n(n + 1) (n + 5) is a multiple of 3 for all n ∈ N.

Let P(n) be the given statement.

Now,

$P(n): n(n+1)(n+5)$ is a multiple of 3 .

Step 1:

$P(1): 1(1+1)(1+5)=12$

It is a multiple of 3 .

Hence, $P(1)$ is true.

Step 2 :

Let $P(m)$ be true.

Then, $m(m+1)(m+5)$ is a multiple of 3 .

Suppose $m(m+1)(m+5)=3 \lambda$, where $\lambda \in N$.

We have to show that $P(m+1)$ is true whenever $P(m)$ is true.

Now,

$P(m+1)=(m+1)(m+2)(m+6)$

$=m(m+1)(m+6)+2(m+1)(m+6)$

$=m(m+1)(m+5+1)+2(m+1)(m+6)$

$=m(m+1)(m+5)+m(m+1)+2(m+1)(m+6)$

$=3 \lambda+(m+1)(m+2 m+6)$$[$ From $P(m)]$

$=3 \lambda+3(m+1)(m+2)$

It is clearly a multiple of 3 .

Thus, $P(m+1)$ is true.

By the principle of mathematical $i$ nduction, $P(n)$ is true for all $n \in N$.