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

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.