Fibonacci numbers with negative indices?

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The Fibonacci sequence can be extended to negative indices using the recurrence relation Fn = F(n+2) - F(n+1) for n < 0. Initial negative terms are identified as F(-1) = -1, F(-2) = -1, and F(-3) = -2. The discussion highlights the need for a unique extension that maintains the original recurrence for all integers. Induction is suggested as a method to prove the formula for negative indices, with a focus on establishing a base case. The exploration of these negative Fibonacci numbers reveals a consistent pattern that aligns with the established sequence.
morbius27
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Homework Statement


Let the Fibonacci sequence Fn be defined by its recurrence relation (1) Fn=F(n-1)+F(n-2) for n>=3. Show that there is a unique way to extend the definition of Fn to integers n<=0 such that (1) holds for all integers n, and obtain an explicit formula for the terms Fn with negative indices n.

The Attempt at a Solution


So I know the solution uses induction, and I think the first few negative terms should be F-1=-1, F-2=-1, F-3=-2 etc. So for the negative integers, Fn=F(n+1) + F(n+2) for n<0, but if the formula is supposed to extend to all integers n, that formula doesn't work...am I thinking about this problem wrong?
 
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Ok, so i worked on this a bit more, and found that the formula I'm trying to prove is Fn=F(n+2)-F(n+1), since this generates the negative terms...would the base case be n=1 and the induction hypothesis prove n=k-1?
 
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