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 f is the golden ratio):
1. If the sequence is geometric, then the ratio of the first two terms is given by X₁:X₀ = f or X₀:X₁=-f
2. If the ratio of the first two terms is X₁:X₀ = f or X₀:X₁=-f then the sequence is geometric.

1. If a sequence is geometric, then its terms are of the form:

X0, X1=rX0, X2=r2X0, ... Xn=rnX0.

Now, I have to find r, if the sequence is Fibonacci-type. Using the definition of a geometric sequence, I obtain:

rn+2 X0 = rn+1 X0 + rn X0 rn+2 = rn+1 + rn .

I see that r must be different from 0, so I could divide both sides by r which leads to the quadratic equation:

r2 = r + 1 r2 - r - 1 = 0 r1,2 = 1±Ö5 2 .

So

X1 X0 = 1+Ö5 2  or  1-Ö5 2 = f or  -1 f .

which completes the proof of (1).

2. If the ratio of the first two terms is given by

X1= 1+Ö5 2 X0

then, using the recursive formula for the sequence, I obtain:

X2=X0+X1=X0 + 1+Ö5 2 X0 = 3+Ö5 2 .

I observe that:

3+Ö5 2 = 6+ 2Ö5 4 = 5+1+2Ö5 4 =( 1+Ö5 2 )2.

So this gives X₂=f² X₀=f X₁. and I have shown 1+f = f².

I calculate X₃: X₃ = X₂ + X₁ = f² X₀ + fX₀ = f(f+1)X₀=f³ X₀.

But this is not enough, I have to utilise induction to show the sequence is geometric, that is X_n=fⁿX₀ for all n. In the general case, I have, using the recurrence formula for the Fibonacci sequence:

Xn+2 = Xn+1 + Xn = fn+1X0 + fnX0 = fn(f+1)X0 = fn+2X0 .

So by the axiom of induction, as I have shown X_n=fⁿX₀ is true for n=1 and 2, it is true for all n and so the sequence is geometric.

A similar proof works when X₀=-fX₁.