We say a rational function $F(z)$ is positive if the coefficients of its Maclaurin expansion, in the variable $z$, are non-negative.

In this context, let 
$$F_r(z):=\frac{1 - 2z + z^r - (1 - z)^r}{(1 - z)^{r - 1}(1 - 2z)}.$$

Is the following true? The cases $r=2$ and $3$ are easier to manage. 

>**QUESTION.** For $r\geq4$, each of the rational functions $F_r(z)$ is positive.