Examples

Chapter 7 Class 11 Binomial Theorem
Serial order wise

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Question 2 Show that the middle term in the expansion of (1 + x)2n is (1 . 3 . 5 …. (2𝑛 − 1))/𝑛! 2n xn, where n is a positive integer. Given Number of terms = 2n which is even So, Middle term = (2n/2 + 1)th term = (n + 1)th term Hence, we need to find Tn + 1 We know that general term of (a + b)nis Tr + 1 = nCr an – r br For Tn + 1 , Putting n = 2n , r = n , a = 1 & b = x Tn+1 = 2nCn (1)2n – n (x)n = (2𝑛)!/𝑛!(2𝑛 −𝑛)! . (1)n . xn = (2𝑛)!/(𝑛! 𝑛!) . xn = (2𝑛(2𝑛 − 1)(2𝑛 − 2) ……. 4 × 3 × 2 × 1)/(𝑛! 𝑛!) xn = ([(2𝑛 − 1)(2𝑛 − 3)….…. × 5 × 3 × 1] [(2𝑛)(2𝑛 − 2)… × 4 × 2])/(𝑛! 𝑛!) xn We need to show (1 . 3 . 5 …. (2𝑛 − 1))/𝑛! 2n xn = ([1 × 3 × 5 × …… × (2𝑛 − 3)(2𝑛 − 1)] [2 × 4 × 6….. × (2𝑛 − 2) × 2𝑛])/(𝑛! 𝑛!) xn = ([1 × 3 × 5……. × (2𝑛−3)(2𝑛−1)] [(2 × 1) ×(2 × 2) ×(2 × 3) × ….. × 2 (𝑛−1) × 2(𝑛)])/(𝑛! 𝑛!) xn = ([1 × 3 × 5……. × (2𝑛−3)(2𝑛−1)] (2 × 2 × 2 × 2 ……..× 2) [1 × 2 × 3 ….. (𝑛−1) 𝑛])/(𝑛! 𝑛!) xn = ([1 × 3 × 5……. × (2𝑛 − 3)(2𝑛 − 1)] 2𝑛 [1 × 2 × 3 ….. (𝑛 − 1) 𝑛])/𝑛!𝑛! xn = ([1 × 3 × 5……. × (2n − 3)(2n − 1)] 2n (n!))/(n! n!) . xn = (𝟏 × 𝟑 × 𝟓……. × (𝟐𝐧 − 𝟏))/(𝐧! ) 2n . xn Hence middle term of expansion (1 + x)2n is (1 . 3 . 5……. ….(2n−1))/(n! ) 2n . xn Hence proved

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.