Question

If the coefficient of x in ${(x^2+\frac{k}{x})}^5$ is 270, then k =

• A. 3
• B. 4
• C. 5
• D. None

Answer 1

The correct option is A

3

$T_{r+1}$=${}^5{C}_r\left(x^2{)}^{5-r}\right(\frac{k}{k}{)}^r$

=${}^5{C}_r$$k^r{x}^{10-3r}$.

For coefficient of x, 10-3r=1⟹ r=3.

∴ coefficient of x=${}^5{C}_3{k}^r$ =270(Given)

⇒ $k^3$ = $\frac{270}{10}$⇒ k=3.