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

For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3. 

So, the given statement is true for n = 1, i.e. P(1) is true. 

Let P(k) be true. Then, 

P(k): k(k + 1)(k + 5) is a multiple of 3

⇒ K(k + 1)(k + 5) = 3m for some natural number m, ... (i) 

Now, (k + 1l)(k + 2)(k + 6) = (k + 1)(k + 2)k + 6(k + 1)(k + 2) 

                             = k(k + 1)(k + 2) + 6(k + 1)(k + 2) 

                             = k(k + 1)(k + 5 – 3) + 6(k + 1)(k + 2) 

                             = k(k + 1)(k + 5) – 3k(k + 1) + 6(k + 1)(k + 2) 

                             = k(k + 1)(k + 5) + 3(k + 1)(k +4) [on simplification] 

                             = 3m + 3(k + 1 )(k + 4) [using (i)] 

                             = 3[m + (k + 1)(k + 4)], which is a multiple of 3

⇒ P(k + 1): (k + 1 )(k + 2)(k + 6) is a multiple of 3

⇒ P(k + 1) is true, whenever P(k) is true. 

Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. 

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.