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How to proof; a sumformula

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Proof that [tex]\sum a, m=0, [/tex] (a-m)! / m!(a-2m)! = the fibonacci sequence.
    2. Relevant equations
    Fibonacci: 1, 1, 2, 3, 5, ... (but I think everyone knows that one!)
    3. The attempt at a solution
    Let Xm = [tex]\suma m=0[/tex] (a-m)! / m!(a-2m)!
    I think I better proof Xm+2 = Xm+1 + Xm (follows from the fibonacci), so I can conclude that Xm = Fm+1.
    I think using induction is the best way to go: Prooving it for X0 and X1, maybe even X2.
    My attempt:
    [tex]\sum a+2, m=0, [/tex] (a+2-m)! / m!(a+2-2m)! = [tex]\sum a+1, m=0, [/tex] (a+1-m)! / m!(a+1-2m)! + [tex]\sum a, m=0, [/tex] (a-m)! / m!(a-2m)!
    =[tex]\sum a+1, m=0, [/tex] (a+1-m)! / m!(a+1-2m)! + [tex]\sum a+1, m=1, [/tex] (a-m-1)! / (m-1)!(a-m-1-m+1)! ... but I think I am making big mistakes here, because whatever I do, I get stuck.
    I just don't see it!

    Vince, freshmen physics, sorry but I don't know how to use the symbols and stuff!
     
  2. jcsd
  3. Nov 15, 2009 #2
    It's difficult to read what you have written. Try editing it by using the proper LaTeX. Just click on my output below to see the way to do it.
    [tex]\sum_{n=0}^\infty a_n[/tex] for sums and subscripts
    [tex]\frac{(a-m)!}{m!(a-2m)!}[/tex] for fractions
    Also, what is Xm?
     
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