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'Golden Fibs' printed from https://nrich.maths.org/
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 $X_0=-\phi X_1$.