Bonjour ,

n ∈ IN donc n=3k ou n=3k+1 ou n=3k+2

1)

n=3k :

n(n+1)(n+2)/3=3k(3k+1)(3k+2)/3=k(3k+1)(3k+2) ∈ IN

2)

n=3k+1 :

n(n+1)(n+2)/3=(3k+1)(3k+2)(3k+3)/3=(3k+1)(3k+2)(3)(k+1)/3

=(3k+1)(3k+2)(k+1) ∈ IN

n=3k+2 :

n(n+1)(n+2)/3=(3k+2)(3k+3)(3k+4)/3=(3k+2)(3)(k+1)(3k+2)/3

=(3k+2)(k+1)(3k+2) ∈ IN