1
$\begingroup$

Using an analog to Newton's binomial theorem with negative exponents, is it true that

$$ \begin{align} \left(\sum_{k=0}^mx^k\right)^{-n} & = \sum_{0\le i_0+...+i_m=n\lt\infty}\binom{-n}{i_0,i_1,...,i_m}1^{i_0}x^{i_1}...x^{mi_m} \\ & =\sum_{0\le i_0+...+i_m=n\lt\infty}\binom{-n}{i_1,...,n-i_0-...-i_{m-1}}x^{i_1}...x^{m(n-i_1-...-i_{m-1})} \\ & = \sum_{0\le i_0+...+i_m=n\lt\infty}\frac{-n(-n-1)...(-n-i_0-...-i_m+1)}{i_0!i_1!...i_{m-1}!}x^\alpha \\ & = \sum_{0\le i_0+...+i_m=n\lt\infty}(-1)^{i_0+...+i_{m-1}}\frac{n(n+1)...(n+i_0+...+i_m-1)}{i_0!i_1!...i_{m-1}!}x^\alpha \\ & = \sum_{0\le i_0+...+i_m=n\lt\infty}(-1)^\beta\frac{(n+i_0+...+i_m-1)!}{i_0!i_1!...i_{m-1}!(n-1)!}x^\alpha \\ & = \sum_{0\le i_0+...+i_m=n\lt\infty}(-1)^\beta\binom{n+i_0+...+i_m-1}{i_0,...,i_{m-1},n-1}x^\alpha \\ \end{align} $$ where $\beta=i_0+...+i_{m}$ and $\alpha=mn-mi_0-(m-1)i_1-...-i_{m-1}$? I would like to keep the summation to "1" Sigma and a single coefficient (multinomial) instead of the product of multiple binomials, so i thought this might be a way to accomplish that...

EDIT: I know that we can use the Binomial theorem in order to get an expression, but I'm looking to see if i can write something "simpler" and with less notational devices.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.