NCERT solution Class 11 principle of mathematical induction

NCERT Solutions

Class 11 Maths

Principle of Mathematical Induction

NCERT Questions

Ex.4.1 Q.1
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Ex.4.1 Q.11
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Ex.4.1 Q.20
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Ex.4.1 Q.24

Ex.4.1 Q.20

Prove the following by using the principle of mathematical induction for all n ∈ N:

10^{2n – 1} + 1 is divisible by 11.

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.

So, 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

= 10²(10^{2k + 2 - 1} + 1 - 1) + 1

= 10²(10^{2k + 2 - 1} + 1) - 10² + 1

= 10^2. 11m – 100 + 1

= 100 × 11m – 100 + 1

= 100 × 11m – 99

= 11(100m – 9)

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

Therefore, 10^{2(k + 1) - 1} + 1is 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.

Principle of Mathematical Induction Class 11 Maths NCERT Chapter 4 NCERT Solutions

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