1. Class 11
2. Important Question for exams Class 11

Transcript

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

Class 11
Important Question for exams Class 11

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.