We can write the given statement as

P (n): n (n + 1) (n + 5), which is a multiple of 3

If n = 1 we get

1 (1 + 1) (1 + 5) = 12, which is a multiple of 3

Which is true.

Consider P (k) be true for some positive integer k

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

k (k + 1) (k + 5) = 3m, where m ∈ N …… (1)

Now let us prove that P (k + 1) is true.

Here

(k + 1) {(k + 1) + 1} {(k + 1) + 5}

We can write it as

= (k + 1) (k + 2) {(k + 5) + 1}

By multiplying the terms

= (k + 1) (k + 2) (k + 5) + (k + 1) (k + 2)

So we get

= {k (k + 1) (k + 5) + 2 (k + 1) (k + 5)} + (k + 1) (k + 2)

Substituting equation (1)

= 3m + (k + 1) {2 (k + 5) + (k + 2)}

By multiplication

= 3m + (k + 1) {2k + 10 + k + 2}

On further calculation

= 3m + (k + 1) (3k + 12)

Taking 3 as common

= 3m + 3 (k + 1) (k + 4)

We get

= 3 {m + (k + 1) (k + 4)}

= 3 × q where q = {m + (k + 1) (k + 4)} is some natural number

(k + 1) {(k + 1) + 1} {(k + 1) + 5} is a multiple of 3

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

Therefore, by the principle of mathematical induction, statement P (n) is true for all natural numbers i.e. n.

Answered by Abhisek | 1 year agoGiven an example of a statement P (n) such that it is true for all n ϵ N.

a + (a + d) + (a + 2d) + … + (a + (n-1)d) = \( \dfrac{n}{2}\) [2a + (n-1)d]

7^{2n} + 2^{3n – 3}. 3n – 1 is divisible by 25 for all n ϵ N

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

(ab)^{ n} = a^{n} b^{n} for all n ϵ N