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Divisible

Ex 4.1, 20
Deleted for CBSE Board 2023 Exams

Example 4 Important Deleted for CBSE Board 2023 Exams

Ex 4.1, 23 Important Deleted for CBSE Board 2023 Exams

Ex 4.1, 22 Deleted for CBSE Board 2023 Exams

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Ex 4.1, 21 Important Deleted for CBSE Board 2023 Exams You are here

Ex 4.1, 19 Deleted for CBSE Board 2023 Exams

Last updated at May 29, 2018 by Teachoo

Ex 4.1,21 Prove the following by using the principle of mathematical induction for all n β N: x2n β y2n is divisible by π₯ + π¦. Let P(n): x2n β y2n = (x + y) Γ d, where d β N For n = 1 L.H.S = x2 Γ 1 β y2 Γ 1 = x2 β y2 = (x + y) (x β y) = R.H.S β΄ P(n) is true for n = 1 Assume P(k) is true x2k β y2k = m (x + y), where m β 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 β y2k . y2 = x2 (m(x + y) + y2k) β y2k . y2 = x2 (m (x + y)) + x2 y2k β y2k . Y2 = x2 (m (x + y)) + y2k (x2 β y2) = x2 (m (x + y)) + y2k (x + y) (x β y) = (x + y) [m.x2 + y2k (x β y)] = (x + y) Γ r where r = m.x2 + y2k (x - y) is a natural number β΄ P(k + 1) is true whenever P(k) is true. β΄ By the principle of mathematical induction, P(n) is true for n, where n is a natural number