If rth term in the expansion of

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 $\mathrm{x}$.

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