Middle term

Chapter 8 Class 11 Binomial Theorem (Deleted)
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 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

#### Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.