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How do I prove my result?

  1. Feb 24, 2009 #1
    I posted a bit about the characteristic value of a Fibonacci series without proof but nowI wonder about whether it could be valid. A remarkable result of it applies to the binominal coefficients. The sum of following Binominal coefficients times corresponing Fibonacci numbers are in them selves terms from the Fibonnacci series. In other words, each of the following sums is a Fibonacci number. "(n,i)" is used her to represent the the binominal coefficients

    A. [tex]\sum_{i=0}^{n}F_{i-1}(n,i)[/tex]

    B. [tex]\sum_{i=0}^{n}F_{i-2}(n,i)[/tex]

    For intance;
    [tex]F_{-1}*1 + F_{0}* 3 + F_{1}*3 + F_{2}*1 = F_{5}[/tex]

    [tex]F_{-2}*1 + F_{-1}*3 + F_{0}*3 + F_{1}*1 = F_{4}[/tex]

    [tex]F_{-1}*1 + F_{0}*4 + F{1}*6 + F_{2}*4 + F_{3}*1 = F_{7}[/tex]

    I could go on but that in it self does not prove the result.
  2. jcsd
  3. Feb 25, 2009 #2


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    Gold Member

    The only thing I can advise you is to check in Combinatorial Identities of John Riordan and another one by Henri Gould.
  4. Feb 28, 2009 #3
    Thanks but I thought of a proof by Induction
    To Prove

    [tex]F_{x+2n) = \sum_{i=0}^{n}F_{x+i}nCi[/tex]

    It is easily verified for n=1

    If true for any positive n then it is true for n+1

    [tex]F_{(x+2*(n+1))} = F_{(x+2n)} + F_{(x+2n+1)}[/tex]

    =[tex]\sum_{i=0}^{n}F_{x+i}nCi + F_{x+1+i}nCi[/tex]

    Combining like F terms and using (n+1)Ci = nC(i-1)+nCi gives the required result
  5. Feb 28, 2009 #4
    How about using Pascel's triangle.
  6. Feb 28, 2009 #5
    Anyway, I don't think it should be hard to prove using induction given that Fibonacci series is defined recursively and polynomial multiplication can be done via convolution.
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