   1. Chapter 8 Class 11 Binomial Theorem
2. Serial order wise
3. Miscellaneous

Transcript

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] ﷐24﷮4﷯ = 6 There is no remainder, Hence, 4 is a factor of 24 We can write 24 = 4 × 6 Similarly, If (a – b) is a factor of an – bn then 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 [b + (a – b)]n = nCo bn (a – b)0 + nC1 bn – 1 (a – b) + nC2 bn–2 (a – b)2 +…….+ nCn (a – b)n b0 (a)n = nCo bn 1 + nC1 bn – 1 (a – b) + nC2 bn–2 (a – b)2 +…….+ nCn (a – b)n 1 (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 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

Miscellaneous 