For all natural numbers n, (n+1)(n+2)(n+3) is divisible by
6
P(n):(n+1)(n+2)(n+3)P(1):2×3×4=24→divisible by 6P(n):(n+1)(n+2)(n+3) is divisible by 6
P(1) is true. Assume P(k) is true
⇒(k+1)(k+2)(k+3)=6mP(k+1):(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)+3(k+2)(k+3)
Since (k+2) and (k+3) are two consecutive natural numbers, one of them must be even i.e. their product is even, say 2q
∴(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)+3.2q=6m+6q=6(m+q)→divisible by 6⇒P(k+1) is true⇒P(n) is true