Ex 4.1, 7

Transcript

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 − 1))/3 Let P(n) : 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) = (𝑛(4𝑛2 + 6𝑛 − 1))/3 For n = 1, L.H.S = 1.3 = 3 R.H.S = (1(4.12 + 6.1 − 1))/3 = (4 + 6 − 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) = (𝑘(4𝑘2 + 6𝑘 − 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) = (𝑘 + 1)(4(𝑘 + 1)^2 + 6(𝑘 + 1) − 1 )/3 1.3 + 3.5 + 5.7 + … + (2k + 2 – 1).(2k + 2 + 1) = (𝑘 + 1)(4(𝑘^2 + 1 + 2𝑘)+ 6𝑘 + 6 − 1)/3 1.3 + 3.5 + 5.7 + … + (2k + 1).(2k + 3) = (𝑘 + 1)(4𝑘^2 +4(1) +4(2𝑘) + 6𝑘 + 6 − 1)/3 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1).(2k + 3) = (𝑘 + 1)(4𝑘^2 + 4 + 8𝑘 + 6𝑘 + 6 − 1)/3 = (𝑘 + 1)(4𝑘^2 +14𝑘 + 9)/3 = ((𝑘(4𝑘^2 +14𝑘 + 9)+ 1(4𝑘^2 +14𝑘 + 9)))/3 = ((4𝑘^3 +18𝑘^2 + 23𝑘 + 9))/3 Thus, P(k +1) :1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1).(2k + 3) = ((4𝑘^3 +18𝑘^2 + 23𝑘 + 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) = (𝑘(4𝑘2 + 6𝑘 − 1))/3 Adding (2k+1).(2k+3) both sides 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1).(2k + 3) = (𝑘(4𝑘2 + 6𝑘 − 1))/3 + (2k + 1).(2k + 3) = (𝑘(4𝑘2 + 6𝑘 − 1) + 3(2𝑘 + 1)(2𝑘 + 3))/3 = (𝑘(4𝑘2 + 6𝑘 − 1) + 3(2𝑘(2𝑘 + 3) + 1(2𝑘 + 3)))/3 = (𝑘(4𝑘2 + 6𝑘 − 1) + 3(2𝑘(2𝑘) +2𝑘(3) + 2𝑘 + 3))/3 = (𝑘(4𝑘2 + 6𝑘 − 1) + 3(4𝑘^2+ 6𝑘 + 2𝑘 + 3))/3 = (𝑘(4𝑘2 + 6𝑘 − 1) + 3(4𝑘^2+8𝑘 + 3))/3 = (𝑘(4𝑘2 + 6𝑘 − 1) + (3(4𝑘^2 ) +3(8𝑘) + 3(3)))/3 = (𝑘(4𝑘2 + 6𝑘 − 1) + (12𝑘^2 + 24𝑘 + 9))/3 = (4𝑘3 + 6𝑘^2 − 𝑘 + (12𝑘^2 + 24𝑘 + 9))/3 = (4𝑘3 + 6𝑘^2 + 12𝑘^2 − 𝑘 + 24𝑘 + 9)/3 = ((4𝑘^3 +18𝑘^2 + 23𝑘 + 9))/3 Thus, 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1).(2k + 3) = ((4𝑘^3 +18𝑘^2 + 23𝑘 + 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