Combinatorics identity; proof of $\sum_{n≥0}\left[\binom{n}{m}(z^n)\right]=\frac{z^m}{(1-z)^{m+1}}$

Asked Nov 28 '19 at 02:39

Active Nov 28 '19 at 04:15

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I am working on a proof and I came across a combinatorial identity on the internet that would really help me, but I cannot prove it. The identity is:

$$\sum_{n≥0}\left[\binom{n}{m}(z^n)\right]=\frac{z^m}{(1-z)^{m+1}}$$

I imagine it has something to do with the binomial theorem, but I just cannot figure it out. So, I am hoping that someone can either tell me what this identity is or help me prove it.

edited Nov 28 '19 at 04:15

k76u4vkweek547v7

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asked Nov 28 '19 at 02:39

Mike

1

It's the binomial theorem for negative integer exponents. – Angina Seng Nov 28 '19 at 02:42
Note that $$\sum \binom{n}{m} z^n = \sum \binom{n + m}{m} z^{n + m} = z^m \sum \binom{n + m}{m} z^{n}.$$ – Trevor Gunn Nov 28 '19 at 04:40

Combinatorics identity; proof of $\sum_{n≥0}\left[\binom{n}{m}(z^n)\right]=\frac{z^m}{(1-z)^{m+1}}$

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