Calculate the closed form of the following series $$\sum_{m=r}^{\infty}\binom{m-1}{r-1}\frac{1}{4^m}$$
The answer given is $$\frac{1}{3^r}$$ I tried expanding the expression so it becomes $$\sum_{m=r}^{\infty}\frac{(m-1)!}{(r-1)!(m-r)!}\frac{1}{4^m}$$but I do not know how to follow.
Any help will be appreciated, thanks.

Say we have an unfair coin that yields head with probability $\frac{3}{4}$ and we toss it repeatedly until we get $r$ heads. The probability that the $r$-th head is from the $m$-th toss is given by the following expression:
$$
\binom{m-1}{r-1}\left(\frac{1}{4}\right)^{m-r}\left(\frac{3}{4}\right)^{r}
$$
If we sum the probability for all possible $m$ then we’ll get one
$$
\begin{align}
1&=\sum_{m=r}^{\infty}{\binom{m-1}{r-1}\left(\frac{1}{4}\right)^{m-r}\left(\frac{3}{4}\right)^{r}}\\
\\
&=3^{r}\sum_{m=r}^{\infty}{\binom{m-1}{r-1}\left(\frac{1}{4}\right)^{m}}\\
\\
\\
\text{or }&\text{equivalently}\\
\\
\\
\left(\frac{1}{3}\right)^{r}&=\sum_{m=r}^{\infty}{\binom{m-1}{r-1}\left(\frac{1}{4}\right)^{m}}\\
\end{align}
$$
For a general case in which probability of getting head is $1-x$ where $0\leq x< 1$:
$$
\begin{align}
1&=\sum_{m=r}^{\infty}{\binom{m-1}{r-1}x^{m-r}\left(1-x\right)^{r}}\\
\\
&=\left(\frac{1-x}{x}\right)^{r}\sum_{m=r}^{\infty}{\binom{m-1}{r-1}x^{m}}
\end{align}
$$
