# I Summation for extended binomial coefficients

1. Aug 2, 2016

### Astudious

Is there a way of writing summation(s) to obtain the extended binomial coefficients?

i.e., Considering the expansion of $$(1+x+x^2+x^3+...+x^N)^M$$

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. Aug 2, 2016

3. Aug 2, 2016

### Ssnow

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 ...

4. Aug 2, 2016

### Svein

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

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

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