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

Ex 4.1, 7 - Chapter 4 Class 11 Mathematical Induction - Part 2
Ex 4.1, 7 - Chapter 4 Class 11 Mathematical Induction - Part 3
Ex 4.1, 7 - Chapter 4 Class 11 Mathematical Induction - Part 4
Ex 4.1, 7 - Chapter 4 Class 11 Mathematical Induction - Part 5

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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) = ( (4 2 + 6 1))/3 Let P(n) : 1.3 + 3.5 + 5.7 + + (2n 1) (2n + 1) = ( (4 2 + 6 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) = ( (4 2 + 6 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) = ( + 1)(4( + 1)^2 + 6( + 1) 1 )/3 1.3 + 3.5 + 5.7 + + (2k + 2 1).(2k + 2 + 1) = ( + 1)(4( ^2 + 1 + 2 )+ 6 + 6 1)/3 1.3 + 3.5 + 5.7 + + (2k + 1).(2k + 3) = ( + 1)(4 ^2 +4(1) +4(2 ) + 6 + 6 1)/3 1.3 + 3.5 + 5.7 + + (2k 1) (2k + 1) + (2k + 1).(2k + 3) = ( + 1)(4 ^2 + 4 + 8 + 6 + 6 1)/3 = ( + 1)(4 ^2 +14 + 9)/3 = (( (4 ^2 +14 + 9)+ 1(4 ^2 +14 + 9)))/3 = ((4 ^3 +18 ^2 + 23 + 9))/3 Thus, P(k +1) :1.3 + 3.5 + 5.7 + + (2k 1) (2k + 1) + (2k + 1).(2k + 3) = ((4 ^3 +18 ^2 + 23 + 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) = ( (4 2 + 6 1))/3 Adding (2k+1).(2k+3) both sides 1.3 + 3.5 + 5.7 + + (2k 1) (2k + 1) + (2k + 1).(2k + 3) = ( (4 2 + 6 1))/3 + (2k + 1).(2k + 3) = ( (4 2 + 6 1) + 3(2 + 1)(2 + 3))/3 = ( (4 2 + 6 1) + 3(2 (2 + 3) + 1(2 + 3)))/3 = ( (4 2 + 6 1) + 3(2 (2 ) +2 (3) + 2 + 3))/3 = ( (4 2 + 6 1) + 3(4 ^2+ 6 + 2 + 3))/3 = ( (4 2 + 6 1) + 3(4 ^2+8 + 3))/3 = ( (4 2 + 6 1) + (3(4 ^2 ) +3(8 ) + 3(3)))/3 = ( (4 2 + 6 1) + (12 ^2 + 24 + 9))/3 = (4 3 + 6 ^2 + (12 ^2 + 24 + 9))/3 = (4 3 + 6 ^2 + 12 ^2 + 24 + 9)/3 = ((4 ^3 +18 ^2 + 23 + 9))/3 Thus, 1.3 + 3.5 + 5.7 + + (2k 1) (2k + 1) + (2k + 1).(2k + 3) = ((4 ^3 +18 ^2 + 23 + 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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.