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

Solution:

We can write

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

We note that

P (1) : 1 (1 + 1)(1 + 5) = 1.2.6 = 12, which is a multiple of 3

Thus P (n) is true for n = 1

Let P (k) be true for some natural number k.

i.e., P (k) : k (k + 1)(k + 5) is a multiple of 3

We can write

k (k + 1)(k + 5) = 3a ....(1)

where a ∈ N

Now, we will prove that

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

Now,

(k + 1) [(k + 1) + 1] [(k + 1) + 5]

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

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

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

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

= [3a + 2 (k + 1)(k + 5)] + (k + 1)(k + 2) (from (1))

= 3a + (k + 1) [2 (k + 5) + (k + 2)]

= 3a + (k + 1)[2k + 10 + k + 2]

= 3a + (k + 1)[3k + 12]

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

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

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

Hence, from the principle of mathematical induction, the statement P (n) is true for all natural numbers i.e., n ∈ N .

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NCERT Solutions Class 11 Maths Chapter 4 Exercise 4.1 Question 19

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

Summary:

We have proved that n (n + 1)(n + 5) is a multiple of 3 by using the principle of mathematical induction for all n ∈ N