Fibonacci IdentityDate: 12/10/2001 at 17:49:43 From: Doug Moll Subject: Form of Fibonacci numbers I am trying to create an inductive proof for the particular identity of Fibonacci numbers that: F(n-1) * F(n+1) = (-1)^n + (Fn)^2 I know I do a base case of n being 0, and then an inductive step of n being n+1, but beyond that I just can't get the math to work out. Is this a valid identity, and if so, how would I work out the math to inductively show this? Thanks much. Doug Date: 12/11/2001 at 06:28:07 From: Doctor Floor Subject: Re: Form of Fibonacci numbers Hi, Thanks for your question. Well, I suppose you have been able to show that this works for n = 0. Now we suppose that F(n-1)*F(n+1) = (-1)^n + (F(n))^2 holds for n. We can rewrite this to F(n-1) * F(n+1) - (F(n))^2 = (-1)^n [*] Now we will replace all n by n+1 at the left-hand side of the equation, and show that it yields (-1)^(n+1). For that we use truth we suppose of [*] as well as application of the Fibonacci rule F(m+2) = F(m)+F(m+1) for all m: F(n)*F(n+2) - (F(n+1))^2 = F(n)*[F(n+1) + F(n)] - F(n+1)*[F(n) + F(n-1)] = F(n)*F(n+1) + (F(n))^2 - F(n)*F(n+1) - F(n+1)*F(n-1) = (F(n))^2 - F(n+1)*F(n-1) = -(-1)^n = (-1)^(n+1) And we get where we want to be. See for an approach without induction, from the Dr. Math archive Fibonacci Trick http://mathforum.org/dr.math/problems/roberts.6.01.99.html If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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