Mathematical Induction
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Ex 4.1, 19 - Prove: n (n + 1) (n + 5) is a multiple of 3 - Ex 4.1

Ex 4.1, 19 - Chapter 4 Class 11 Mathematical Induction - Part 2

Ex 4.1, 19 - Chapter 4 Class 11 Mathematical Induction - Part 3 Ex 4.1, 19 - Chapter 4 Class 11 Mathematical Induction - Part 4 Ex 4.1, 19 - Chapter 4 Class 11 Mathematical Induction - Part 5

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Transcript

Question19 Prove the following by using the principle of mathematical induction for all n ∈ N: n (n + 1) (n + 5) is a multiple of 3. Introduction If a number is multiple of 3, then it will come in table of 3 3 × 1 = 3 3 × 2 = 6 3 × 3 = 9 Any number multiple of 3 = 3 × Natural number Question19 Prove the following by using the principle of mathematical induction for all n ∈ N: n (n + 1) (n + 5) is a multiple of 3. Let P(n): n (n + 1) (n + 5) = 3d, where d ∈ N For n = 1 , L.H.S = 1 (1 + 1) (1 + 5) = 1.(2).(6) = 12 = (3) × 4 = R.H.S , ∴P(n) is true for n = 1 Assume P(k) is true k (k + 1) (k + 5) = 3m , where m ∈ N ((k(k + 1)) (k + 5)= 3m (k2 + k) (k + 5) = 3m k2(k + 5) + k(k + 5) = 3m k3 + 5k2 + k2 + 5k =3m k3 + 6k2 + 5k =3m We will prove that P(k + 1) is true L.H.S = (k+1) ((k+1)+1) ((k+1)+5) = (k+1) (k+2) (k+6) = ((k + 1) (k + 2)) (k + 6) = ( k(k + 2) + 1(k + 2)) (k + 6) = ( k2 + 2k + k + 2) (k + 6) = (k + 6) ( k2 + 3k +2) = k (k2 + 3k +2) + 6 (k2 + 3k +2) = k3 + 3k2 +2k + 6k2 + 6 × 3k + 6 × 2 = k3 + 3k2 +2k + 6k2 + 18k + 12 = k3 + 9k2 + 20k +12 = (3m – 6k2 – 5k ) + 9k2 + 20k +12 = 3m – 6k2 + 9k2 – 5k + 20k +12 = 3m + 3k2 + 15k + 12 = 3 (m + k2 + 5k + 4) = 3 r , where r = m + k2 + 5k +4 r is a natural number ∴ 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.