For $n=2,3,\dots$ and $k=1,2,\dots,n$, I would like to prove that $$ \sum_{\ell=k+1}^{n}\frac{2^{2\ell}}{\ell}{2n-2\ell\choose n-\ell}\left(1+\frac{2n-2\ell+1}{\ell-1}\right)=\frac{2^{2k}(2n-2k+1){2n-2k\choose n-k}}{k}. $$
I have checked it for small $n$ and $k$. However, I am unable to come up with a method to prove it. I am wondering if it is possible to (simplify) deal with $$ \sum_{\ell=k+1}^{n}\frac{2^{2\ell}}{\ell}{2n-2\ell\choose n-\ell}. $$
I would be appreciate it if someone could give me some help. Thanks a lot.