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  1. Chapter 8 Class 11 Binomial Theorem
  2. Concept wise

Transcript

Ex 8.2, 7 Find the middle terms in the expansions of ("3 โ€“ " ๐‘ฅ3/6)^7 Number of terms = n = 7 Since n is odd there will be two middle termx ((๐‘› + 1)/2)^๐‘กโ„Ž term & ((๐‘› + 1)/2+1)^๐‘กโ„Ž term Hence we need to find 4th and 5th term i.e. T4 and T5 We know that general term of (a + b)n is Tr + 1 = nCr an โ€“ r br = ((7 + 1)/2)^๐‘กโ„Ž term = 4th term = ((7 + 1)/2)^๐‘กโ„Ž term = 4th term For ("3 โ€“ " ๐’™๐Ÿ‘/๐Ÿ”)^๐Ÿ• Putting a = 3 , b = ((โˆ’๐‘ฅ3)/6) , n = 7 Tr + 1 = 7Cr (3)7 โ€“ r ("โ€“" ๐‘ฅ3/6)^๐‘Ÿ Tr + 1 = 7Cr (3)7 โ€“ r (โ€“1)r (๐‘ฅ3/6)^๐‘Ÿ Tr + 1 = 7Cr (3)7 โ€“ r (โ€“1)r (๐‘ฅ3/6)^๐‘Ÿ Tr + 1 = 7Cr (3)7 โ€“ r (โ€“1)r (1/6)^๐‘Ÿ ๐‘ฅ^3๐‘Ÿ Finding 4th term Tr + 1 = 7Cr (3)7 โ€“ r (โ€“1)r (1/6)^๐‘Ÿ ๐‘ฅ^3๐‘Ÿ Finding 4th term i.e. T4 = T3 + 1 , Hence r = 3 T3+1 = 7C3 (3)7 โ€“ 3 (โ€“1)3 (1/6)^3 ๐‘ฅ^(3(3)) T4 = 7C3 . 34 . (โˆ’1) . 1/63 . (x)9 = โ€“ 7!/3!(7 โˆ’ 3)! . 34. 1/((2 ร— 3)3) . ๐‘ฅ9 = โ€“ 7!/(3! 4!) . 34 . 1/(23 . 33) . ๐‘ฅ9 = โ€“ ((7 ร— 6 ร— 5 ร— 4!)/(3 ร— 2 ร— 1 ร— 4!)) . 34/33 . ๐‘ฅ9/23 = (โˆ’๐Ÿ๐ŸŽ๐Ÿ“)/๐Ÿ– x9 Finding 5th term Tr + 1 = 7Cr (3)7 โ€“ r (โ€“1)r (1/6)^๐‘Ÿ ๐‘ฅ^3๐‘Ÿ Finding 5th term i.e. T5 = T4 + 1 , Hence r = 4 T4+1 = 7C4 (3)7 - 4 (โ€“1)4 (1/6)^4 ๐‘ฅ^(3(4)) = 7C4 . 33 . 1/64 . (x)12 = 7!/4!(7 โˆ’ 4)! . 33 . 1/(2 ร— 3)4 . x12 = 7!/(3! 4!) . 33 . 1/(24 . 34) . ๐‘ฅ12 = ((7 ร— 6 ร— 5 ร— 4!)/(3 ร— 2 ร— 1 ร— 4!)) . 34/34 . ๐‘ฅ12/24 = ๐Ÿ‘๐Ÿ“/๐Ÿ’๐Ÿ– x12 Hence, the middle terms are (โˆ’๐Ÿ๐ŸŽ๐Ÿ“)/๐Ÿ– x9 & ๐Ÿ‘๐Ÿ“/๐Ÿ’๐Ÿ– x12

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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.