Example 12 - Chapter 8 Class 11 Binomial Theorem

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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 – 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 Tn+1 = nCr an– rbr 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