what series the function $1/(1-ax)^r, a,r\in N$ generats.

I want to kmow what series the function $$1/(1-ax)^r, a,r\in N$$ generats. I thoghut about doing this:

lets name y=ax now we have $$1/(1-y)^r, r\in N$$ and we know $$1/(1-y)^r= \sum_{n=0}^{\infty}{n+r-1\choose r-1}y^n$$

now lets put back y=ax so $$1/(1-ax)^r = \sum_{n=0}^{\infty}{n+r-1\choose r-1}a^nx^n$$

does is make sense?

The series is $$S=(1-ax)^{-r}=\sum_{k=0}^{\infty}(-1)^k {-r \choose k} (ax)^k=\sum_{k=0}^{\infty} {r+k-1 \choose k} (ax)^k.$$ Which is valid for $$|x|
• Good question, note that ${s \choose k}=\frac{s(s-1)(s-2)(s-3)....(s-k+1)}{k!}$, in this s may not be positive integer. It can be negative integer or non-integer or even complex. Check that ${-1 \choose k}=(-1)^k$, ${-3 \choose 2}=-3(-3-1)/2$ etc. You can check the identity thar ${-r \choose k}=(-1)^k { r+k-1 \choose k.}$ Check ${-4 \choose 3}-=-20$ – Dr Zafar Ahmed DSc Jun 12 at 17:37
• @KIMKES1232 One more interesting way to calculate ${-4 \choose 3}=\frac{(-4)!}{(-7)! 3!}.$ In this $(-4)!$= product of all integers from $-\infty$ to -4. Similarly $(-7)!$ means product of all numbers from $-\infty$ up to $-7$. do the cancellations to get -20\ – Dr Zafar Ahmed DSc Jun 12 at 18:16