Slide17.JPG

Slide18.JPG
Slide19.JPG

Subscribe to our Youtube Channel - https://you.tube/teachoo

  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise

Transcript

Ex 8.2, 8 Find the middle terms in the expansions of (๐‘ฅ/3+9๐‘ฆ)^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 = ๐‘ฅ/3 & b = 9y and r = 1 in (1) T5 + 1 = 10C5 (๐‘ฅ/3)^(10 โˆ’ 5) . (9y)5 T6 = 10C5 (๐‘ฅ/3)^5 . (9)5 . y5 = 10!/5!(10 โˆ’ 5)! . (๐‘ฅ/3)^5. (32) 5 . y5 = 10!/5!5! . ๐‘ฅ5/35 . 310 . y5 = (10 ร— 9 ร— 8 ร— 7 ร— 6 ร— 5!)/(5 ร— 4 ร— 3 ร— 2 ร— 1 ร—5!) . 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 ร—35)/(5 ร— 4 ร— 3 ร— 2 ร— 1 ) . x5 . y5 = 61236 x5 y5 Hence, middle term is 61236 x5 y5

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.