2010/039) A problem in fibonacci sequence

show that
Σ (n = 1 to ∞) F(n)/x^n = x/(x^2-x-1) where F(n) is n th Fibonacci number

Where "x" is any integer >= 2

proof:
this is based on Geometric series and not on relationship in Fibonacci sequence which can be found at http://en.wikipedia.org/wiki/Fibonacci_number#Power_series

We know

F(n) = (p^n – (1-p)^n)/ sqrt(5) where p is phi (value = (1+ sqrt(5))/2)

So sqrt(5) F(n)/x^n = (p/x)^n – (1-p)/x)^n

So Σ (n = 1 to ∞) sqrt(5) F(n)/x^n = Σ (n = 1 to ∞)
(p/x)^n – Σ (n = 1 to ∞) (1-p)/x)^n

As x is positive
The sum converges when (p/x) < 1 and (1-p)/x < 1

Or x > p that is x > the golden ratio and need not be integer.

The sum = 1/(1-(p/x)) – 1/(1-(1-p)/x)
= x/(x-p) – x/(x-1+p)
= x((1/(x-p) – 1/(x-1+p))
= x(x-1+p) –(x-p))/)(x^2- x –(p(p-1))

As p(p-1) = 1 this is x^2-x-1
we get
So Σ (n = 1 to ∞) sqrt(5) F(n)/x^n = x(2p-1)/ (x^2-x-1)

As p = (1+ sqrt(5))/2) so (2p-1) = sqrt(5)

So Σ (n = 1 to ∞) sqrt(5) F(n)/x^n = x sqrt(5)/ (x^2-x-1)

Or Σ (n = 1 to ∞) F(n)/x^n = x/ (x^2-x-1)
Proved