m en n zijn natuurlijke getallen met inbegrip van 0
\(X_1\ f(m,0) = 0^mX_2\ f(m,n+1)=f(m,n) + (n+1)^m\)
Voorbeeld f(0,n)\(f(0,0)=0^0f(0,1)=0^0 + 1^0f(0,2)=0^0 + 1^0 + 2^0f(0,3)=0^0 + 1^0 + 2^0 + 3^0\ldots\ldotsf(0,n)=0^0 + 1^0 + 2^0 + 3^0 + 4^0+\ldots + n^0\)
Of f(3,n)\(f(3,0)=0^3f(3,1)=0^3 + 1^3f(3,2)=0^3 + 1^3 + 2^3f(3,3)=0^3 + 1^3 + 2^3 + 3^3\ldots\ldotsf(3,n)=0^3 + 1^3 + 2^3 + 3^3 + 4^3+\ldots + n^3\)
Er geldt:\(f(0,n)=n+1\)
\(f(1,n)=\frac{1}{2}n^2+\frac{1}{2}n\)
\(f(2,n)=\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n\)
\(f(3,n)=\frac{1}{4}n^4+\frac{1}{2}n^3+\frac{1}{4}n^2\)
\(f(4,n)=\frac{1}{5}n^5+\frac{1}{2}n^4+\frac{1}{3}n^3-\frac{1}{30}n\)
\(f(5,n)=\frac{1}{6}n^6+\frac{1}{2}n^5+\frac{5}{12}n^4-\frac{1}{12}n^2\)
etc.Kan dit nu beschreven worden in 1 formule? Dus f(m,n)=….
