Using the principal of mathematical induction, prove for n belonging to N:

n(n+1)(n+2) is a multiple of 6

P(n) : n(n+1)(n+2) is a multiple of 6

P(1) = 1(1+1) (1+2) = 6

P(1) is true.

let P(k) be also true.

k(k+1)(k+2)

= (k²+k)(k+2)

= k³ +3k²+2k

now,

(k+1)³ + 3 (k+1)² + 2(k+1)

= k³ +1+ 3k²+3k +3k²+6k+3 +2k+2

= k³+6k²+11k+6

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

= (k+1) ((k+1)+1) ((k+1)+2)

P(k+1) is also true.

therefore , P(n) is true for all values of n belonging to N , by PMI.