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Question 19 Deleted for CBSE Board 2024 Exams You are here
Last updated at April 16, 2024 by Teachoo
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