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Misc 4 - Prove that a - b is a factor of a^n - b^n - Binomial Theorem

Misc  4 - Chapter 8 Class 11 Binomial Theorem - Part 2
Misc  4 - Chapter 8 Class 11 Binomial Theorem - Part 3

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Misc 4 (Introduction) If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer. [Hint: write an = (a – b + b)n and expand] As 4 divides 24, 4 is a factor of 24 We can write 24 = 4 × 6 Similarly, If (a – b) is a factor of an – bn then we can write an – bn = (a – b) k where k is a natural number Misc 4 If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer. [Hint: write an = (a – b + b)n and expand] We can write an as an = [a – b + b]n = [b + (a – b)]n = nCo bn + nC1 bn – 1 (a – b) + nC2 bn–2 (a – b)2 +…….+ nCn (a – b)n We know that (a + b)n = nC0 an + nC1 an – 1 b 1 + nC2 an – 2 b2 + ….. + nCn bn Putting a = b ,b = a – b (a)n = 1 × bn + nC1 bn – 1 (a – b) + nC2 bn–2 (a – b)2….. + 1 × (a – b)n an = bn + nC1 bn – 1 (a – b) + nC2 bn–2 (a – b)2………. + (a – b)n an – bn = nC1 bn – 1 (a – b) + nC2 bn–2 (a – b)2………. + (a – b)n Taking (a − b) common an – bn = (a – b) [nC1 bn – 1 + nC2 bn–2 (a – b)1………. + (a – b)n-1 ] an – bn = (a – b) [k] where k = [nC1 bn – 1 + nC2 bn–2 (a – b)1………. + (a – b)n-1 ] is a natural number Hence, (a – b) is a factor of an – bn

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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.