Fibonacci numbers with negative indices?

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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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