# Closed form for $G(x) = \sum_{n=0}^{\infty}\binom{k}{n}x^n$

I want to find the generating function of the following recurrence relation

$$a_n = \binom{k}{n}$$

I already know that in the case of $a_n=\binom{n}{k}$, the generating function $A(x)$ can be given as

\begin{align*} A(z)&=&\sum_{n=0}^{\infty}\binom{n}{k}x^n\\ &=&\frac{z^k}{(1-z)^{k+1}} \end{align*}

How do I do this in the case that I am choosing n element from k?

Thanks!!

• Unless $k$ varies as well, it will just be a polynomial (since $\binom{k}{n}=0$ for $n>k$). – Clayton May 1 '18 at 23:06
• Well how did you obtain the generating function in the other case? Try examining the method used in that case and see how to modify it for this case. – Dave May 1 '18 at 23:06

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