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)

= (k^{2}+k)(k+2)

= k^{3} +3k^{2}+2k

now,

(k+1)^{3} + 3 (k+1)^{2 }+ 2(k+1)

= k^{3} +1+ 3k^{2}+3k +3k^{2}+6k+3 +2k+2

= k^{3}+6k^{2}+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.