Check sibling questions

Ex 8.2, 8 - Find middle terms of (x/3 + 9y)10 - Binomial

Ex 8.2,8 - Chapter 8 Class 11 Binomial Theorem - Part 2
Ex 8.2,8 - Chapter 8 Class 11 Binomial Theorem - Part 3


Transcript

Ex 8.2, 8 Find the middle terms in the expansions of (𝑥/3+9𝑦)^10 Number of terms n = 10 Since n is even. There will be one middle term Middle term = 𝑛/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 = 𝑥/3 & b = 9y and r = 1 in (1) T5 + 1 = 10C5 (𝑥/3)^(10 − 5) . (9y)5 T6 = 10C5 (𝑥/3)^5 . (9)5 . y5 = 10!/5!(10 − 5)! . (𝑥/3)^5. (32) 5 . y5 = 10!/5!5! . 𝑥5/35 . 310 . y5 = (10 × 9 × 8 × 7 × 6 × 5!)/(5 × 4 × 3 × 2 × 1 ×5!) . x5 .310/35 . y5 = (10 × 9 × 8 × 7 × 6)/(5 × 4 × 3 × 2 × 1 ) . x5 . 35 . y5 = (10 × 9 × 8 × 7 × 6 ×35)/(5 × 4 × 3 × 2 × 1 ) . x5 . y5 = 61236 x5 y5 Hence, middle term is 61236 x5 y5

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.