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