TITLE: How find this sum $\sum_{k=0}^{n}\binom{n}{k}|n-2k|$ closed form or asymptotic behaviour?
QUESTION [5 upvotes]: Find the following series   closed form or asymptotic behaviour
$$\dfrac{\displaystyle \sum_{k=0}^{n}\binom{n}{k}|n-2k|}{2^n}$$
I use wolfram can't give the  closed form: see wolfram ,so I think maybe can find the asymptotic expansion?
I think this problem is equivalent find following closed( or asymptotic behaviour)
\begin{align*}
&\sum_{k=0}^{[n/2]}\binom{n}{k}(n-2k)+\sum_{k=[n/2]+1}^{n}\binom{n}{k}(2k-n)\\
&=n\left(\sum_{k=0}^{[n/2]}\binom{n}{k}-\sum_{k=[n/2]+1}^{n}\binom{n}{k}\right)-2\left(\sum_{k=1}^{[n/2]}k\binom{n}{k}-\sum_{k=[n/2]+1}^{n}k\binom{n}{k}\right)\\
&=-2\left(\sum_{k=1}^{[n/2]}k\binom{n}{k}-\sum_{k=[n/2]+1}^{n}k\binom{n}{k}\right)\\
\end{align*}

REPLY [8 votes]: $$\begin{align*}
\sum_{k=0}^n\binom{n}k|n-2k|&=2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}k(n-2k)\\\\
&=2\left(n\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}k-2\sum_{k=0}^{\lfloor n/2\rfloor}k\binom{n}k\right)\\\\
&=2n\left(\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}k-2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-1}{k-1}\right)\\\\
&=2n\left(\sum_{k=0}^{\lfloor n/2\rfloor}\left(\binom{n}k-\binom{n-1}{k-1}\right)-\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-1}{k-1}\right)\\\\
&=2n\left(\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-1}k-\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-1}{k-1}\right)\\\\
&=2n\left(\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-1}k-\sum_{k=0}^{\lfloor n/2\rfloor-1}\binom{n-1}k\right)\\\\
&=2n\binom{n-1}{\lfloor n/2\rfloor}\;.
\end{align*}$$
It doesn’t matter whether $n$ is odd or even: if $n$ is even, the $k=\frac{n}2$ term is zero anyway.