Prove that f(n+1)*f(n-1)-f(n)^2=(-1)^n?

Question

for every positive integer n

where (n+1) and so on are indices.

Answer 1

The numbers 1/1,2/1,3/2,5/3,...,f(n)/f(n-1), f(n+1)/f(n) , ...
converge to the golden ratio.
Convergents in a continued fraction expansion obey the
rule pq' - p'q = +/- 1 where p/q ,p'/q' are successive
convergents. Checking the sign: 2/1-1/1=1. Therefore
f(n+1)f(n-1) -f(n)^2 = (-1)^n. I have assumed f means
fibonacci.

Answer 2

Here f(n) must be the nth Fibonacci number.
You should have specified this in your question.
I will work out a proof for you with Binet's formula
and come back here. No time right now, sorry!
BTW: Binet's formula states that
f(n) = 1/√5(a^n - b^n),
where a = (1+ √5).2 and b = (1-√5)/2.

Answer 3

What is the initial conditions? i.e. f(0) or f(1). For any recursive relations, we need at least as many initial conditions as the order of the recursion.