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Last updated at Feb. 15, 2020 by Teachoo

Transcript

Ex 4.1,1 Prove the following by using the principle of mathematical induction for all n N: 1 + 3 + 32+ + 3n 1 = ((3 1))/2 Let P (n) : 1 + 3 + 32+ + 3n 1 = ((3 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 = ((3 1))/2 We will prove that P(k + 1) is true. 1 + 3 + 32 + ..+ 3(k + 1) 1 = ((3^( +1) 1))/2 1 + 3 + 32 + ..3(k 1) + 3(k) = ((3^( +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 = ((3 1))/2 Adding 3k both sides 1 + 3 + 32 + ..+ 3k 1 + 3k = ((3 1))/2 + 3k = ((3 1) + 2(3^ ))/2 = (3 1 + 2(3^ ))/2 = ( 3(3^ ) 1)/2 = (3^( +1) 1)/2 Thus, 1 + 3 + 32 + ..3(k 1) + 3(k) = ((3^( +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

Ex 4.1

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.