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Showing a Sequence Also Satisfies the Fibonacci Sequence

  1. Jan 17, 2013 #1
    1. The problem statement, all variables and given/known data

    The problem:
    Let r satisfy r2= r + 1. Show that the sequence an = Arn, where A is constant, satisfies the Fibonacci sequence an = an-1 + an-2 for n > 2.

    2. Relevant equations

    The given equations above are the only relevant equations.

    3. The attempt at a solution

    I think have to show that Arn = Arn-1 + Arn-2, but I'm not sure how to manipulate the given equations to achieve such a thing. I can factor one side so that Arn = A(rn-1 + rn-2), and then say that rn = rn-1 + n-2, but I have no idea what to do from here.
     
  2. jcsd
  3. Jan 17, 2013 #2

    Dick

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    Divide both sides by r^(n-2).
     
  4. Jan 17, 2013 #3
    So, I'm guessing I can manipulate the exponents as you would normal ones, and dividing both sides by rn-2 would result in r2 = rn-1, correct?
     
  5. Jan 17, 2013 #4

    Dick

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    No? What are r^n/r^(n-2), r^(n-1)/r^(n-2) and r^(n-2)/r^(n-2)??
     
  6. Jan 17, 2013 #5
    rn/rn-2 = rn-1/rn-2 + 1, which simplifies to r2 = r + 1.

    I'm still unsure of how this can be used to prove that Arn also satisfies the Fibonacci sequence.

    EDIT: Does showing that the two set equal to each other simplifies to r2 = r + 1 prove that the two are equal? If so, why is that? This question really seems to be going right over my head.
     
    Last edited: Jan 17, 2013
  7. Jan 17, 2013 #6

    Dick

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    It doesn't prove the two are equal. It proves that if r^2=r+1 then Ar^n=Ar^(n-1)+Ar^(n-2). All of your steps are reversible. Multiply both sides of r^2=r+1 by Ar^(n-2).
     
  8. Jan 17, 2013 #7
    Thank you for the explanation! I was thinking in just one way rather than looking at the problem from multiple angles.
     
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