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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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo