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  4. If the coefficient of x in (x2+k / x)5 is 270, then the value of k is

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

Binomial Theorem

Solution:

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