# 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) : 41^{n} - 14^{n }is a multiple of 27

We note that

P (1) : 41^{1} - 14^{1} = 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 .

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.

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