Divisible

Chapter 4 Class 11 Mathematical Induction
Concept wise

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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