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Fibonacci Identity

Date: 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.

Date: 12/11/2001 at 06:28:07
From: Doctor Floor
Subject: Re: Form of Fibonacci numbers


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

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
Associated Topics:
High School Fibonacci Sequence/Golden Ratio
High School Number Theory

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