# Example 2 - Chapter 4 Class 11 Mathematical Induction

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 2 Prove that 2𝑛>𝑛 for all positive integers n. Let P(n) : 2𝑛>𝑛 for all positive n For n = 1 L.H.S = 2𝑛 = 21 = 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. 2k> k We will prove that P(k + 1) is true. i.e 2𝑘 + 1> k + 1 From (1) 2k> k Multiplying by 2 on both sides. 2k × 2 > 2 × k 2.2k > 2 k 2k + 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) 2k + 1 > k + k and k + k > k + 1 Hence 2k + 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.