斐波那契数列性质总结

upd.2026.1.11

Fibonacci Sequence

托更了2周。

1. Definition:

定义斐波那契数列:

\(f_{n}:f_1=f_2=1,f_i=f_{i-1}+f_{i-2}(i \geq3)\)

2.通项求法:

1.特征方程:


\(x^2=x+1\)

解得
\(x_1=\frac{1+\sqrt{5}}{2},x_2=\frac{1-\sqrt{5}}{2}\)

待定系数:
\(f_i=c_1x_1^n+c_2x_2^n\)

\[\begin{equation}\begin{cases}c_1x_1 = f_1 \\c_2x_2 = f_2\end{cases}\end{equation} \]

(1)解得通项公式:
\(f_i=\frac{1}{\sqrt5}[(\frac{1+\sqrt{5}}{2})^i-(\frac{1-\sqrt{5}}{2})^i]\)

2.生成函数:

构造生成函数\(G(x)=\sum_{i=1}f_ix^i\)

\[\begin{equation}\begin{cases}x^2G(x) = \sum f_ix^{i+2} ① \\xG(x) = \sum f_ix^{i+1}②\\G(x)=\sum f_ix^i③\end{cases}\end{equation} \]

(2)中\(③-②-①\)得:

\((1-x-x^2)G(x)=\sum f_i(x^{i+2}-x^{i+1}-x)=\sum_{i=3}x^i(f_i-f_{i-1}-f_{i-2})+x^2(f_2-f_1)+xf_1=0+x^2*0+x=0\)

整理得:

\(G(x)=\frac{-x}{x^2-x-1}\)

\(x^2-x-1=0\)两根为\(x_1,x_2\),得:

\(G(x)=\frac{-x}{(x-x_1)(x-x_2)}=\frac{x}{(1-x_1x)(1-x_2x)}\)

利用级数公式\(\frac{1}{1-x}=\sum x\)

\(G(x)=\sum x^i(\frac{1}{\sqrt5}[(\frac{1+\sqrt{5}}{2})^i-(\frac{1-\sqrt{5}}{2})^i])\),系数即为通项

3.特殊性质

性质1: \(\sum^n_1 f_{2i}=f_{2n+1}-1\)

证明:

\(\sum^n_1f_{2i}+1=f_1+f_2+f_4+...+f_{2n-2}+f_{2n}=f_3+f_4+...+f_{2n-2}+f_{2n}=f_{2n+1}\)

\(\sum^n_1f_{2i}=f_{2n+1}-1\)

性质2:\(\sum^{n-1}_1 f_{2i-1}=f_{2n}\)

证明:
\(\sum^{n-1}_1 f_{2i-1}=f_1+f_3+...+f_{2n-3}+f_{2n-1}=f_2+f_3+...+f_{2n-3}+f_{2n-1}=f_{2n}\)

性质3: \(\sum^n_1f_i^2=f_nf_{n+1}\)

证明:

\(f_nf_{n+1}=f_n(f_n+f_{n-1})=f_n^2+f_nf_{n-1}\),递归得到:

\(f_nf_{n+1}=f_n^2+f_{n-1}^2+f_{n-2}^2+...+f_2^2+f_1^2=\sum^n_1f_i^2\)

性质4: \(\frac{f_{2n}}{f_n}=f_{n-1}+f_{n+1}\)

证明:
\(f_{2n}=f_{2n-1}+f_{2n-2}=2f_{2n-2}+f_{2n-3}=3f_{2n-3}+2f_{2n-4}\),由于系数满足斐波那契递推关系,递归得到:

\(f_{2n}=f_nf_{n-1}+f_nf_{n+1}\)

整理得:

\(\frac{f_{2n}}{f_n}=f_{n-1}+f_{n+1}\)

4.相关拓展

1.Lucas Sequence:

定义如下:

\(L_n=L_{n-1}+L_{n-2}(n \geq 3),L_1=1,L_2=3\)

有通项公式:

\(L_n=(\frac{1+\sqrt5}{2})^n+(\frac{1-\sqrt5}{2})^n\)

且与Fibonacci Sequence有如下性质:

\(L_nf_n=f_{2n}\)

\(L_n=f_{n+1}+f_{n-1}\)

部分斐波那契数列有的性质Lucas数列都有

2.Fib-Lucas Sequence

不难发现斐波那契数列和卢卡斯数列区别来自前两项

我们将满足递推关系\(a_n=a_{n-1}+a_{n-2}(n \geq 3)\)的数列叫做fib-lucas 数列

记作\(a[a_1,a_2]\)

且对于所有的fib-lucas数列\({a_n},{b_n}有如下性质:\)

1.\(\{a_n-b_n\},\{a_n+b_n\}\)都为fib-lucas数列

2.\(\{a_n\}\)可由斐波那契数列的有限项之和表示

3.\(|a_i^2-a_{i-1}{i+1}|=a_1^2\)

posted @ 2026-01-11 00:57  ZzhAllen  阅读(7)  评论(0)    收藏  举报