Fibonacci Induction Proof in terms of Phi

Question

I am trying to prove this equation with an induction proof:

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

where $\ \phi = \frac{\sqrt{5} - 1}{2} $

I have started off by proving two base cases: $\ F(1)$ and $\ F(2):$

$$\ F(1) = \frac{(1 + \phi)^1 - (-\phi)^1}{\sqrt{5}} = \frac{(1 + \phi + \phi)}{\sqrt{5}} = \frac{1 + 2\phi}{\sqrt{5}} = \frac{1 + \sqrt{5} - 1}{\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{5}} = 1 $$

$$\ F(2) = \frac{(1 + \phi)^2 - (-\phi)^2}{\sqrt{5}} = \frac{(1 + 2\phi + \phi^2 - \phi^2)}{\sqrt{5}} = \frac{(1 + 2\phi)}{\sqrt{5}} = \frac{(1 + \sqrt{5} - 1)}{\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{5}} = 1$$

However, I do not know how to go about proving the inductive step: $\ F(n + 1) $. Can someone show me how to proceed from here?

Answer 1

Let us assume this true for $<n$

$$\ F(n) = \frac{(1 + \phi)^n - (-\phi)^n}{\sqrt{5}}$$ Note that $$F(n)=F(n-1)+F(n-2)$$ $$F(n)=\frac{(1 + \phi)^{n-1} - (-\phi)^{n-1}}{\sqrt{5}}+\frac{(1 + \phi)^{n-2} - (-\phi)^{n-2}}{\sqrt{5}}$$ $$F(n)=\frac{(1 + \phi)^{n-1} - (-\phi)^{n-1}+{(1 + \phi)^{n-2} - (-\phi)^{n-2}}}{\sqrt{5}}$$ $$=\frac{(1+\phi)^{n-2}(1+\phi+1)-(-\phi)^{n-2}(-\phi+1)}{\sqrt5}$$ $$=\frac{(1+\phi)^{n-2}(1+\phi)^{2}-(-\phi)^{n-2}(-\phi)^2}{\sqrt5}$$ $$=\frac{(1+\phi)^{n}-(-\phi)^{n}}{\sqrt5}$$ Note that ,I used $$1-\phi=(\phi)^2=(-\phi)^2$$ and similarly$$2+\phi=(\phi)^2+2\phi+1=(1+\phi)^2$$

Answer 2

Using a compact notation and ignoring the factor $\sqrt5$,

$$F_{n+2}\propto a^2a^n+b^2b^n$$

while

$$F_{n+1}+F_n\propto a\,a^n+b\,b^n+a^n+b^n.$$

Then by grouping and identifying,

$$a^2=a+1,\\b^2=b+1,$$

which are both true for the given coefficients.

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Notice that the $\phi$ in the initial expression is the Golden section minus one.

Rewriting with $\varphi=\phi+1$ and noting that $\varphi\,\phi=1$, you find the usual Binet formula

$$F_n=\frac{\varphi^n-(-\varphi)^{-n}}{\sqrt5}.$$