# Generating Functions and the Negative Binom

I'm reading from 3 different sets of notes on generating functions and having a little trouble integrating their approaches.

First, I'm used to working with the following definitions:

$$c(n,k)=\binom{n}{k}=\frac{n!}{k!(n-k)!}$$ and $$D(n,k)=\binom{n-1+k}{k}=\binom{n-1+k}{n-1}$$

One set of notes says for any positive $n,k$ then $$c(-n,k)=(-1)^{k}\binom{n-1+k}{k}$$ gives the $k^{\text th}$ coefficient for $(1+x)^{-n}$

The other 2 sets give me the equation: $$\frac{1}{(1-x)^{n}}=\sum_{k=0}^\infty\binom{n-1+k}{n-1}x^k=\sum_{k=0}^\infty\binom{n-1+k}{k}x^k$$

He then goes on to say this last is the generating function for the series of negative binom coefficients and that it is exactly $D(n,k)$. How can both these be correct? Either the $k^{\text th}$ coefficient is $(-1)^{k}D(n,k)$ or it's $D(n,k)$ but how can it be both?

I'm obviously missing something trivial here I think...

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## 1 Answer

Silly me, it was something trivial.

In the first equation we're looking at $(1+x)^{-n}$ and in the second $(1-x)^{-n}$. If we make $x$ negative in the second equation then it's sign will alternate the same way as $(-1)^k$ in the first equation.

$$(1-(-x))^{-n}=\sum_{k=0}^\infty D(n,k)(-x)^k=\sum_{k=0}^\infty (-1)^kD(n,k)x^k$$

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