Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 4.1,19 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 Ex 4.1,19 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

Ex 4.1

Ex 4.1, 1
Important

Ex 4.1, 2

Ex 4.1, 3

Ex 4.1, 4

Ex 4.1, 5

Ex 4.1, 6

Ex 4.1, 7 Important

Ex 4.1, 8

Ex 4.1, 9

Ex 4.1, 10

Ex 4.1, 11 Important

Ex 4.1, 12

Ex 4.1, 13 Important

Ex 4.1, 14

Ex 4.1, 15

Ex 4.1, 16

Ex 4.1, 17

Ex 4.1, 18

Ex 4.1, 19 You are here

Ex 4.1, 20

Ex 4.1, 21 Important

Ex 4.1, 22

Ex 4.1, 23 Important

Ex 4.1, 24

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.