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

Ex 8.1,9 - Chapter 8 Class 11 Binomial Theorem - Part 2
Ex 8.1,9 - Chapter 8 Class 11 Binomial Theorem - Part 3


Transcript

Ex 8.1, 9 Using Binomial Theorem, evaluate (99)5 (99)5 = (100 – 1)5 We know that (a + b)n = nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ….…. + nCn – 1 a1 bn – 1 + nCn bn Hence (a + b)5 = = 5!/0!( 5 − 0)! a5 + 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 = 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!)/(2 ×1 ×3!) a2b3 + (5 × 4!)/4! ab4 + 5!/(5! ) b5 = 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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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.