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  1. Chapter 4 Class 11 Mathematical Induction
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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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