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Golden Fibs

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

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 $X_0=-\phi X_1$.