Last updated at March 9, 2017 by Teachoo

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Ex 4.1,1Prove the following by using the principle of mathematical induction for all n ∈ N: 1 + 3 + 32+……+ 3n – 1 = ((3𝑛 − 1))/2 Let P (n) : 1 + 3 + 32+……+ 3n – 1 = ((3𝑛 − 1))/2 For n = 1, L.H.S = 1 R.H.S = ((31 − 1))/2 = ((3 − 1))/2 = ((2))/2 = 1 L.H.S. = R.H.S ∴P(n) is true for n = 1 Assume that P(k) is true 1 + 3 + 32 +…..+ 3k – 1 = ((3𝑘 − 1))/2 We will prove that P(k + 1) is true. 1 + 3 + 32 +…..+ 3(k + 1) – 1 = ((3^(𝑘+1) − 1))/2 1 + 3 + 32 +…..3(k – 1) + 3(k) = ((3^(𝑘+1) − 1))/2 We have to prove P(k+1) from P(k) i.e. (2) from (1) From (1) 1 + 3 + 32 +…..+ 3k – 1 = ((3𝑘 − 1))/2 Adding 3k both sides 1 + 3 + 32 +…..+ 3k – 1 + 3k = ((3𝑘 − 1))/2 + 3k = ((3𝑘 − 1) + 2(3^𝑘 ))/2 = (3𝑘 − 1 + 2(3^𝑘 ))/2 = ( 3(3^𝑘 )− 1)/2 = (3^(𝑘+1) − 1)/2 Thus, 1 + 3 + 32 +…..3(k – 1) + 3(k) = ((3^(𝑘+1) − 1))/2 ∴ P(k+1) is true when 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 is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.