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Ex 4.1, 23 - Prove: 41n - 14n is a multiple of 27 - Class 11

Ex 4.1, 23 - Chapter 4 Class 11 Mathematical Induction - Part 2
Ex 4.1, 23 - Chapter 4 Class 11 Mathematical Induction - Part 3
Ex 4.1, 23 - Chapter 4 Class 11 Mathematical Induction - Part 4

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Ex 4.1,23 Prove the following by using the principle of mathematical induction for all n N: 41n 14n is a multiple of 27. Introduction If a number is multiple of 27, then it will come in table of 27 27 1 = 27 27 2 = 54 27 3 = 71 Any number multiple of 27 = 27 Natural number Ex 4.1,23 Prove the following by using the principle of mathematical induction for all n N: 41n 14n is a multiple of 27. Let P(n):41n 14n = 27d , where d N For n=1, L.H.S = 411 141 = 41 14 = 27 = 27 1 = R.H.S P(n) is true for n = 1 Assume P(k) is true 41k 14k = 27m, where m N We will prove that P(k + 1) is true L.H.S = 41k+1 14k+1 = 41k . 411 14k . 141 = 41k . 41 14k . 14 = (27m + 14k) 41 14k . 14 = 41 27m + 41 14k 14k . 14 = 41 27m + 14k (41 14) = 41 27m + 14k (27) = 27 (41m 14k) = 27 r, where r = (41m 14k ) is a natural number P(k + 1) is true whenever P(k) is true. By the principle of mathematical induction, P(n) is true for n, where n is a natural number

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.