Prove the following by using the principle of mathematical induction for all n ∈ N :
41ⁿ - 14ⁿ is a multiple of 27

Solution:

We can write

P (n) : 41ⁿ - 14ⁿ is a multiple of 27

We note that

P (1) : 41¹ - 14¹ = 41 - 14 = 27, which is a multiple of 27

Thus, P (n) is true for n = 1

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

i.e., P (k) : 41^k - 14^k is a multiple of 27.

We can write

41^k - 14^k = 27a .... (1)

where a ∈ N .

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

Now,

41^{k + 1} - 14^{k + 1}

= 41.41^k - 14.14^k

= 41.(41^k - 14^k + 14^k ) - 14.14^k (added and subtracted 14^k)

= 41.(41^k - 14^k ) + 41.14^k - 14.14^k

= 41.27a + 14^k (41 - 14) .... [from (1)]

= 41.27a + 14^k.27

= 27 (41a + 14^k ), which is a multiple of 27.

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 23

Prove the following by using the principle of mathematical induction for all n ∈ N : 41ⁿ - 14ⁿ is a multiple of 27.

Summary:

We have proved that 41ⁿ - 14ⁿ is a multiple of 27 by using the principle of mathematical induction for all n ∈ N.