Question

The value of the term independent of x in the expansion of (x2−1x)9 is:

• A. (A) 9
• B. (B) 18
• C. (C) 48
• D. (D) 32

Answer 1

Answer: (D)

(D) 32

Answer 2

Explanation:
Concept:In the binomial expansion of (a+b)n, the term which does not involve any variable is said to be an independent term.The general term in the binomial expansion of (a+b)n is given by: Tr+1=nCr×an−r×brGiven: (x2−1x)9Let (r+1)th be the independent term in the expansion of (x2−1x)9.We know that the general term in the binomial expansion of (a+b)n is given by:Tr+1=nCr×an−r×brHere, a=x2,n=9 and b=1x.⇒Tr+1=9Cr×x2(9−r)×(1x)r=9Cr×x18−3r∵ The (r+1)th term is the independent term in the expansion of (x2−1x)9⇒x18−3r =x0⇒18−3r=0⇒r=6⇒7th term in the expansion of (x2−1x)9 is the independent term.We have to find the value of the 7th term in the expansion of (x2−1x)9 ⇒T7=9C6×1=84Hence, the correct option is (D).