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Last updated at Jan. 29, 2020 by Teachoo

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Ex 8.2, 8 Find the middle terms in the expansions of (๐ฅ/3+9๐ฆ)^10 Number of terms n = 10 Since n is even. There will be one middle term Middle term = ๐/2 + 1 = 10/2 + 1 = 5 + 1 = 6th term. Hence we need to find 6th term. i.e. T6 We know that general term of (a + b)n is Tr + 1 = nCr (a)n-r . (b)r Finding 6th term i.e. T6 = T5 + 1 , Hence r = 5 Putting n = 5 , a = ๐ฅ/3 & b = 9y and r = 1 in (1) T5 + 1 = 10C5 (๐ฅ/3)^(10 โ 5) . (9y)5 T6 = 10C5 (๐ฅ/3)^5 . (9)5 . y5 = 10!/5!(10 โ 5)! . (๐ฅ/3)^5. (32) 5 . y5 = 10!/5!5! . ๐ฅ5/35 . 310 . y5 = (10 ร 9 ร 8 ร 7 ร 6 ร 5!)/(5 ร 4 ร 3 ร 2 ร 1 ร5!) . x5 .310/35 . y5 = (10 ร 9 ร 8 ร 7 ร 6)/(5 ร 4 ร 3 ร 2 ร 1 ) . x5 . 35 . y5 = (10 ร 9 ร 8 ร 7 ร 6 ร35)/(5 ร 4 ร 3 ร 2 ร 1 ) . x5 . y5 = 61236 x5 y5 Hence, middle term is 61236 x5 y5

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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.