Fibonacci Sequence PropertyDate: 11/29/2001 at 05:35:36 From: Enrique Somolinos Subject: Demonstration of a Fibonacci sequence property Hi, I have to demonstrate a particular property of the Fibonacci sequence: Let F(k) be the k number of the Fibonacci sequence. I have to prove that F(k) is a divisor of F(nk), where n is a natural number (so, F(nk) = A*F(k) where A is a natural number). I have tried my best but I can't find a way to demonstrate this. Can you help me please? Thank you in advance, Enrique Date: 11/29/2001 at 10:58:17 From: Doctor Rob Subject: Re: Demonstration of a Fibonacci sequence property Thanks for writing to Ask Dr. Math, Enrique. You can prove this by using the addition formula for the Fibonacci numbers: F(r+s) = F(r-1)*F(s) + F(r)*F(s+1), together with the Principle of Mathematical Induction on the variable k. If you don't know how to prove the above formula, you can do that by fixing s and using induction on r. Alternatively, you can use the Binet Formula for the Fibonacci numbers: F(t) = (a^t-b^t)/(a-b), a = (1+sqrt[5])/2, b = (1-sqrt[5])/2, a^2 = a + 1, b^2 = b + 1. When this is substituted in the addition formula, and everything is simplified, you will see that it is reduced to an identity. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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