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I discovered a very interesting thing recently, and proved it by induction.

I found that the $n$th power of $\phi={1+\sqrt5 \over2}$, the Golden Ratio, is the product of $\phi$ with the $n$th Fibonacci number added to the $(n-1)$th Fibonacci number.

However, I'm not able to prove it by any other method. Can anyone help?

$\text{Theorem}:\; \phi^n=F_n\phi+F_{n-1}\qquad[F_0=0]$

$\text{Proof}:$

$\text{Trivial for }n=1.$

$\text{Let this be true for all }n\text{ till }k.$

$\text{Then,}$

$\begin{align}\phi^{k+1}&=\phi^k\phi\\&=(F_n\phi+F_{n-1})\phi\\&=F_n\phi^2+F_{n-1}\phi\\&=F_n(\phi+1)+F_{n-1}\phi\\&=F_n\phi+F_{n-1}\phi+F_n\\&=(F_n+F_{n-1})\phi+F_n\\&=F_{n+1}\phi+F_{(n+1)-1}\end{align}$

$\text{Therefore Proved by Induction Hypothesis}.$

So, is there an alternative proof? I tried taking $\phi^2=\phi+1$ and branching outwards, but got nowhere.

DynamoBlaze
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1 Answers1

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Fibonacci number can be defined using the recurrence relation $$F_n=F_{n-1}+F_{n-2}$$ with $$F_0=0,F_1=1$$

Solving the relation will give $$F_n=\frac{1}{\sqrt5}(\phi^n-\varphi^n)$$

where $$\phi=\frac{1+\sqrt5}{2},\varphi=\frac{1-\sqrt5}{2}$$

You can use this expression to prove your finding.

\begin{align} F_n\phi+F_{n-1}=&=\frac{1}{\sqrt5}\left(\phi^n-\varphi^n\right)\phi+\frac{1}{\sqrt5}\left(\phi^{n-1}-\varphi^{n-1}\right)\\ &=\frac{1}{\sqrt5}\left(\phi^{n+1}+\phi^{n-1}-\phi\varphi^n-\varphi^{n-1}\right)\\ &=\frac{1}{\sqrt5}\left(\phi^{n+1}+\phi^{n-1}-\phi\varphi\varphi^{n-1}-\varphi^{n-1}\right)\\ &=\frac{1}{\sqrt5}\left(\phi^{n+1}+\phi^{n-1}\underbrace{-\left(-1\right)\varphi^{n-1}-\varphi^{n-1}}_{=0}\right)\\ &=\frac{1}{\sqrt5}\left(\phi^{n+1}+\phi^{n-1}\right)\\ &=\frac{\phi^{n}}{\sqrt5}\left(\phi+\frac{1}{\phi}\right)\\ &=\frac{\phi^{n}}{\sqrt5}\left(\sqrt5\right)\\ &=\phi^n \end{align}

Karn Watcharasupat
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