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Fibonacci Proofs via Induction

  1. Feb 3, 2010 #1
    So I am looking at the following two proofs via induction, but I have not a single idea where to start.
    The First is:
    1. Suppose hat F1=1, F2=1, F3=2, F4=3, F5=5 where Fn is called a Fibonacci number and in general:
    Fn=Fn-1+Fn-2 for n>/= 3. Prove that:

    Secondly is:
    2. Prove that F1+F2+F5+...+F2n-1=F2n

    Any help. I am looking for a proof via induction with a base case and induction step.
  2. jcsd
  3. Feb 3, 2010 #2
    You could start by trying a base case. Let n=3. What can you conclude from this? Is the statement valid?
  4. Feb 3, 2010 #3
    The base case part I think is more straight foward, but when I try to see if this hold for some k+1, things start to get tricky...
  5. Feb 3, 2010 #4
    I find it easier to work in summation notation here to keep it neat.
    Take the sum from i=1 to n+1 of fi. You can take fn+1 out of the sum and then apply the induction hypothesis to the other term. Then use the recursion formula for the fibonacci sequence to simplify.
  6. Feb 3, 2010 #5
    So we I seperate out the summation and apply the induction hypothesis to the other side I get:
    [tex]\Sigma[/tex]Fj+Fk+Fk-1=(Fk+2 + Fk+1) -1
    So I basically want to move the Fk+1 from the RHS to the LHS, but I would have to subtract it, and that gets me nowhere...
  7. Feb 3, 2010 #6
    Simplify fn+1 + fn+2
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