Mathematical Induction
Serial order wise

Ex 4.1, 11 - Prove 1/1.2.3 + 1/2.3.4 + 1/3.4.5 .. + 1/n(n+1)(n+2) - Equal - 1 upon addition

Ex 4.1, 11 - Chapter 4 Class 11 Mathematical Induction - Part 2
Ex 4.1, 11 - Chapter 4 Class 11 Mathematical Induction - Part 3 Ex 4.1, 11 - Chapter 4 Class 11 Mathematical Induction - Part 4 Ex 4.1, 11 - Chapter 4 Class 11 Mathematical Induction - Part 5 Ex 4.1, 11 - Chapter 4 Class 11 Mathematical Induction - Part 6

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Transcript

Question11 Prove the following by using the principle of mathematical induction for all n ∈ N: 1﷮1.2.3﷯ + 1﷮2.3.4﷯ + 1﷮3.4.5﷯ + ……..+ 1﷮𝑛 𝑛 + 1﷯(𝑛 + 2)﷯ = 𝑛(𝑛 + 3)﷮4(𝑛 + 1)(𝑛 + 2)﷯ Let P (n) : 1﷮1.2.3﷯ + 1﷮2.3.4﷯ + 1﷮3.4.5﷯ + ……..+ 1﷮𝑛 𝑛 + 1﷯(𝑛 + 2)﷯ = 𝑛(𝑛 + 3)﷮4(𝑛 + 1)(𝑛 + 2)﷯ For n = 1, L.H.S = 1﷮1.2.3﷯ = 1﷮6﷯ R.H.S = 1.(1 + 3)﷮4(1 + 1)(1 + 2)﷯ = 1.4﷮4.2.3﷯ = 1﷮2.3﷯ = 1﷮6﷯ Hence, L.H.S. = R.H.S , ∴ P(n) is true for n = 1 Assume P(k) is true 1﷮1.2.3﷯ + 1﷮2.3.4﷯ + 1﷮3.4.5﷯ + ……..+ 1﷮𝑘 𝑘 + 1﷯(𝑘 + 2)﷯ = 𝑘(𝑘 + 3)﷮4(𝑘 + 1)(𝑘 + 2)﷯ We will prove that P(k + 1) is true. R.H.S = 𝑘 + 1﷯ 𝑘 + 1﷯+ 3﷯﷮4 𝑘 + 1﷯+ 1﷯ 𝑘 + 1﷯+ 2﷯﷯ L.H.S = 1﷮1.2.3﷯ + 1﷮2.3.4﷯ + 1﷮3.4.5﷯ + ……..+ 1﷮ 𝑘 + 1﷯ 𝑘 + 1﷯+ 1﷯( 𝑘 + 1﷯+ 2)﷯ Rough Factorizing 𝐤﷮𝟑﷯+ 𝟒+ 𝟗𝐤+ 𝟔𝐤﷮𝟐﷯ Let f(k) = k﷮3﷯+ 4+ 9k+ 6k﷮2﷯ Putting k = − 1 f(−1) = (−1)﷮3﷯+ 4+9 −1﷯+6 (−1)﷮2﷯ = −1 + 4− 9+6 = 0 Thus, (k + 1) is a factor of f(k) ∴ f(k) = (𝑘+1)( 𝑘﷮2﷯+5𝑘+4)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.