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Ex 4.1, 21 - Prove x2n - y2n is divisible by x + y - Class 11 - Divisible

  1. Chapter 4 Class 11 Mathematical Induction
  2. Serial order wise
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Ex 4.1,21 Prove the following by using the principle of mathematical induction for all n โˆˆ N: x2n โ€“ y2n is divisible by ๐‘ฅ + ๐‘ฆ. Let P(n): x2n โ€“ y2n = (x + y) ร— d, where d โˆˆ 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 โˆด P(n) is true for n = 1 Assume P(k) is true x2k โ€“ y2k = m (x + y), where m โˆˆ 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 โˆด P(k + 1) is true whenever P(k) is true. โˆด By the principle of mathematical induction, P(n) is true for n, where n is a natural number

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