Slide57.JPG

Slide58.JPG
Slide59.JPG Slide60.JPG Slide61.JPG

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

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

Transcript

Example 15 Find the term independent of x in the expansion of (โˆ›๐‘ฅ " + " 1/(2 โˆ›๐‘ฅ))^18, x > 0. Calculating general term of expansion We know that general term of (a + b)n is Tr+1 = nCr (a)nโ€“r . (a)n For general term of expansion (โˆ›๐‘ฅ " + " 1/(2 โˆ›๐‘ฅ))^18 Putting n = 18 , a = โˆ›๐‘ฅ , b = 1/(2 โˆ›๐‘ฅ) โˆด Tr + 1 = 18Cr (โˆ›๐‘ฅ)18 โ€“ r (1/(2 โˆ›๐‘ฅ))^๐‘Ÿ = 18Cr (ใ€–๐‘ฅ")" ใ€—^(1/3 ร— (18 โˆ’ ๐‘Ÿ)) (1/2 "ร—" 1/โˆ›๐‘ฅ)^๐‘Ÿ = 18Cr (ใ€–๐‘ฅ")" ใ€—^((18 โˆ’ ๐‘Ÿ)/3 ) (1/2)^๐‘Ÿ (1/โˆ›๐‘ฅ)^๐‘Ÿ = 18Cr (ใ€–๐‘ฅ")" ใ€—^((18 โˆ’ ๐‘Ÿ)/3 ) 1/2^๐‘Ÿ (1/๐‘ฅ^(1/3) )^๐‘Ÿ = 18Cr (ใ€–๐‘ฅ")" ใ€—^((18 โˆ’ ๐‘Ÿ)/3 ) 1/2^๐‘Ÿ 1/๐‘ฅ^(๐‘Ÿ/3) = 18Cr (ใ€–๐‘ฅ")" ใ€—^((18 โˆ’ ๐‘Ÿ)/3 ) 1/2^๐‘Ÿ ๐‘ฅ^((โˆ’๐‘Ÿ)/3) = 18Cr (ใ€–๐‘ฅ")" ใ€—^((18 โˆ’ ๐‘Ÿ)/3 ) ๐‘ฅ^((โˆ’๐‘Ÿ)/3) 1/2^๐‘Ÿ = 18Cr (ใ€–๐‘ฅ")" ใ€—^((18 โˆ’ ๐‘Ÿ)/3 โˆ’ ๐‘Ÿ/3) 1/2^๐‘Ÿ = 18Cr (ใ€–๐‘ฅ")" ใ€—^((18 โˆ’ ๐‘Ÿ โˆ’ ๐‘Ÿ)/3 ) 1/2^๐‘Ÿ = 18Cr (ใ€–๐‘ฅ")" ใ€—^((18 โˆ’ 2๐‘Ÿ)/3 ) 1/2^๐‘Ÿ โˆด Tr + 1 = 18Cr (ใ€–๐‘ฅ")" ใ€—^((18 โˆ’ 2๐‘Ÿ)/3 ) 1/2^๐‘Ÿ We need to find the term independent of x So, power of x is 0 ๐‘ฅ^((18 โˆ’2๐‘Ÿ)/3) = x0 Comparing power (18 โˆ’ 2๐‘Ÿ)/3 = 0 18 โ€“ 2r = 0 18 = 2r 18/2 = r 9 = r r = 9 Putting r = 9 in (1) Tr+1 = 18Cr (ใ€–๐‘ฅ")" ใ€—^((18 โˆ’ 2๐‘Ÿ)/3 ) 1/2^๐‘Ÿ T9+1 = 18C9 .๐‘ฅ^((18 โˆ’3(9))/3). 1/29 = 18C9 .x0. 1/29 = 18C9 . 1/29 Hence, the term which is independent of x is 10th term = T10 = 18C9 . ๐Ÿ/๐Ÿ๐Ÿ—a

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.