Misc 4 - Chapter 7 Class 11 Binomial Theorem

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