# Misc 7 - Chapter 8 Class 11 Binomial Theorem

Last updated at Dec. 24, 2019 by Teachoo

Last updated at Dec. 24, 2019 by Teachoo

Transcript

Misc 7 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 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 (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 = 0.951 So, approximate value of (0.99)5 = 0.951

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.