Class XI Chapter 4 – Principle of Mathematical Induction Maths ______________________________________________________________________________ 2 2k1 10 10 1 1 1 2 2 k 10 10 1 1 10 2 1 2 10 .**11**m 100 1 100**11**m 99 **11** 100m 9 **11** r, where r 100m 9 Therefore, Thus, P k 1 2 k1 1 10 1 [Using (1)] is some natural number is divisible by **11**. 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. Question 21: Prove the following by using the principle of mathematical induction for all x y 2n 2n Solution 21: is divisible by x y. Let the given statement be 2n 2 : P n x y It can be observed that This is so because Let x Pk y 2k 2k Let n , P n i.e., is divisible by x y . Pn is true for n = 1. 21 21 2 2 x y x y x y x y be true for some positive integer k, i.e., is divisible by x y. 2k 2k x y m x y We shall now prove that Consider x 2k1 2 k1 y x x y y 2k 2 2k 2 , where P k 1 x x y y y y 2 2 k 2 k 2 k 2 k 2 x m x y y y y 2 2 k 2 k 2 2 2k 2 2k 2 m x y x y x y y m N ..... 1 is true whenever P(k) is true. x y is divisible by [Using (1)] n N: .

Class XI Chapter 4 – Principle of Mathematical Induction Maths ______________________________________________________________________________ 2 2 k 2 2 m x y x y x y 2 2k m x y x y x y x y 2 2k x y mx y x y x y , which is a factor of Thus, Pk 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. . Question 22: Prove the following by using the principle of mathematical induction for all is divisible by 8. Solution 22: Let the given statement be P(n), i.e., 2 2 P n n :3 8n 9 is divisible by 8. It can be observed that P(n) is true for n = 1 Since 2 1 2 3 819 64, which is divisible by 8. Let P(k) be true for some positive integer k 2k 2 , i.e., 3 8k9 2k2 3 8k 9 8 m; We shall now prove that Consider 2 k1 2 3 8 k 1 9 2k 2 2 3 3 8k 8 9 is divisible by 8. where m N....... 1 P k 1 2 2k2 3 3 8k 9 8k 9 8k 17 k 2 2 2 2 3 3 8k 9 3 8k 9 8k 17 9.8m 9 8k 9 8k 17 9.8m 72k 818k 17 9.8m 64k 64 8 9m 8k 8 8, r Therefore, where r 9m 8k 8 is true whenever is a natural number 2 k1 2 3 8 k 1 9 is divisible by 8. Pk is true. n N n 2n2 :3 8 9

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