Examples

Chapter 7 Class 11 Binomial Theorem
Serial order wise

### Transcript

Question 2 Show that the middle term in the expansion of (1 + x)2n is (1 . 3 . 5 ā¦. (2š ā 1))/š! 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 ā r br For Tn + 1 , Putting n = 2n , r = n , a = 1 & b = x Tn+1 = 2nCn (1)2n ā n (x)n = (2š)!/š!(2š āš)! . (1)n . xn = (2š)!/(š! š!) . xn = (2š(2š ā 1)(2š ā 2) ā¦ā¦. 4 Ć 3 Ć 2 Ć 1)/(š! š!) xn = ([(2š ā 1)(2š ā 3)ā¦.ā¦. Ć 5 Ć 3 Ć 1] [(2š)(2š ā 2)ā¦ Ć 4 Ć 2])/(š! š!) xn We need to show (1 . 3 . 5 ā¦. (2š ā 1))/š! 2n xn = ([1 Ć 3 Ć 5 Ć ā¦ā¦ Ć (2š ā 3)(2š ā 1)] [2 Ć 4 Ć 6ā¦.. Ć (2š ā 2) Ć 2š])/(š! š!) xn = ([1 Ć 3 Ć 5ā¦ā¦. Ć (2šā3)(2šā1)] [(2 Ć 1) Ć(2 Ć 2) Ć(2 Ć 3) Ć ā¦.. Ć 2 (šā1) Ć 2(š)])/(š! š!) xn = ([1 Ć 3 Ć 5ā¦ā¦. Ć (2šā3)(2šā1)] (2 Ć 2 Ć 2 Ć 2 ā¦ā¦..Ć 2) [1 Ć 2 Ć 3 ā¦.. (šā1) š])/(š! š!) xn = ([1 Ć 3 Ć 5ā¦ā¦. Ć (2š ā 3)(2š ā 1)] 2š [1 Ć 2 Ć 3 ā¦.. (š ā 1) š])/š!š! xn = ([1 Ć 3 Ć 5ā¦ā¦. Ć (2n ā 3)(2n ā 1)] 2n (n!))/(n! n!) . xn = (š Ć š Ć šā¦ā¦. Ć (šš§ ā š))/(š§! ) 2n . xn Hence middle term of expansion (1 + x)2n is (1 . 3 . 5ā¦ā¦. ā¦.(2nā1))/(n! ) 2n . xn Hence proved