How to prove the closed form for the generating function $g_r(x) = \sum_{n=0}^\infty {n \choose r} x^n$ possibly using Newton's Binomial Theorem

Question

define the generating function $g_r(x) = \sum_{n=0}^\infty {n \choose r} x^n$

how do you find the closed form for this function?

I got $x^r{(1-x)}^{-(r+1)}$ but I am not sure how to prove it. I think it relates to Newton's Binomial Theorem and I know we have $(1-z)^{-n} = \sum_{k=0}^\infty {n+k-1 \choose k} z^k$ where $|z| < 1$, but again, not sure about the right steps to take.

any help is appreciated! thank you

Answer 1

Actually you have found all results needed for the proof. It remains only to order them correctly. In what follows I assume that $r$ is a non-negative integer.

We have by the extended binomial theorem: $$ (1-x)^{-r}=\sum_{k=0}^\infty\binom{-r}k (-x)^k=\sum_{k=0}^\infty\binom{r+k-1}k x^k. $$

Therefore: $$ x^r(1-x)^{-r-1}=\sum_{k=0}^\infty\binom{r+k}k x^{r+k}=\sum_{k=0}^\infty\binom{r+k}r x^{r+k}=\sum_{n=r}^\infty\binom{n}r x^{n}, $$ where in the last step we reindexed the summation: $k\mapsto n-r$.