Question

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

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (A)

$3$

The general term is
$t_{r+1}{=}^5{C}_r.(x^2{)}^r.(\frac{k}{x}{)}^{5-r}$
If the power of $x$ in above general term should be $1$, we have,
$2r-(5-r)=1$ which gives $r=2$.
Putting this value of $r$ in the general term expression, we get the coefficient of $x$.
The coefficient is $10k^3$.
This is given to be equal to $270$.
Equating this, we get the value of $k$ as $3$.