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# CLASS_11_MATHS_SOLUTIONS_NCERT

## Class XI Chapter 4 –

Class XI Chapter 4 – Principle of Mathematical Induction Maths ______________________________________________________________________________ We shall now prove that Pk 1 is true whenever P(k) is true. Consider 1 2 .... k k 1 2k 1 2 k 1 1 2 1 8 1 8 2 k k 1 4 2 k 4 k 1 8 k 8 8 1 4 2 k 12 k 9 8 1 2k 3 8 2 1 2 8 1 1 k 2 1 8 1 2 1 2 3 .... k k 1 2k 1 k 1 8 Hence, Thus, P k 1 is true whenever P(k) is true. [Using (1)] Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N. Question 19: Prove the following by using the principle of mathematical induction for all 1n 5 n n is a multiple of 3. Solution 19: Let the given statement be P(n), i.e., P n : n n 1 n 5 , It can be noted that Pn which is a multiple of 3. is true for n 1 since Let P(k) be true for some positive integer k, i.e., 1k 5 k k is a multiple of 3. k k 1 k 5 3 m, We shall now prove that Consider k k k where P k 1 1 1 1 1 5 m N .... 1 1 11 1 5 12 is true whenever P(k) is true. n N: , which is a multiple of 3.

Class XI Chapter 4 – Principle of Mathematical Induction Maths ______________________________________________________________________________ k k k 1 2 5 1 k 1k 2k 5 k 1k 2 k k 1k 5 2k 1k 5 k 1k 2 3m k 1 2 k 5 k 2 3m k 1 2k 10 k 2 3m k 1 3k 12 3m 3 k 1 k 4 Therefore, 3 m k 1 k 4 3 q, where Thus, P k 1 k 1 k 1 1 k 1 5 is true whenever Pk q m k 1 k 4 is a multiple of 3. is true. is some natural number. Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N. Question 20: Prove the following by using the principle of mathematical induction for all 2 1 10 n 1 is divisible by 11. Solution 20: Let the given statement be P(n), i.e., 2n 1 P n :10 1 It can be observed that Since Let i.e., 2.11 is divisible by 11. Pn P 1 10 1 11 Pk is true for n = 1 , which is divisible by 11. be true for some positive integer k, 2k 1 10 1 is divisible by 11. 2k1 10 1 11 m, where m N .... 1 We shall now prove that Pk 1 is true whenever P(k) is true. Consider 2 k1 1 10 1 2k 21 10 1 2k 1 10 1 n N:

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Class XI Chapter 2 - Relations and

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Class XI Chapter 11 - Conic Section

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Class XI Chapter 12 - Introduction

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Class XI Chapter 13 - Limits and De

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Class XI Chapter 14 - Mathematical

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Class XI Chapter 15 - Statistics Ma

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