Closed Form for $\sum\limits_{n=-2a}^\infty(n+a){2a\choose-n}^4$

When $2a$ is a non-negative integer: $$S=\sum_{n=-2a}^{\infty} (n+a) {2a \choose -n}^4= \sum_{n=0}^{2a}(-n+a) {2a \choose n}^4=\sum_{n=0}^{2a}~(-(2a-n)+a)~{2a\choose n}^4$$ $$=\sum_{n=0}^{2a} (n-a) {2a \choose n}^4~ =-S.$$ So $$S=-S \implies S=0$$.Here we have used : ${n \choose-k}= 0~$ if $k\in I^+$, $\sum_{k=0}^{n} f(k) =\sum_{k=0}^n f(n-k)$, and ${n \choose k}={n \choose n-k}$