Prove the following by using the principle of mathematical induction for all n ∈ N:  $n(n+1)(n+5)$ is a multiple of 3.

Let the given statement be $P(n)$, i.e.,
$P(n): n(n+1)(n+5)$ is a multiple of 3.
It can be noted that P(n) is true for $n=1$ since $1(1+1)(1+5)=12$, which is a multiple of 3.
Let $P(k)$ be true for some positive integer $k$, i.e.,
$K(k+1)(k+5)$ is a multiple of 3.
$\therefore k(k+1)(k+5)=3m$, where $m \in N$ ----------(1)
We shall now prove that $P(k+1)$ is true whenever $P(k)$ is true.
Consider
$(k+1) \{(k+1)+1\}\{(k+1)+5\}$
$= (k+1)(k+2) \{ (k+5)+1\}$
$= (k+1)(k+2)(k+5)+(k+1)9k+2)$
$=\{ k(k+1)(k+5)+2(k+1)(k+5) \}+(k+1)(k+2)$
$= 3m+(k+1)\{ 2 (k+5)+(k+2) \}$
$= 3m +(k+1) \{2k+10+k+2\}$
$= 3m +(k+1) (3k+12)$
$= 3m+3(k+1)(k+4)$
$= 3\{ m+(k+1)(k+4)\}= 3 \times q$, where $q=\{ m+(k+1)(k+4) \}$ is some natural number
Therefore, $(k+1) \{ (k+1)+1\}\{(k+1)+5\}$ is a multiple of 3.
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.