Prove the following by using the principle of mathematical induction for all n ∈ N :
3²ⁿ⁺² - 8n - 9 is divisible by 8

Solution:

We can write,

P (n) : 3^{2n + 2} - 8n - 9 is divisible by 8

We note that

P (1) : 3^{2.1 + 2} - 8n - 9 = 3⁴ - 8 - 9 = 81 - 17 = 64, which is divisible by 8.

Thus P (n) is true for n = 1

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

i.e., P (k) : 3^{2k + 2} - 8k - 9 is divisible by 8

We can write

3^{2k + 2} - 8k - 9 = 8a ....(1)

where a ∈ N .

Now, we will prove that P (k + 1) is true whenever P (k) is true.

Now,

3^{2(k+ 1) + 2} - 8(k + 1) - 9

= 3^{2k + 4} - 8k - 8 - 9

= 3².3^{2k+ 2} - 8k - 17

= 3² (3^{2k+ 2} - 8k - 9 + 8k + 9) - 8k - 17 (added and subtracted 8k and 9)

= 3² (3^{2k+ 2} - 8k - 9) + 3² (8k + 9) - 8k - 17

= 3².8a + 72k + 81- 8k - 17 .... [from (1)]

= 9.8a + 64k + 64

= 8(9a + 8k + 8)

From the last line, we see that

8(9a + 8k + 8) is divisible by 8.

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 22

Prove the following by using the principle of mathematical induction for all n ∈ N : 3²ⁿ⁺² - 8n - 9 is divisible by 8

Summary:

We have proved that 3²ⁿ⁺² - 8n - 9 is divisible by 8 by using the principle of mathematical induction for all n ∈ N