Summation for extended binomial coefficients

[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?

 

[chiro]

Hey Astudious.

You probably want to look at the multinomial theorem and do a bit more work to get some useful results.

https://en.wikipedia.org/wiki/Multinomial_theorem

 

[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 ...

 

[Svein]

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

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