$P\left(n\right):(n+1)(n+2)(n+3)P\left(1\right):2\times 3\times 4=24\to \text{divisible by 6}P\left(n\right):(n+1)(n+2)(n+3)\text{is divisible by 6}$
P(1) is true. Assume P(k) is true
$\Rightarrow (k+1)(k+2)(k+3)=6mP(k+1):(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)+3(k+2)(k+3)$
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
$\therefore (k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)+3.2q=6m+6q=6(m+q)\to \text{divisible by 6}\Rightarrow \text{P(k+1) is true}\Rightarrow \text{P(n) is true}$