negative binomial theorem

$ (1-x)^{-d} = \sum_{n=0}^\infty \binom{n+d-1}{d-1}x^n

other form:

$ 1/(1-x)^{d+1} = \sum_{n=0}^\infty \binom{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)\binom{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}\binom{n+d}{d-1}x^n

Definition of [binomial coefficient

$ \binom{n}{k} = \frac{n!}{k!(n-k)!}

from (e.g. time, place, numerical quantity)

$ \binom{n+d}{d-1} = \frac{(n+d)!}{(d-1)!(n+1)!}

therefore

$ \frac{n+1}{d}\binom{n+d}{d-1} = \frac{(n+d)!(n+1)}{d (d-1)! (n+1)!} = \frac{(n+d)!}{d! n!} = \binom{n + d}{d}

To recap.

$ (1-x)^{-(d+1)} = \sum_{n=0}^\infty \binom{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 \binom{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

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