Example 6 - Chapter 8 Class 11 Binomial Theorem (Deleted)
Last updated at Jan. 29, 2020 by Teachoo
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Example 6 Show that the middle term in the expansion of (1 + x)2n is (1 . 3 . 5 β¦. (2π β 1))/π! 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 β r br For Tn + 1 , Putting n = 2n , r = n , a = 1 & b = x Tn+1 = 2nCn (1)2n β n (x)n = (2π)!/π!(2π βπ)! . (1)n . xn = (2π)!/(π! π!) . xn = (2π(2π β 1)(2π β 2) β¦β¦. 4 Γ 3 Γ 2 Γ 1)/(π! π!) xn = ([(2π β 1)(2π β 3)β¦.β¦. Γ 5 Γ 3 Γ 1] [(2π)(2π β 2)β¦ Γ 4 Γ 2])/(π! π!) xn We need to show (1 . 3 . 5 β¦. (2π β 1))/π! 2n xn = ([1 Γ 3 Γ 5 Γ β¦β¦ Γ (2π β 3)(2π β 1)] [2 Γ 4 Γ 6β¦.. Γ (2π β 2) Γ 2π])/(π! π!) xn = ([1 Γ 3 Γ 5β¦β¦. Γ (2πβ3)(2πβ1)] [(2 Γ 1) Γ(2 Γ 2) Γ(2 Γ 3) Γ β¦.. Γ 2 (πβ1) Γ 2(π)])/(π! π!) xn = ([1 Γ 3 Γ 5β¦β¦. Γ (2πβ3)(2πβ1)] (2 Γ 2 Γ 2 Γ 2 β¦β¦..Γ 2) [1 Γ 2 Γ 3 β¦.. (πβ1) π])/(π! π!) xn = ([1 Γ 3 Γ 5β¦β¦. Γ (2π β 3)(2π β 1)] 2π [1 Γ 2 Γ 3 β¦.. (π β 1) π])/π!π! xn = ([1 Γ 3 Γ 5β¦β¦. Γ (2n β 3)(2n β 1)] 2n (n!))/(n! n!) . xn = (π Γ π Γ πβ¦β¦. Γ (ππ§ β π))/(π§! ) 2n . xn Hence middle term of expansion (1 + x)2n is (1 . 3 . 5β¦β¦. β¦.(2nβ1))/(n! ) 2n . xn Hence proved