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Example 2 - Prove 2n > n - Chapter 4 Mathematical Induction - Inequality

Example 2 - Chapter 4 Class 11 Mathematical Induction - Part 2
Example 2 - Chapter 4 Class 11 Mathematical Induction - Part 3 Example 2 - Chapter 4 Class 11 Mathematical Induction - Part 4


Transcript

Example 2 Prove that 2﷮𝑛﷯>𝑛 for all positive integers n. Let P(n) : 2﷮𝑛﷯>𝑛 for all positive n For n = 1 L.H.S = 2﷮𝑛﷯ = 2﷮1﷯ = 1 R.H.S = n = 1 Since 2 > 1 L.H.S > R.H.S ∴ P(n) is true for n = 1. Assume that P(k) is true for all positive integers k i.e. 2﷮k﷯> k We will prove that P(k + 1) is true. i.e 2﷮𝑘 + 1﷯> k + 1 From (1) 2﷮k﷯> k Multiplying by 2 on both sides. 2﷮k﷯ × 2 > 2 × k 2.2﷮k﷯ > 2 k 2﷮k + 1﷯ > k + k Now, k is positive We have proved P(1) is true So we have to prove for k > 1 k > 1 Adding k both sides k + k > k + 1 From (2) and (3) 2﷮k + 1﷯ > k + k and k + k > k + 1 Hence 2﷮k + 1﷯ > k + 1 ∴ L.H.S > R.H.S ∴ P (k + 1) is true whenever p(k) is true. ∴ By the principal of mathematical induction, P(n) is true for n, is a positive integer.

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Davneet Singh

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.