\[\sum_{j=1}^{d}e_{j,d}^{2}\frac{(n+m+2j-2p)!}{(n-m+2j-2p)!}\frac{1}{2n+4j-4p+1}%
=\frac{(n+m)!}{(n-m)!}\frac{1}{2n+1}.\]