Mathematics - Binomial Theorem Question with Solution | TestHub

For ${\left(2x^3+\frac{3}{x^k}\right)}^{12}$

$T_{r+1}={}^{12}C_r{\left(2x^3\right)}^{12-r}·{\left(\frac{3}{x^k}\right)}^r={}^{12}C_r·2^{12-r}·3^r·x^{36-3r-kr}$

For the term to be independent of $x$,

$36-3r-kr=0$ $\Rightarrow r=\frac{36}{k+3}$ 

Here $k$ can be $0,1,3,6,9$ for $r$ to be a whole number.

But for ${}^{12}C_r·2^{12-r}·3^r$ to be in the form of $2^8·l$

Only $k=3, 6$ satisfies the above relation.