General Term of Binomial Theorem

Chapter 7 Class 11 Binomial Theorem
Serial order wise

### Transcript

Question 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 term & ((๐ + 1)/2+1)^๐กโ 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 term = 4th term = ((7 + 1)/2)^๐กโ term = 4th term For ("3 โ " ๐๐/๐)^๐ Putting a = 3 , b = ((โ๐ฅ3)/6) , n = 7 Tr + 1 = 7Cr (3)7 โ r ("โ" ๐ฅ3/6)^๐ Tr + 1 = 7Cr (3)7 โ r (โ1)r (๐ฅ3/6)^๐ Tr + 1 = 7Cr (3)7 โ r (โ1)r (๐ฅ3/6)^๐ Tr + 1 = 7Cr (3)7 โ r (โ1)r (1/6)^๐ ๐ฅ^3๐ Finding 4th term Tr + 1 = 7Cr (3)7 โ r (โ1)r (1/6)^๐ ๐ฅ^3๐ 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 . (โ1) . 1/63 . (x)9 = โ 7!/3!(7 โ 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 = (โ๐๐๐)/๐ x9 Finding 5th term Tr + 1 = 7Cr (3)7 โ r (โ1)r (1/6)^๐ ๐ฅ^3๐ 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 โ 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๐/๐๐ x12 Hence, the middle terms are (โ๐๐๐)/๐ x9 & ๐๐/๐๐ x12

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.