Q_m,1 = Q_m-1,2
Q_m,n = Q_m,n-1 + C_m-1,n+1

\(Q_{3,n}=\dfrac{1}{6}n(n+4)(n+5)\)

\(Q_{4,n}=\displaystyle{\sum_{k=1}^n\dfrac{1}{6}(k+1)(k+5)(k+6)}\)

(k+1) (k+5) (k+6) = k (k+1) (k+2) + 9k (k+1) + 30 (k+1)
を用いて、
\(\displaystyle{\sum_{k=1}^n(k+1)(k+5)(k+6)}\)
\(=\dfrac{1}{4}n(n+1)(n+2)(n+3)+3n(n+1)(n+2)+15n(n+1)+30n\)
\(=\dfrac{1}{4}n(n^3+18n^2+107n+210)\)
\(=\dfrac{1}{4}n(n+5)(n+6)(n+7)\)
よって、 \(Q_{4,n}=\dfrac{1}{24}n(n+5)(n+6)(n+7)\)