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Binomial identity.

  1. Oct 1, 2007 #1
    prove that:
    [tex]\forall n \in N \forall 0<=k<=2^{n-1} (C(2^n,k)=\sum_{j=0}^{k}C(2^{n-1},j)C(2^{n-1},k-j))[/tex]

    attempt at solution
    induction seems to be too long im opting for a shorter solution, so the sum that it's wrriten in the rhs is the square of the sum of the term C(2^(n-1),j) but other than that don't know how to procceed.

    any advice?

    thanks in advance.
  2. jcsd
  3. Oct 1, 2007 #2
    C is the binomial coefficient.
  4. Oct 3, 2007 #3
    so i guess no one here is that good with B.Cs?
  5. Oct 4, 2007 #4
    Have you tried using the fact that [tex]2^n=2^{n-1}+2^{n-1}[/tex]
    Then expand [tex](1+x)^{2^n}[/tex] in two different ways, equating coefficients of [tex]x^k[/tex]?
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