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


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: $k\mapsto n-r$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.