Question

If the coefficient of $x$ in ${\left(x^2+k/x\right)}^5$ is $270$, then the value of $k$ is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B)

$T_{r+1}={}^5{C}_r{\left(x^2\right)}^{5-r}(k/x{)}^r={}^5{C}_r{k}^r{x}^{10-3r}$
For coefficient of $x,10-3r=1$
$\Rightarrow r=3$
coefficient of $x={}^5{C}_3{k}^3=270$
$\Rightarrow k^3=\frac{270}{10}=27$
$\therefore k=3$