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Ex 4.1, 20 - Prove: 102n-1 + 1 is divisble by 11 - Chapter 4 - Divisible

Ex 4.1, 20 - Chapter 4 Class 11 Mathematical Induction - Part 2
Ex 4.1, 20 - Chapter 4 Class 11 Mathematical Induction - Part 3 Ex 4.1, 20 - Chapter 4 Class 11 Mathematical Induction - Part 4

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Ex 4.1,20 Prove the following by using the principle of mathematical induction for all n ∈ N: 102n – 1 + 1 is divisible by 11. Introduction If a number is divisible by 11, 22 = 11 × 2 = 11 × 7 = 11 × 9 Any number divisible by 11 = 11 × Natural number Ex 4.1,20 Prove the following by using the principle of mathematical induction for all n ∈ N: 102n – 1 + 1 is divisible by 11. Let P(n): 102n – 1 + 1 = 11d where d ∈ N For n=1, L.H.S = 102.1 – 1 + 1 = 101 + 1 = 10 + 1 = 11 = 11 × 1 = R.H.S ∴P(n) is true for n = 1 Assume P(k) is true 102k – 1 + 1 = 11m, where m ∈ N We will prove that P(k + 1) is true. L.H.S = 102(k+1) – 1 + 1 = 102k+2 - 1 + 1 = 10( 2k - 1) + 2 + 1 = 10(2k - 1) .102 + 1 = (11m – 1) .102 + 1 = (11m – 1) × 100 + 1 = 100 × 11m – 100 + 1 = 100 × 11m – 99 = 100 × 11m – 9 × 11 = 11 (100m – 9) = 11r ,where r = (100m – 9) is some 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.