\sum^{m}_{k=0}\frac{1}{2^k}{{m+1+k}\choose k}=\sum^{\rho-1}_{k=0}\frac{1}{2^k}{{\rho+k}\choose k}=2^\rho-\frac{1}{2^\rho}{{2\rho}\choose \rho}=2^{m+1}-{{2m+2}\choose {m+1}}