斐波那契通项推导

题面

对于数列\(\{F_i\}\)，有\(F_1=F_2=1,F_{n+2}=F_{n+1}+F_n,n\in N^*\)

推导

初等代数解法

\[\begin{align} &我们令F_{n+1}+\lambda F_{n}=(1+\lambda)F_{n}+F_{n-1}，等式两边为一阶递推式\\ &可列\frac{1}{\lambda}=1+\lambda，易得\lambda=\frac{-1+\sqrt5}{2}\frac{-1-\sqrt5}{2}\\ &我们令\lambda=\frac{-1+\sqrt5}{2}\\ &易得F_{n+1}+\frac{-1+\sqrt5}{2}F_{n}=\frac{1+\sqrt5}{2}F_{n}+F_{n-1}\\ &令a_n=F_{n+1}+\frac{-1+\sqrt5}{2}F_{n}，则a_1=\frac{1+\sqrt5}{2}，a_n=\frac{1+\sqrt5}{2}a_{n-1}\\ & 可得a_n=(\frac{1+\sqrt5}{2})^n，F_{n+1}=\frac{1-\sqrt5}{2}F_{n}+(\frac{1+\sqrt5}{2})^n\\ &即\frac{F_{n+1}}{(\frac{1-\sqrt5}{2})^{n+1}}=\frac{F_{n}}{(\frac{1-\sqrt5}{2})^{n}}+\frac{-1-\sqrt5}2\times\Big(\frac{{-3-\sqrt5}}{2}\Big)^n\\ &令b_n=\frac{F_{n}}{(\frac{1-\sqrt5}{2})^{n}},则b_{n+1}=b_{n}+\frac{-1-\sqrt5}2\times\Big(\frac{{-3-\sqrt5}}{2}\Big)^n,b_1=\frac{-1-\sqrt5}2\\ &则b_n=b_1+\frac{-1-\sqrt5}2\sum_{i=1}^{n-1}\Big(\frac{{-3-\sqrt5}}{2}\Big)^i=\frac{-1-\sqrt5}2\sum_{i=0}^{n-1}\Big(\frac{{-3-\sqrt5}}{2}\Big)^i\\&=\frac{-1-\sqrt5}{-5-\sqrt5}\Bigg[\Big(\frac{{-3-\sqrt5}}{2}\Big)^n-1\Bigg]=\frac{\sqrt5}5\Bigg[\Big(\frac{{-3-\sqrt5}}{2}\Big)^n-1\Bigg]\\ &则F_n=\frac{\sqrt5}5\Bigg[\Big(\frac{{-3-\sqrt5}}{2}\Big)^n-1\Bigg](\frac{1-\sqrt5}{2})^{n}=\frac{\sqrt5}5\Bigg[\bigg(\frac{1+\sqrt5}2\bigg)^n-\bigg(\frac{1-\sqrt5}2\bigg)^n\Bigg] \end{align} \]

线性代数解法

\[\begin{align} &线性递推数列的特征方程为：x^2=x+1\\ &解得x_1=\frac{1+\sqrt5}{2},x_2=\frac{1-\sqrt5}{2}则F_n=C_1x_1^n+C_2x_2^n\\ &\because F_1=F_2=1,\therefore C_1x_1+C_2x_2=C_1x_1^2+C_2x_2^2=1\\ &解得C_1=C_2=\frac{\sqrt5}5\\ &\therefore F_n=\frac{\sqrt5}5\Big[\big(\frac{1+\sqrt5}2\big)^n-\big(\frac{1-\sqrt5}2\big)^n\Big] \end{align} \]

生成函数解法

\[\begin{align} &令S(x)=F_1x+F_2x^2+\cdots+F_nx^n+\cdots\\ &那么有S(x)(1-x-x^2) =F_1x+(F_2-F-1)x^2+\cdots+(a_n-a_{n-1}-a_{n-2})x^n+\cdots=x\\ &因此S(x)=\frac{x}{1-x-x^2}=\frac{x}{(1-\frac{1+\sqrt5}{2}x)(1-\frac{1-\sqrt5}{2}x)}\\ &令S(x)=\frac{A}{(1-\frac{1+\sqrt5}{2}x)}+\frac B{(1-\frac{1-\sqrt5}{2}x)}=\frac{A(1-\frac{1-\sqrt5}{2}x)+B(1-\frac{1+\sqrt5}{2}x)}{(1-\frac{1+\sqrt5}{2}x)(1-\frac{1-\sqrt5}{2}x)}\\ &可列\begin{cases}A+B=0\\-\frac{1-\sqrt5}{2}A-\frac{1+\sqrt5}{2}B=1\end{cases} 解得\begin{cases}A=\frac{\sqrt5}5\\B=-\frac{\sqrt5}5\end{cases}\\ &由\frac{1}{1-kx}=1+kx+k^2x^2+\cdots+k^nx^n+\cdots\\ &得F_n=\frac{\sqrt5}5\Big[\big(\frac{1+\sqrt5}2\big)^n-\big(\frac{1-\sqrt5}2\big)^n\Big] \end{align} \]