Question

The term containing $x$ in the expansion of ${\left(x^2+\frac{1}{x}\right)}^5$ is -

• A. $2^{\text{nd }}$
• B. $3^{\text{rd }}$
• C. $4^{\text{th }}$
• D. $5^{\text{th }}$

Answer 1

Answer: (C)

Let its come in $T_{r+1}$
then $T_{r+1}={}^5{C}_r{\left(x^2\right)}^{5-r}\cdot x^{-r}$
$\Rightarrow 10-3r=1$
$3r=9$
$r=3$
Hence $T_{3+1}=T_4$