∑(2⁢n)!(2⁢j)!⁢(2⁢k)!⁢(2⁢ℓ)!⁢B2⁢j⁢B2⁢k⁢B2⁢ℓ=(n-1)⁢(2⁢n-1)⁢B2⁢n+n⁢(n-12)⁢B2⁢n-2,