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.
= k3 +3k2+2k
(k+1)3 + 3 (k+1)2 + 2(k+1)
= k3 +1+ 3k2+3k +3k2+6k+3 +2k+2
= (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.