$$ {{\sum_{k<\frac{n}{2}}(n-2k){n \choose k}} \over {2^{n}-1}} $$
This problem should be solved using the Vandermonde Identity. But I am not able to solve it. I don't understand how to apply the identity to solve this problem. Are there any other ways to solve this problem ?
I am ignoring $\frac{1}{2^n-1}$ for now. We can add that back in the final result.
$ \displaystyle \sum \limits_{k=0}^{\lfloor \frac{n}{2} \rfloor} (n-2k) {n\choose k} = n\sum \limits_{k=0}^{\lfloor \frac{n}{2} \rfloor} {n\choose k} - 2\sum \limits_{k=0}^{\lfloor \frac{n}{2} \rfloor} k{n\choose k}$
$= n\sum \limits_{k=0}^{\lfloor \frac{n}{2} \rfloor} {n\choose k} - 2n\sum \limits_{k=1}^{\lfloor \frac{n}{2} \rfloor} {n-1 \choose k-1}$
If $n$ is odd,
$= \frac{n}{2} \sum \limits_{k=0}^{n} {n\choose k} - n\sum \limits_{k=1}^{n} {n-1 \choose k-1} + n {n-1 \choose {\lfloor \frac{n}{2} \rfloor}}$
$= \frac{n}{2} \sum \limits_{k=0}^{n} {n\choose k} - n\sum \limits_{k=0}^{n-1} {n-1 \choose k} + (n - {\lfloor \frac{n}{2} \rfloor}) {n \choose {\lfloor \frac{n}{2} \rfloor}}$
$ = ({\lfloor \frac{n}{2} \rfloor} + 1) {n \choose {\lfloor \frac{n}{2} \rfloor} + 1}$
$ = m {n \choose m} \, \,$ where $m = {\lfloor \frac{n}{2} \rfloor} + 1$
If $n$ is even,
$= \frac{n}{2} \sum \limits_{k=0}^{n} {n\choose k} + \frac{n}{2} {n \choose {\lfloor \frac{n}{2} \rfloor}} - n\sum \limits_{k=0}^{n-1} {n-1 \choose k}$
$= \frac{n}{2} {n \choose {\lfloor \frac{n}{2} \rfloor}} = (\lfloor \frac{n}{2} \rfloor + 1) {n \choose {\lfloor \frac{n}{2} \rfloor}+1}$
$ = m {n \choose m} \, \,$ where $m = {\lfloor \frac{n}{2} \rfloor} + 1$
So your answer is $\displaystyle \frac {m {n \choose m}}{2^n - 1} \,$ where $m = {\lfloor \frac{n}{2} \rfloor} + 1$