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I Summation for extended binomial coefficients

  1. Aug 2, 2016 #1
    Is there a way of writing summation(s) to obtain the extended binomial coefficients?

    i.e., Considering the expansion of [tex](1+x+x^2+x^3+...+x^N)^M[/tex]

    can we write expressions (presumably involving summation and/or product notation) for the coefficients (on x^j in the expansion of the above, for each integer j from j=0 to j=NM, i.e. each of the NM+1 non-0 coefficients) without expanding the polynomial by hand?
     
  2. jcsd
  3. Aug 2, 2016 #2

    chiro

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  4. Aug 2, 2016 #3

    Ssnow

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    Hi, you can use the formula

    ##(x_{0}+\cdots +x_{N})^{M}=\sum_{\alpha_{0}+\ldots +\alpha_{N}=M}\frac{N!}{\alpha_{0}!\cdots \alpha_{N}! }x_{0}^{\alpha_{0}}\cdots x_{N}^{\alpha_{N}}##

    now setting ##x_{i}=x^{i}## for every ##i=0,...,N## you obtain the expansion and you can see if it is possible to simplify the index notation in order to find the coefficients ...
     
  5. Aug 2, 2016 #4

    Svein

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    I can at least start you in a direction that might take you where you want to go:

    Consider [itex]
    2^{n}=(1+1)^{n}=\sum_{k=0}^{n}\begin{pmatrix}
    n \\
    k\\
    \end{pmatrix}1^{n-k}1^{k}=\sum_{k=0}^{n}\begin{pmatrix}
    n \\
    k\\
    \end{pmatrix}
    [/itex]. Thus, putting x=1 in your formula would give you a start?
     
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