Question:
If $r$ th term in the expansion of $\left(2 x^{2}-\frac{1}{x}\right)^{12}$ is without $x$, then $r$ is equal to
(a) 8
(b) 7
(c) 9
(d) 10
Solution:
(c) 9
$r$ th term in the given expansion is ${ }^{12} C_{r-1}\left(2 x^{2}\right)^{12-r+1}\left(\frac{-1}{x}\right)^{r-1}$
$=(-1)^{r-1}{ }^{12} C_{r-1} 2^{13-r} x^{26-2 r-r+1}$
For this term to be independent of $x$, we must have :
$27-3 r=0$
$\Rightarrow r=9$
Hence, the 9 th term in the expansion is independent of $\mathrm{x}$.