Discussion Forum



1)

The sum of the coefficients of all odd degree terms in the expansion is

(x+√x3−1)5+(x−√x3−1)5,(x>1)is


A) -1

B) 0

C) 1

D) 2

Answer:

Option D

Explanation:

Key Idea =  (a+b)n+(a−b)n

    = 2(nC0an+nC2an−2b2+nC4an−4b4.....)

We have,

           (x+√x3−1)5+(x−√x3−1)5,x>1

           = 2(5C0x5+5C2x3(√x3−1)2+5C4x(√x3−1)4)

           = 2(x5+10x3(x3−1)+5x(x3−1)2)

           = 2(x5+10x6−10x3+5x7−10x4+5x)

  Sum of coefficients of all odd degree terms is 2 (1-10+5+5)=2