∑l(-1)l+j+j1+j2⁢1l⁢(l+1)⁢{j1j2lj2j1j}=1j1⁢(j1+1)-j2⁢(j2+1)⁢((2⁢j1-j)!⁢(2⁢j2+j+1)!(2⁢j2-j)!⁢(2⁢j1+j+1)!)12,