Answer by Angina Seng for Combinatorial Identity with powers of 2

Your $f(X)$ is$$\sum_{i=m}^\infty{i\choose m}\frac{X^{2i+1}}{(1-X)^{2i+2}}=\frac{X}{(1-X)^2}\frac{(X^2/(1-X)^2)^m}{(1-X^2/(1-X)^2)^{m+1}}$$(using the principle in your previous line). This simplifies to$$\frac{X^{2m+1}}{(1-2X)^{m+1}}.$$