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

Ex 4.1, 21 - Chapter 4 Class 11 Mathematical Induction - Part 2
Ex 4.1, 21 - Chapter 4 Class 11 Mathematical Induction - Part 3

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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.