Ex 7.1

Chapter 7 Class 11 Binomial Theorem
Serial order wise

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### Transcript

Ex 7.1, 7 Using Binomial Theorem, evaluate (102)5 (102)5 = (100 + 2)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 × 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!) ab4 + 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!)/4! ab4 + 5!/(5! ) b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find (100 + 2)5 Putting a = 100 & b = 2 (100 + 2)5 = (100)5 + 5 (100)4 (2) + 10 (100)3 (2)2 + 10 (100)2 (2)3 + 5(100) (2)4 + (2)5 (102)5 = 10000000000 + 10 (100000000) + 40(100000) (8) + 5 (100) (16) + 32 = 10000000000 + 1000000000 + 40000000 + 8000 + 32 = 11040808032 Hence, (102)5 = 11040808032

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.