Example 12 - Chapter 8 Class 11 Binomial Theorem (Deleted)
Last updated at Jan. 29, 2020 by Teachoo
Last updated at Jan. 29, 2020 by Teachoo
Transcript
Example 12 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
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