You may also like

Converging Product

In the limit you get the sum of an infinite geometric series. What about an infinite product (1+x)(1+x^2)(1+x^4)... ?

Circles Ad Infinitum

A circle is inscribed in an equilateral triangle. Smaller circles touch it and the sides of the triangle, the process continuing indefinitely. What is the sum of the areas of all the circles?

Binary Squares

If a number N is expressed in binary by using only 'ones,' what can you say about its square (in binary)?

Golden Fibs

Age 16 to 18
Challenge Level

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 $X_0=-\phi X_1$.