Mathematical Induction - Questions and Solutions

Mathematical Induction
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Question20 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 Question20 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

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.