negative binomial theorem

$(1-x{)}^{-d}={\sum }_{n=0}^{\infty }(\genfrac{}{}{0ex}{}{n+d-1}{d-1})x^n$ other form:

• $1/(1-x{)}^{d+1}={\sum }_{n=0}^{\infty }(\genfrac{}{}{0ex}{}{n+d}{d})x^n$

proof To use mathematical induction, we assume that it is established at d and show that it is established at d+1

Formal derivative of a formal power series $-d\cdot (1-x{)}^{-d-1}\cdot (-1)={\sum }_{n=0}^{\infty }(n+1)(\genfrac{}{}{0ex}{}{n+d}{d-1})x^n$ Divide both sides by d to organize $(1-x{)}^{-(d+1)}={\sum }_{n=0}^{\infty }\frac{n+1}{d}(\genfrac{}{}{0ex}{}{n+d}{d-1})x^n$

• Definition of [binomial coefficient
• $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$ from (e.g. time, place, numerical quantity) $\left(\genfrac{}{}{0ex}{}{n+d}{d-1}\right)=\frac{(n+d)!}{(d-1)!(n+1)!}$ therefore $\frac{n+1}{d}\left(\genfrac{}{}{0ex}{}{n+d}{d-1}\right)=\frac{(n+d)!(n+1)}{d(d-1)!(n+1)!}=\frac{(n+d)!}{d!n!}=\left(\genfrac{}{}{0ex}{}{n+d}{d}\right)$ To recap. $(1-x{)}^{-(d+1)}={\sum }_{n=0}^{\infty }(\genfrac{}{}{0ex}{}{n+(d+1)-1}{(d+1)-1})x^n$ Therefore, it is shown to hold at d+1

For d=1 $(1-x{)}^{-1}={\sum }_{n=0}^{\infty }(\genfrac{}{}{0ex}{}{n}{0})x^n={\sum }_{n=0}^{\infty }{x}^n$ This was shown in [infinite sum compression using the inverse of a formal power series#5f0a99d3aff09e00008d4555

[Polynomials and Formal Power Series (3) Linear Asymptotes and Formal Power Series | maspy’s HP https://maspypy.com/%e5%a4%9a%e9%a0%85%e5%bc%8f%e3%83%bb%e5%bd%a2%e5%bc%8f%e7%9a%84%e3%81%b9% e3%81%8d%e7%b4%9a%e6%95%b0%ef%bc%88%ef%bc%93%ef%bc%89%e7%b7%9a%e5%bd%a2%e6%bc%b8%e5%8c%96%e5%bc%8f%e3%81%a8%e5%bd%a2%e5%bc%8f]

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