Miscellaneous

Chapter 7 Class 11 Binomial Theorem
Serial order wise

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### Transcript

Misc 4 Find an approximation of (0.99)5 using the first three terms of its expansion. (0.99)5 = (1 – 0.01)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 = = a5 + 5!/1!(5 − 1)! a4 b1 + 5!/2!(5 − 2)! a3 b2 + 5!/3!(5 − 3)! a2b3 + 5!/4!(5 − 4)! a b4 + b5 = a5 + 5!/(1! × 4!) a4 b + 5!/(2! 3!) a3 b2 + 5!/(3! 2!) a2b3 + 5!/(4! 1!) a b4 + b5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 We need to find (1 – 0.01)5 Putting a = 1 & b = –0.01 (1 – 0.01)5 = (1)5 + 5(1)4 (–0.01) + 10(1)3 (–0.01)2 + 10 (1)2 (–0.01)3 + 5 (1) (–0.01)4 + (–0.01)5 (0.99)5 = 1 + 5 (1) (–0.01) + 10 (1) (0.0001) + 10(1) (–0.000001) + 5(1) (0.00000001) + (–0.0000000001) Using first three terms, (0.99)5 = 1 – 0.05 + 0.001 = 1.001 – 0.050 = 0.9510 So, approximate value of (0.99)5 = 0.9510

#### 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.