Ex 4.1, 5 - Prove 1.3 + 2.32 + 3.33 + ... n.3n = (2n-1) 3n+1 - Equal - Addition

  1. Chapter 4 Class 11 Mathematical Induction
  2. Serial order wise
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Ex 4.1, 5: Prove the following by using the principle of mathematical induction for all n โˆˆ N: 1.3 + 2.32 + 3.33 + โ€ฆ.. n.3n = ((2๐‘› โˆ’ 1) 3^(๐‘› + 1) + 3 )/4 Let P(n) : 1.3 + 2.32 + 3.33 + โ€ฆ.. n.3n = ((2๐‘› โˆ’ 1) 3^(๐‘› + 1) + 3 )/4 For n = 1, we have L.H.S =1.3 = 3 R.H.S = ((2.1 โˆ’ 1) 3^(1+1) + 3)/4 = (1 ร— 3^2 + 3)/4 = (9 + 3)/4 = 12/4 = 3 Hence, L.H.S. = R.H.S โˆด P(n) is true for n = 1 Assume P(k) is true 1.3 + 2.32 + 3.33 + โ€ฆ.. k.3k = ((2๐‘˜ โˆ’ 1) 3^(๐‘˜ + 1) + 3 )/4 We will prove that P(k + 1) is true. 1.3 + 2.32 + 3.33 + โ€ฆ.. + (k + 1)3k + 1 = ((2(๐‘˜ + 1)โˆ’1) 3^((๐‘˜+1) + 1) + 3 )/4 1.3 + 2.32 + 3.33 + โ€ฆ.. + (k + 1)3k + 1 = ((2๐‘˜ + 2 โˆ’ 1) 3^(๐‘˜ + 2) + 3 )/4 1.3 + 2.32 + 3.33 + โ€ฆ.. + (k + 1)3k + 1 = ((2๐‘˜ + 1) 3^(๐‘˜ + 2) + 3 )/4 1.3 + 2.32 + 3.33 + โ€ฆ.. + k3k + (k + 1)3k + 1 = ((2๐‘˜ + 1) 3^(๐‘˜ + 2) + 3 )/4 We have to prove P(k+1) from P(k) i.e. (2) from (1) From (1) 1.3 + 2.32 + 3.33 + โ€ฆ.. k.3k = ((2๐‘˜ โˆ’ 1) 3^(๐‘˜ + 1) + 3 )/4 Adding (k + 1) 3k+1both sides 1.3 + 2.32 + 3.33 + โ€ฆ.. k.3k + (k + 1)3k + 1 = ((2๐‘˜ โˆ’ 1) 3^(๐‘˜ + 1) + 3 )/4 + (k + 1)3k + 1 = ((2๐‘˜ โˆ’ 1) 3^(๐‘˜ + 1) + 3 + 4(๐‘˜ + 1) 3^(๐‘˜+1))/4 = ((2๐‘˜ใ€– (3ใ€—^(๐‘˜ + 1)) โˆ’ 3^(๐‘˜ + 1) ) + 3 + 4(๐‘˜)3^(๐‘˜+1)+4(3^(๐‘˜+1) ))/4 = (2๐‘˜ใ€– (3ใ€—^(๐‘˜ + 1)) + 4(๐‘˜) 3^(๐‘˜+1) โˆ’ 3^(๐‘˜ + 1) + 4(3^(๐‘˜+1) ) + 3)/4 = (6๐‘˜ใ€– (3ใ€—^(๐‘˜ + 1)) + 3(3^(๐‘˜+1) ) + 3)/4 = ((3^(๐‘˜+1) )(6๐‘˜ + 3) + 3)/4 = ((3^(๐‘˜+1) )3(2๐‘˜ + 1) + 3)/4 = ((3^1 ร— 3^(๐‘˜+1) )(2๐‘˜ + 1) + 3)/4 = ((3^(๐‘˜+2) )(2๐‘˜ + 1) + 3)/4 Thus, 1.3 + 2.32 + 3.33 + โ€ฆ.. + k3k + (k + 1)3k + 1 = ((2๐‘˜ + 1) 3^(๐‘˜ + 2) + 3 )/4 which is the same as P(k + 1) โˆด P(k + 1) is true whenever P(k) is true. โˆดBy the principle of mathematical induction, P(n) is true for n, where n is a natural number

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