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