Fibonacci SequenceDate: 01/29/2001 at 22:05:04 From: Alisha Winters Subject: Fibonacci Sequence I'm taking a class for teaching math to elementary students, and we did a problem with the Fibonacci sequence. Afterward our teacher challenged each of us to find out if there is a formula to find the first "n" terms of the sequence. Math is not my strongest subject so I've come to you to see if you can help. Thanks! Alisha Date: 01/29/2001 at 22:30:30 From: Doctor Rob Subject: Re: Fibonacci Sequence Thanks for writing to Ask Dr. Math, Alisha. Yes, there is a formula for the n-th Fibonacci number. It is: F(n) = [([1+sqrt(5)]/2)^n - ([1-sqrt(5)]/2)^n]/sqrt[5]. Now it may seem impossible that this complicated formula involving sqrt(5) would evaluate to an integer, but it does. F(0) = [([1+sqrt(5)]/2)^0 - ([1-sqrt(5)]/2)^0]/sqrt[5], = [1 - 1]/sqrt[5] = 0. F(1) = [([1+sqrt(5)]/2)^1 - ([1-sqrt(5)]/2)^1]/sqrt[5], = [1/2+sqrt(5)/2-1/2+sqrt(5)/2]/sqrt[5], = sqrt[5]/sqrt[5] = 1. F(2) = [([1+sqrt(5)]/2)^2 - ([1-sqrt(5)]/2)^2]/sqrt[5], = [(3+sqrt[5])/2 - (3-sqrt[5])/2]/sqrt[5], = sqrt[5]/sqrt[5] = 1. You can see that the square roots always cancel out, which seems a bit miraculous, but true! - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/