Calculation of the moments using Hypergeometric distribution

Let vector $a\in 2n $ is such that first $l$ of its coordinates are $1$ and the rest are $0$ ($a=(1,\ldots, 1,0, \ldots, 0)$). Let $\pi$ be $k$-th permutation of set $\{1, \ldots, 2n\}$.
Define
$$g=\left|\sum_{i=1}^n a_{\pi(i)}-\sum_{i=n+1}^{2n}a_{\pi(i)}\right|.$$

Using Hypergeometric distribution calculate /approximate the $q$-th moment $E|g|^q,$ for any $q\ge 2$.

I’ve got that the $q$-th moment is
$$
E|g|^q=\sum_{k=0}^l\frac{{l \choose k}{2n-l \choose n-k}(2k-l)^q}{{2n\choose n}}.
$$
But now I am stuck…

Thank you for your help.

Solutions Collecting From Web of "Calculation of the moments using Hypergeometric distribution"

By comparing the last expression to the probability function of the hypergeometric distribution, you see that $E|g|^q=E(2X−l)^q$, where $X$ is $\rm{Hypergeometric}(2n,l,n).$

Therefore $E(X)=\frac{nl}{2n}=l/2=:\mu$. Thus
$$E|g|^q=E(2X−l)^q={2}^qE(X-l/2)^q=2^qE(X-\mu)^q.$$

Expressed in words, $E|g|^q$ is $2^q$ times the $q$:th central moment of $X$.

The central moments of the hypergeometric distribution are known and can be computed (preferably not by hand…).