Conjectured form of $\sum_{r=0}^{\infty}\binom{2r}{r}\frac{r^{n}}{a^{r}}$

Question

I was able to conjecture this result, $$\sum_{r=0}^{\infty}\binom{2r}{r}\frac{r^{n}}{a^{r}}=(-1)^n\left[\left(x\cdot \frac{d}{dx}\right)^n\left(\frac{1}{\sqrt{1-\frac{4}{x}}}\right)\right]_{x=a}$$ Is this result known in literature and if yes then I would love to see a formal proof if possible.

Answer 1

Hint: Use the binomial theorem to prove $$(1-4x)^{-1/2} = \sum_{r=0}^\infty\binom{2r}{r}x^r$$ and work from there.