Prove the following by using the principle of mathematical induction for all n ∈ N :
10²ⁿ ⁻ ¹ + 1 is divisible by 11

Solution:

We can write

P (n) : 10^{2n - 1} + 1 is divisible by 11

We note that

P (1) : 10^{2.1 - 1} + 1 = 10 + 1, which is divisible by 11.

Thus P (n) is true for n = 1

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

i.e., P (k) : 10^{2k - 1} + 1 is divisible by 11

We can write

10^{2k - 1} + 1 = 11a .... (1)

where a ∈ N

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

Now,

10 ^{2(k + 1) - 1} + 1

= 10^{2k + 1} + 1

= 10² (10^{2k - 1} ) + 1

= 10² (10^{2k - 1} + 1 - 1) + 1 (added and subtracted 1)

= 10² (10^{2k - 1} + 1) - 10² + 1

= 10².11a - 100 + 1 ... [from (1)]

= 10².11a - 99

= 11(100a - 9), which is divisible by 11.

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 20

Prove the following by using the principle of mathematical induction for all n ∈ N : 10²ⁿ ⁻ ¹ + 1 is divisible by 11

Summary:

We have proved that 10²ⁿ ⁻ ¹ + 1 is divisible by 11 by using the principle of mathematical induction for all n ∈ N