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


Transcript

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 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.