Question

The value of the term independent of $x$ in the expansion of ${(x^2-\frac{1}{x})}^9$ is:

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

Answer 1

The Correct Option is D

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: $T_{r+1}={}^n{C}_r\times a^{n-r}\times b^r$Given: ${(x^2-\frac{1}{x})}^9$Let $(r+1{)}^{\text{th }}$ be the independent term in the expansion of ${(x^2-\frac{1}{x})}^9$.We know that the general term in the binomial expansion of $(a+b{)}^n$ is given by:$T_{r+1}={}^n{C}_r\times a^{n-r}\times b^r$Here, $a=x^2,n=9$ and $b=\frac{1}{x}$.$\Rightarrow T_{r+1}={}^9{C}_r\times x^{2(9-r)}\times {\left(\frac{1}{x}\right)}^r={}^9{C}_r\times x^{18-3r}$$\because$ The $(r+1{)}^{\text{th }}$ term is the independent term in the expansion of ${(x^2-\frac{1}{x})}^9\Rightarrow x^{18-3r}$ $=x^0$$\Rightarrow 18-3r=0\Rightarrow r=6$$\Rightarrow 7^{\text{th }}$ term in the expansion of ${(x^2-\frac{1}{x})}^9$ is the independent term.We have to find the value of the $7^{\text{th }}$ term in the expansion of ${(x^2-\frac{1}{x})}^9$ $\Rightarrow T_7={}^9{C}_6\times 1=84$Hence, the correct option is (D).