How find this sum $\sum_{k=0}^{n}\binom{n}{k}|n-2k|$ closed form or asymptotic behaviour?

Question

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*}

Answer 1

Here is a very basic asymptotic behavior: Let $X_{1}, X_{2}, \cdots$ be i.i.d. symmetric Bernoulli trials with $\Bbb{P}(X_{i} = 1) = \Bbb{P}(X_{i} = -1) = 1/2$. Then for $S_{n} = X_{1} + \cdots + X_{n}$ we have

$$ \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} |n - 2k| = \Bbb{E}|S_{n}|. $$

From the central limit theorem, it follows that $S_{n}/\sqrt{n} \Rightarrow Z$ for a standard normal random variable $Z \sim \mathcal{N}(0, 1)$. So we have the following asymptotic relation

$$ \Bbb{E}|S_{n}| \sim \sqrt{n}\, \Bbb{E}|Z| = \sqrt{\frac{2n}{\pi}}. $$

Answer 2

$$\begin{align*} \sum_{k=0}^n\binom{n}k|n-2k|&=2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}k(n-2k)\\\\ &=2\left(n\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}k-2\sum_{k=0}^{\lfloor n/2\rfloor}k\binom{n}k\right)\\\\ &=2n\left(\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}k-2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-1}{k-1}\right)\\\\ &=2n\left(\sum_{k=0}^{\lfloor n/2\rfloor}\left(\binom{n}k-\binom{n-1}{k-1}\right)-\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-1}{k-1}\right)\\\\ &=2n\left(\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-1}k-\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-1}{k-1}\right)\\\\ &=2n\left(\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-1}k-\sum_{k=0}^{\lfloor n/2\rfloor-1}\binom{n-1}k\right)\\\\ &=2n\binom{n-1}{\lfloor n/2\rfloor}\;. \end{align*}$$

It doesn’t matter whether $n$ is odd or even: if $n$ is even, the $k=\frac{n}2$ term is zero anyway.