Question 20:

Prove the following by using the principle of mathematical induction for all n ∈ N: 10^{2n − 1} + 1 is divisible by 11.

Answer

Let the given statement be P(n), i.e.,

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

It can be observed that P(n) is true for n = 1 since P(1) = 10^{2.1 – 1} + 1 = 11, which is divisible by 11.

Let P(k) be true for some positive integer k,

i.e., 10^{2k − 1} + 1 is divisible by 11.

∴10^{2k −1} + 1 = 11m, where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

10^2(k+1)−1 + 1

= 10^2k+2−1 + 1

=10^2k+1 + 1

=10²(10^2k−1 + 1 – 1) + 1

=10²(10^2k−1 + 1) – 10² + 1

=10² ∙ 11m – 100 + 1                [Using (1)]

= 100 × 11m – 99

=11(100m – 9)

=11r, where r = (100m – 9) is some natural number

Therefore, 10^{2(k+1) −1} + 1 is divisible by 11.

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

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

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