Relationship between golden ratio powers and Fibonacci series

Question

Can anyone prove the following equation?

($F_n$ is the $n$th element of Fibonacci series and $n \in N$.)

$\phi = 1 \times \phi + 0$

$\phi^2 = 1 \times \phi + 1 $

$\phi^3 = 2 \times \phi + 1 $

$\phi^4 = 3 \times \phi + 2 $

$\phi^5 = 5 \times \phi + 3 $

$\phi^6 = 8 \times \phi + 5 $

...

$\phi^n = F_n \times \phi + F_{n-1} $

(However for $ \phi ^ 1 $ , $F_{1-1}=F_0$ is not defined.)

Answer 1

You can easily prove that by induction: if $\phi^n=F_{n-1}+\phi F_n$, then $$ \phi^{n+1}=\phi^n\phi=(F_{n-1}+\phi F_n)\phi=\phi F_{n-1}+\phi^2 F_n=\phi F_{n-1}+(1+\phi) F_n=F_n+\phi(F_{n-1}+F_n)=F_n+\phi F_{n+1}. $$

Answer 2

Recall that :

$$ F_n=\frac{{\phi}^{n}-{(-\frac{1}{\phi})}^{n}}{\sqrt{5}} $$

So:

$$ {F}_{n-1}+\phi F_n=\frac{{\phi}^{n-1}-{(-\frac{1}{\phi})}^{n-1}}{\sqrt{5}}+\phi \frac{{\phi}^{n}-{(-\frac{1}{\phi})}^{n}}{\sqrt{5}}\\ =\frac{{\phi}^{n-1}-{(-\frac{1}{\phi})}^{n-1}+{\phi}^{n+1}-\phi {(-\frac{1}{\phi})}^{n}}{\sqrt{5}}\\ =\frac{{\phi}^{n-1}(1+{\phi}^{2})-{(-\frac{1}{\phi})}^{n-1}(1+\phi{(-\frac{1}{\phi})})}{\sqrt{5}}\\ ={\phi}^{n-1}\frac{1+{(\frac{1+\sqrt{5}}{2})}^{2}}{\sqrt{5}}\\ ={\phi}^{n-1}\frac{1+\frac{6+2\sqrt{5}}{4}}{\sqrt{5}}\\ ={\phi}^{n-1}\frac{1+\frac{3+\sqrt{5}}{2}}{\sqrt{5}}\\ ={\phi}^{n-1}\frac{5+\sqrt{5}}{2\sqrt{5}}\\ ={\phi}^{n-1}\sqrt{5}(\frac{1+\sqrt{5}}{2\sqrt{5}})\\ ={\phi}^{n-1}\phi\\ ={\phi}^{n} $$