt inline sum 斐波那契数列 math n+1
\(F_n=F_{n-1}+F_{n-2}\)
\(\frac{1}{-k}=\frac{1-k}{1}\)
\(k=\frac{1+\sqrt{5}}{2}\)
\(F_n-\frac{1+\sqrt{5}}{2}F_{n-1}=\frac{1-\sqrt{5}}{2}(F_{n-1}-\frac{1+\sqrt{5}}{2}F_{n-2})\)
\(T_n=F_n-\frac{1+\sqrt{5}}{2}F_{n-1}=\frac{1-\sqrt{5}}{2}T_{n-1}\)
\(T_1=1,T_n=(\frac{1-\sqrt{5}}{2})^{n-1}\)
\(F_{n+1}=F_n+(\frac{1-\sqrt{5}}{2})^n\)
\(F_{n+1}=\sum_{i=0}^n(\frac{1-\sqrt{5}}{2})^i(\frac{1+\sqrt{5}}{2})^{n-i}\)
\(F_{n+1}=\sum_{i=0}^n (-1)^i(\frac{1+\sqrt{5}}{2})^{n-2i}\)
\(q=-(\frac{1-\sqrt{5}}{2})^2=\frac{\sqrt{5}-3}{2}\)
\(F_{n+1}=\frac{2}{5-\sqrt{5}}[(\frac{1-\sqrt{5}}{2})^{n+2}+(\frac{1+\sqrt{5}}{2})^n]\)
\(F_{n+1}=\frac{\sqrt{5}}{5}[(\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1}]\)
\(F_n=\frac{\sqrt{5}}{5}[(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n]\)
标签：span 斐波那契 斐波那契数 sqrt inline sum 斐波那契数列 math n+1
原文地址：https://www.cnblogs.com/mhrlovezmy/p/9865160.html
