Help in summing: $\sum_{k=m} ^{2m} ~k\cdot ~2^{-k} {k \choose m}$

Question

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.

Answer 1

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

$$\bbox[5px,border:2px solid #00A000]{ 2m+1 - \frac{m+1}{2^{2m}} {2m+1\choose m}.}$$

Answer 2

Now, I am able to find this sum by interesting adjustments, I present it below by making use of the result that $$\sum_{k=m}^{2m} 2^{-k} {k \choose m}=1 ~~~~(1) $$ Let $$S=\sum_{k=m}^{2m} 2^{-k} k~ {k \choose m}= \sum_{k=m}^{2m} 2^{-k} \frac{(k+1-1)k!}{m! (k-m)!}=\sum_{k=m}^{2m} (m+1)~ 2^{-k} {k+1 \choose m+1} -\sum_{k=m}^{2m} 2^{-k} {k \choose m}~~~~(2). $$ Using (1) and introducing $k+1=p, m+1=q$, we get $$S=\sum_{p=q}^{2q-1} q ~2^{-(p-1)} {p \choose q} -1= \sum_{p=q}^{2q} 2q ~2^{-p} {p \choose q}-2q~ 2^{-2q} {2q \choose q}-1=2q-1-2q~2^{-2q}{2q \choose q}.$$ $$\implies S= 2m+1-2(m+1) 2^{-2m-2} {2m+2 \choose m+1}.$$ $$\implies \sum_{k=m}^{2m} k~ 2^{-k} {k \choose m}=(2m+1)- 2^{-2m-1}~ (m+1) ~{2m+2 \choose m+1}$$ Other proofs are welcome.