Calculate sum with binom. coef.

Question

I'd like to get a closed form(i.e. formula) for this sum: $$ 2^{-n}\sum_{k=0}^{n}{k \dbinom{n}{\frac{n+k}{2}}},$$ where $\dbinom{n}{\frac{n+k}{2}} = 0$ if $\frac{n+k}{2}$ is not integer. How to do it if it's possible?

Answer 1

We consider the even case $2n$. The odd case can be done similarly.

We obtain \begin{align*} \color{blue}{\sum_{{k=0}\atop{k\equiv 0(2)}}^{2n}}&\color{blue}{k\binom{2n}{n+\frac{k}{2}}}\\ &=\sum_{k=0}^n2k\binom{2n}{n+k}\tag{1}\\ &=2\sum_{k=0}^n(n+k)\binom{2n}{n+k}-2n\sum_{k=0}^n\binom{2n}{n+k}\tag{2}\\ &=4n\sum_{k=0}^n\binom{2n-1}{n+k-1}-2n\left(2^{2n-1}+\frac{1}{2}\binom{2n}{n}\right)\tag{3}\\ &=4n\left(2^{2n-1}+\binom{2n-1}{n-1}\right)-n2^{2n}-n\binom{2n}{n}\tag{4}\\ &\,\,\color{blue}{=n\binom{2n}{n}} \end{align*}

Comment:

• In (1) we replace $k$ with $2k$.

• In (2) we rearrange the sum as preparation for the next step.

• In (3) we apply $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$ to the left-hand sum and simplify the right hand sum.

• In (4) we simplify the left-hand sum.