Example 6 - Chapter 8 Class 11 Binomial Theorem

Transcript

Example 6 Show that the middle term in the expansion of (1 + x)2n is 1 . 3 . 5 . (2 1) ! 2n xn, where n is a positive integer. 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-r br For Tn + 1 , Putting n = 2n , r = n , a = 1 & b = x Tn+1 = 2nCn (1)2n n (x)n = 2 ! ! 2 ! . (1)n . xn = 2 ! ! ! . xn = 2 2 1 2 2 .4 3 2 1 ! ! xn = 2 1 2 3 . . 5 3 1 [ 2 2 2 4 2] ! ! xn = 1 3 5 2 3 2 1 [2 4 6 .. 2 2 2 ] ! ! xn = 1 3 5 . 2 3 2 1 [ 2 1 (2 2) (2 3) .. 2 1 2( )] ! ! xn = 1 3 5 . 2 3 2 1 2 2 2 2 .. 2 [1 2 3 .. 1 ] ! ! xn = 1 3 5 . 2 3 2 1 2 [1 2 3 .. 1 ] ! ! xn = 1 3 5 . 2n 3 2n 1 2n (n!) n! n! . xn = 1 3 5 . 2n 1 n! 2n . xn Hence middle term of expansion (1 + x)2n is 1 . 3 . 5 . . 2n 1 n! 2n . xn Hence proved