# Question 8 - Examples - Chapter 7 Class 11 Binomial Theorem

Last updated at April 16, 2024 by Teachoo

Examples

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Chapter 7 Class 11 Binomial Theorem

Serial order wise

Last updated at April 16, 2024 by Teachoo

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 β 1. Middle term of (1 + x)2n Since 2n is even, Middle term = (2n/2 " +1" )^π‘β = ("n + 1")th term = Tn+1 We know that general term of expansion (a + b)n is Tr + 1 = nCr an β r br For (1 + x)2n , r = n , n = 2n , a = 1 , b = x Putting values Tn+1 = 2nCn (1)2nβn . (x)n Tn+1 = 2nCn.(x)n Hence, Coefficient is 2nCn Middle term of (1 + x)2n β 1 Since 2n β 1 is odd, There will be 2 middle terms ((2n β 1) + 1)/2 and (((2n β 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 anβr br For Tn in (1 + x)2n β 1 Putting r = n β 1 , n = 2n β 1 , a = 1 & b = x Tn β 1 + 1 = 2n β 1Cn β 1 . (1) (2n β 1) β (nβ1) .x n β 1 Tn = 2n β 1Cn β 1 .(x)n β 1 Hence, coefficient is 2n β 1Cn β 1 Similarly, For Tn + 1 in (1 + x)2n β 1 Putting r = n , n = 2n β 1 , a = 1 & b = x Tn+1 = 2n β 1Cn . (1)2n β 1 β n . xn = 2nβ1Cn . xn Hence, Coefficient is 2nβ1Cn Now we have to prove Coefficient of middle term of (1+x)2n = Sum of coefficient of middle term of (1 + x)2nβ1 2nCn = 2n β 1Cn β 1 + 2n β 1Cn Thus, We have to prove 2nCn = 2n β 1Cn β 1 + 2n β 1Cn 2nCn = 2π!/π!(2π β π)! = 2π!/(π!(π)!) 2n β 1Cn β 1 + 2n β 1Cn = ((2π β 1)!)/((π β 1) ![(2π β1) β (π β 1)]!) + (2π β 1)!/(π! (2π β 1 β π)!) = ((2π β 1)!)/((π β 1) !(2π β 1 β π + 1)!) + (2π β1)!/(π! (2π β1 β π)!) = ((2π β 1)!)/((π β1)! (π)!) + (2π β 1)!/(π! (π β1)!) = 2 Γ ((2π β 1)!)/(π! (π β 1)! ) = 2 Γ ((2π β 1)!)/(π! (π β 1)! ) Γ π/π = (2π(2π β 1)!)/(π!π(π β 1)! ) = 2π!/(π! π! ) = L.H.S Hence L.H.S = R.H.S Hence proved