Help in summing: $\sum_{k=m} ^{2m} ~k\cdot ~2^{-k} {k \choose m}$ While solving an interesting problem I require two sums
$$S_1=\sum_{k=m}^{2m} 2^{-k} {k \choose m} ~ \text{and} ~  S_2=\sum_{k=m}^{2m} k \cdot ~ 2^{-k}  {k \choose m}$$
The former is know to be unity see inside the solutions of
How to prove that $\sum_{i=0}^n 2^i\binom{2n-i}{n} = 4^n$.
However, here I require your help in finding $S_2$ by hand.

Starting from
$$\sum_{k=m}^{2m} k 2^{-k} {k\choose m}$$
we have as in the answer that was first to appear
$$\sum_{k=m}^{2m} (k+1) 2^{-k} {k\choose m}
- \sum_{k=m}^{2m} 2^{-k} {k\choose m}
\\ = (m+1) \sum_{k=m}^{2m} 2^{-k} {k+1\choose m+1}
- \sum_{k=m}^{2m} 2^{-k} {k\choose m}.$$
For the first sum term we get
$$(m+1) [z^{m+1}] (1+z) \sum_{k=m}^{2m} 2^{-k} (1+z)^k
\\ = (m+1) 2^{-m} [z^{m+1}] (1+z)^{m+1}
\sum_{k=0}^{m} 2^{-k} (1+z)^k
\\ = (m+1) 2^{-m} [z^{m+1}] (1+z)^{m+1}
\frac{1-((1+z)/2)^{m+1}}{1-(1+z)/2}
\\ = (m+1) \frac{1}{2^{m-1}} [z^{m+1}] (1+z)^{m+1}
\frac{1-((1+z)/2)^{m+1}}{1-z}.$$
This is
$$(m+1) \frac{1}{2^{m-1}}
\sum_{k=0}^{m+1} {m+1\choose k}
- (m+1) \frac{1}{2^{2m}}
\sum_{k=0}^{m+1} {2m+2\choose k}
\\ = 4(m+1) - (m+1) \frac{1}{2^{2m}}
\left(\frac{1}{2} 2^{2m+2} + \frac{1}{2}{2m+2\choose m+1}\right)
\\ = 2(m+1) - (m+1) \frac{1}{2^{2m+1}} {2m+2\choose m+1}.$$
The second one is very similar:
$$[z^{m}] \sum_{k=m}^{2m} 2^{-k} (1+z)^k
\\ =  \frac{1}{2^m} [z^{m}] (1+z)^m \sum_{k=0}^{m} 2^{-k} (1+z)^k
\\ = \frac{1}{2^m} [z^{m}]  (1+z)^m
\frac{1-((1+z)/2)^{m+1}}{1-(1+z)/2}
\\ =  \frac{1}{2^{m-1}}  [z^{m}](1+z)^m
\frac{1-((1+z)/2)^{m+1}}{1-z}$$
This is
$$\frac{1}{2^{m-1}} 2^m
- \frac{1}{2^{2m}} \sum_{k=0}^m {2m+1\choose k}
= 2 - \frac{1}{2^{2m}} \frac{1}{2} 2^{2m+1}
= 2 - 1 = 1.$$
Collecting everything we find
$$\bbox[5px,border:2px solid #00A000]{
2m+1 - \frac{m+1}{2^{2m}} {2m+1\choose m}.}$$
