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Ex 4.1, 7 - Prove 1.3 + 3.5 + 5.7 + .. + (2n-1) (2n+1) - Class 11 - Equal - Addition

  1. Chapter 4 Class 11th Mathematical Induction
  2. Serial order wise
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Ex 4.1, 7: Prove the following by using the principle of mathematical induction for all n ∈ N: 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = (𝑛(4𝑛2 + 6𝑛 βˆ’ 1))/3 Let P(n) : 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = (𝑛(4𝑛2 + 6𝑛 βˆ’ 1))/3 For n = 1, L.H.S = 1.3 = 3 R.H.S = (1(4.12 + 6.1 βˆ’ 1))/3 = (4 + 6 βˆ’ 1)/3 = 9/3 = 3 L.H.S. = R.H.S ∴ P(n) is true for n = 1 Assume P(k) is true 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) = (π‘˜(4π‘˜2 + 6π‘˜ βˆ’ 1))/3 We will prove that P(k + 1) is true. 1.3 + 3.5 + 5.7 + … + (2(k + 1) – 1).(2(k + 1) + 1) = (π‘˜ + 1)(4(π‘˜ + 1)^2 + 6(π‘˜ + 1) βˆ’ 1 )/3 1.3 + 3.5 + 5.7 + … + (2k + 2 – 1).(2k + 2 + 1) = (π‘˜ + 1)(4(π‘˜^2 + 1 + 2π‘˜)+ 6π‘˜ + 6 βˆ’ 1)/3 1.3 + 3.5 + 5.7 + … + (2k + 1).(2k + 3) = (π‘˜ + 1)(4π‘˜^2 +4(1) +4(2π‘˜) + 6π‘˜ + 6 βˆ’ 1)/3 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1).(2k + 3) = (π‘˜ + 1)(4π‘˜^2 + 4 + 8π‘˜ + 6π‘˜ + 6 βˆ’ 1)/3 = (π‘˜ + 1)(4π‘˜^2 +14π‘˜ + 9)/3 = ((π‘˜(4π‘˜^2 +14π‘˜ + 9)+ 1(4π‘˜^2 +14π‘˜ + 9)))/3 = ((4π‘˜^3 +18π‘˜^2 + 23π‘˜ + 9))/3 Thus, P(k +1) :1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1).(2k + 3) = ((4π‘˜^3 +18π‘˜^2 + 23π‘˜ + 9))/3 We have to prove P(k+1) from P(k) i.e. (2) from (1) From (1) 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) = (π‘˜(4π‘˜2 + 6π‘˜ βˆ’ 1))/3 Adding (2k+1).(2k+3) both sides 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1).(2k + 3) = (π‘˜(4π‘˜2 + 6π‘˜ βˆ’ 1))/3 + (2k + 1).(2k + 3) = (π‘˜(4π‘˜2 + 6π‘˜ βˆ’ 1) + 3(2π‘˜ + 1)(2π‘˜ + 3))/3 = (π‘˜(4π‘˜2 + 6π‘˜ βˆ’ 1) + 3(2π‘˜(2π‘˜ + 3) + 1(2π‘˜ + 3)))/3 = (π‘˜(4π‘˜2 + 6π‘˜ βˆ’ 1) + 3(2π‘˜(2π‘˜) +2π‘˜(3) + 2π‘˜ + 3))/3 = (π‘˜(4π‘˜2 + 6π‘˜ βˆ’ 1) + 3(4π‘˜^2+ 6π‘˜ + 2π‘˜ + 3))/3 = (π‘˜(4π‘˜2 + 6π‘˜ βˆ’ 1) + 3(4π‘˜^2+8π‘˜ + 3))/3 = (π‘˜(4π‘˜2 + 6π‘˜ βˆ’ 1) + (3(4π‘˜^2 ) +3(8π‘˜) + 3(3)))/3 = (π‘˜(4π‘˜2 + 6π‘˜ βˆ’ 1) + (12π‘˜^2 + 24π‘˜ + 9))/3 = (4π‘˜3 + 6π‘˜^2 βˆ’ π‘˜ + (12π‘˜^2 + 24π‘˜ + 9))/3 = (4π‘˜3 + 6π‘˜^2 + 12π‘˜^2 βˆ’ π‘˜ + 24π‘˜ + 9)/3 = ((4π‘˜^3 +18π‘˜^2 + 23π‘˜ + 9))/3 Thus, 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1).(2k + 3) = ((4π‘˜^3 +18π‘˜^2 + 23π‘˜ + 9))/3 which is the same as P(k +1) ∴ 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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