Last updated at May 29, 2018 by Teachoo

Transcript

Ex 4.1,21 Prove the following by using the principle of mathematical induction for all n โ N: x2n โ y2n is divisible by ๐ฅ + ๐ฆ. Let P(n): x2n โ y2n = (x + y) ร d, where d โ N For n = 1 L.H.S = x2 ร 1 โ y2 ร 1 = x2 โ y2 = (x + y) (x โ y) = R.H.S โด P(n) is true for n = 1 Assume P(k) is true x2k โ y2k = m (x + y), where m โ N We will prove that P(k + 1) is true. L.H.S = x2(k+1) - y2(k+1) = x2k+2 - y2k+2 = x2k . x2 โ y2k . y2 = x2 (m(x + y) + y2k) โ y2k . y2 = x2 (m (x + y)) + x2 y2k โ y2k . Y2 = x2 (m (x + y)) + y2k (x2 โ y2) = x2 (m (x + y)) + y2k (x + y) (x โ y) = (x + y) [m.x2 + y2k (x โ y)] = (x + y) ร r where r = m.x2 + y2k (x - y) is a natural number โด P(k + 1) is true whenever P(k) is true. โด 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

Ex 4.1, 20

Ex 4.1, 21 Important You are here

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.