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Example 5 - Prove (1 + x)n >= (1 + nx) - Mathematical Induction - Inequality

Example 5 - Chapter 4 Class 11 Mathematical Induction - Part 2
Example 5 - Chapter 4 Class 11 Mathematical Induction - Part 3 Example 5 - Chapter 4 Class 11 Mathematical Induction - Part 4


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Example5 Prove that (1 + x)n ≥ (1 + nx), for all natural number n, where x > – 1. Introduction Since 10 > 5 then 10 > 4 + 1 then 10 > 4 We will use this theory in our question Example 5 Prove that (1 + x)n ≥ (1 + nx), for all natural number n, where x > – 1. Let P(n): (1 + x)n ≥ (1 + nx), for x > – 1. For n = 1, L.H.S = (1 + x)1 = (1 + x) R.H.S = (1 + 1.x) = (1 + x) L.H.S ≥ R.H.S, ∴P(n) is true for n = 1 Assume P(k) is true (1 + x)k ≥ (1 + kx), x > – 1 We will prove that P(k + 1) is true. L.H.S = (1 + x)k + 1 R.H.S = (1 + (k+1)x) L.H.S ≥ R.H.S ∴ 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

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.