Ex 8.1, 9 - Using Binomial Theorem, evaluate (99)5 - Class 11 - Ex 8.1

  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise
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Ex 8.1,9 Using Binomial Theorem, evaluate (99)5 (99)5 = (100 – 1)5 We know that (a + b)n = nC0 an b0 + nC1 an – 1 b1 + nC2 an – 2 b2 + …. …. + nCn – 1 a1 bn – 1 + nCn a0 bn Hence (a + b)5 = = ﷐5!﷮0!﷐ 5 − 0﷯!﷯ a5 × 1 + ﷐5!﷮1!﷐ 5 − 1﷯!﷯ a4 b1 + ﷐5!﷮2!﷐ 5 − 2﷯!﷯ a3 b2 + ﷐5!﷮3!﷐ 5 − 3﷯!﷯ a2b3 + ﷐5!﷮4!﷐ 5 − 4﷯!﷯ a b4 + ﷐5!﷮5!﷐ 5 −5﷯!﷯ b5 × 1 = ﷐5!﷮0! × 5!﷯ a5 + ﷐5!﷮1! × 4!﷯ a4 b + ﷐5!﷮2! 3!﷯ a3 b2 + ﷐5!﷮3! 2!﷯ a2b3 + ﷐5!﷮4! 1!﷯ a b4 + ﷐5!﷮5! 0!﷯ b5 = ﷐5!﷮5!﷯ a5 + ﷐5 × 4!﷮4!﷯ a4 b + ﷐5 × 4 × 3!﷮2! 3!﷯ a3 b2 + ﷐5 × 4 × 3!﷮2 × 1 ×3!﷯ a3b2 + ﷐5 × 4 × 3!﷮3! ×1 ×3!﷯ a2b3 + ﷐5 × 4!﷮4!﷯ ab4 + ﷐5!﷮5! ﷯ b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 Thus, (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find (100 – 1)5 Putting a = 100 & b = –1 (100 – 1)5 = (100)5 + 5(100)4 (–1) + 10 (100)3 (–1)2 + 10(100)2 (–1)3 + 5(100) (–1)4 + (–1)5 (99)5 = 10000000000 – 5(100000000) + 10(1000000) – 10(10000) + 5 (100) –1 = 10000000000 – 500000000 + 100000 – 100000 + 500 – 1 = 9509900499

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