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Chapter 7 Class 11 Binomial Theorem
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Question 8 Show that the coefficient of the middle term in the expansion of (1 + x)2n is equal to the sum of the coefficients of two middle terms in the expansion of (1 + x)2n – 1. Middle term of (1 + x)2n Since 2n is even, Middle term = (2n/2 " +1" )^𝑡ℎ = ("n + 1")th term = Tn+1 We know that general term of expansion (a + b)n is Tr + 1 = nCr an – r br For (1 + x)2n , r = n , n = 2n , a = 1 , b = x Putting values Tn+1 = 2nCn (1)2n–n . (x)n Tn+1 = 2nCn.(x)n Hence, Coefficient is 2nCn Middle term of (1 + x)2n – 1 Since 2n – 1 is odd, There will be 2 middle terms ((2n − 1) + 1)/2 and (((2n − 1) + 1)/2 " + 1" ) term = nth term and (n + 1)th terms. = Tn and Tn+1 We know that general term for expansion (a + b)n Tr+1 = n Cr an–r br For Tn in (1 + x)2n – 1 Putting r = n – 1 , n = 2n – 1 , a = 1 & b = x Tn – 1 + 1 = 2n – 1Cn – 1 . (1) (2n – 1) – (n–1) .x n – 1 Tn = 2n – 1Cn – 1 .(x)n – 1 Hence, coefficient is 2n – 1Cn – 1 Similarly, For Tn + 1 in (1 + x)2n – 1 Putting r = n , n = 2n – 1 , a = 1 & b = x Tn+1 = 2n – 1Cn . (1)2n – 1 – n . xn = 2n–1Cn . xn Hence, Coefficient is 2n–1Cn Now we have to prove Coefficient of middle term of (1+x)2n = Sum of coefficient of middle term of (1 + x)2n–1 2nCn = 2n – 1Cn – 1 + 2n – 1Cn Thus, We have to prove 2nCn = 2n – 1Cn – 1 + 2n – 1Cn 2nCn = 2𝑛!/𝑛!(2𝑛 − 𝑛)! = 2𝑛!/(𝑛!(𝑛)!) 2n – 1Cn – 1 + 2n – 1Cn = ((2𝑛 − 1)!)/((𝑛 − 1) ![(2𝑛 −1) − (𝑛 − 1)]!) + (2𝑛 − 1)!/(𝑛! (2𝑛 − 1 − 𝑛)!) = ((2𝑛 − 1)!)/((𝑛 − 1) !(2𝑛 − 1 − 𝑛 + 1)!) + (2𝑛 −1)!/(𝑛! (2𝑛 −1 − 𝑛)!) = ((2𝑛 − 1)!)/((𝑛 −1)! (𝑛)!) + (2𝑛 − 1)!/(𝑛! (𝑛 −1)!) = 2 × ((2𝑛 − 1)!)/(𝑛! (𝑛 − 1)! ) = 2 × ((2𝑛 − 1)!)/(𝑛! (𝑛 − 1)! ) × 𝑛/𝑛 = (2𝑛(2𝑛 − 1)!)/(𝑛!𝑛(𝑛 − 1)! ) = 2𝑛!/(𝑛! 𝑛! ) = L.H.S Hence L.H.S = R.H.S 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.