Check sibling questions

Example 6 - Show that middle term in expansion of (1 + x)^2n is

Example  6 - Chapter 8 Class 11 Binomial Theorem - Part 2
Example  6 - Chapter 8 Class 11 Binomial Theorem - Part 3

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Example 6 Show that the middle term in the expansion of (1 + x)2n is (1 . 3 . 5 …. (2𝑛 βˆ’ 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 βˆ’π‘›)! . (1)n . xn = (2𝑛)!/(𝑛! 𝑛!) . xn = (2𝑛(2𝑛 βˆ’ 1)(2𝑛 βˆ’ 2) ……. 4 Γ— 3 Γ— 2 Γ— 1)/(𝑛! 𝑛!) xn = ([(2𝑛 βˆ’ 1)(2𝑛 βˆ’ 3)….…. Γ— 5 Γ— 3 Γ— 1] [(2𝑛)(2𝑛 βˆ’ 2)… Γ— 4 Γ— 2])/(𝑛! 𝑛!) xn We need to show (1 . 3 . 5 …. (2𝑛 βˆ’ 1))/𝑛! 2n xn = ([1 Γ— 3 Γ— 5 Γ— …… Γ— (2𝑛 βˆ’ 3)(2𝑛 βˆ’ 1)] [2 Γ— 4 Γ— 6….. Γ— (2𝑛 βˆ’ 2) Γ— 2𝑛])/(𝑛! 𝑛!) xn = ([1 Γ— 3 Γ— 5……. Γ— (2π‘›βˆ’3)(2π‘›βˆ’1)] [(2 Γ— 1) Γ—(2 Γ— 2) Γ—(2 Γ— 3) Γ— ….. Γ— 2 (π‘›βˆ’1) Γ— 2(𝑛)])/(𝑛! 𝑛!) xn = ([1 Γ— 3 Γ— 5……. Γ— (2π‘›βˆ’3)(2π‘›βˆ’1)] (2 Γ— 2 Γ— 2 Γ— 2 ……..Γ— 2) [1 Γ— 2 Γ— 3 ….. (π‘›βˆ’1) 𝑛])/(𝑛! 𝑛!) xn = ([1 Γ— 3 Γ— 5……. Γ— (2𝑛 βˆ’ 3)(2𝑛 βˆ’ 1)] 2𝑛 [1 Γ— 2 Γ— 3 ….. (𝑛 βˆ’ 1) 𝑛])/𝑛!𝑛! xn = ([1 Γ— 3 Γ— 5……. Γ— (2n βˆ’ 3)(2n βˆ’ 1)] 2n (n!))/(n! n!) . xn = (𝟏 Γ— πŸ‘ Γ— πŸ“β€¦β€¦. Γ— (𝟐𝐧 βˆ’ 𝟏))/(𝐧! ) 2n . xn Hence middle term of expansion (1 + x)2n is (1 . 3 . 5……. ….(2nβˆ’1))/(n! ) 2n . xn Hence proved

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.