Buktikan: n(n+1)(n+2) habis dibagi 3

n(n+1)(n+2) habis dibagi 3 
n(n+1)(n+2) = 3m

n = 1 → 6 hbs dibagi 3
n = k → k(k+1)(k+2)  = 3m
n = k+ 1 → (k+1)(k+2)(k+3) = k³ + 6k² + 11 k + 6
(k+1)(k+2)(k+3) = (k³ + 3k² + 2k ) +( 3k² + 9k +6)
(k+1)(k+2)(k+3) = k(k+1)(k+2) + 3(k² + 3k + 2)
(k+1)(k+2)(k+3) = 3m + 3 (k + 1)(k+2)
(k+1)(k+2)(k+3) = 3 (m + (k+1)(k+2)) → m + (k+1)(k+2) bilangan bulat
maka  n(n+1)(n+2) habis dibagi 3 untuk n asli