# Golden Fibs

When is a Fibonacci sequence also a geometric sequence? When the
ratio of successive terms is the golden ratio!

The Fibonacci sequence $F_n$ is defined by the relation $$F_{n+2}=F_n + F_{n+1}$$ where $F_0=0$ and $F_1=1$. Now suppose that we take the same relation and more general sequences $X_n$ with any two starting values $X_0$ and $X_1$. Prove that the sequence is geometric if and only if the first two terms are in the ratio $1 : \pm \phi$ where $\phi$ is the golden ratio $(1+\sqrt 5)/2$.

The Fibonacci recurrence relation gives a relationship between the first three terms of the sequence and, if it is also a geometric sequence with common ratio $r$, then this relation gives a quadratic equation in $r$. The question asks for both the result and also its converse to be proved.

We have to thank Andrei Lazanu of School No. 205, Bucharest, Romania for this solution.

Given a general Fibonacci sequence $X_n$, that is a sequence satisfying the Fibonnaci relation: $X_{n+2}= X_{n+1}+ X_n$, I have to prove the following (where $\phi$ is the golden ratio):

1. If the sequence is geometric, then the ratio of the first two terms is given by $X_1:X_0= \phi$ or $X_0:X_1=-\phi$

2. If the ratio of the first two terms is $X_1:X_0= \phi$ or $X_0:X_1=-\phi$ then the sequence is geometric.

1. If a sequence is geometric, then its terms are of the form: $$X_0,\ X_1=rX_0,\ X_2=r^2X_0,\ ... X_n=r^nX_0.$$

Now, I have to find r, if the sequence is Fibonacci-type. Using the definition of a geometric sequence, I obtain: $$\eqalign{ r^{n+2} X_0 &= r^{n+1} X_0 + r^n X_0 \cr r^{n+2} &= r^{n+1} + r^n}.$$ I see that $r$ must be different from 0, so I could divide both sides by $r$ which leads to the quadratic equation: $$\eqalign{ r^2 &= r + 1\cr r^2 - r - 1 &= 0 \cr r_{1,2} &= {{1\pm \sqrt 5}\over 2}.}$$ So $${X_1\over X_0} = {1+\sqrt5 \over 2}\ {\rm or}\ {1-\sqrt5 \over 2}= \phi \ {\rm or}\ {-1\over \phi}.$$ which completes the proof of (1).

2. If the ratio of the first two terms is given by $$X_1={1+\sqrt 5 \over 2}X_0$$ then, using the recursive formula for the sequence, I obtain: $$ X_2=X_0+X_1=X_0 + {1+\sqrt 5\over 2}X_0 = {3+\sqrt 5\over 2}.$$ I observe that: $${3+\sqrt 5\over 2}={6+ 2\sqrt 5\over 4}={5+1+2\sqrt 5\over 4}=({1+\sqrt5\over 2})^2.$$ So this gives $X_2=\phi^2 X_0=\phi X_1.$ and I have shown $1+\phi = \phi^2$.

I calculate $X_3$: $X_3 = X_2 + X_1 = \phi^2 X_0 + \phi X_0 = \phi(\phi+1)X_0=\phi^3 X_0.$

But this is not enough, I have to utilise induction to show the sequence is geometric, that is $X_n=\phi^nX_0$ for all $n$. In the general case, I have, using the recurrence formula for the Fibonacci sequence: $$\eqalign{ X_{n+2} &= X_{n+1} + X_n \cr &= \phi^{n+1}X_0 + \phi^nX_0 \cr &= \phi^n(\phi+1)X_0 \cr &= \phi^{n+2}X_0}.$$ So by the axiom of induction, as I have shown $X_n=\phi^nX_0$ is true for $n=1$ and $2$, it is true for all $n$ and so the sequence is geometric.

A similar proof works when $X_0=-\phi X_1$.

This is straightforward but care must be taken to explain the significance of both roots of the quadratic equation. Also you must prove both the 'if' and the 'only if', that is, for general Fibonacci sequences, 'if the first two terms are in the golden ratio then the sequence is geometric' and 'if the sequence is geometric then the ratio of successive terms is the golden ratio'.

Induction is required