Question:
The term without $x$ in the expansion of $\left(2 x-\frac{1}{2 x^{2}}\right)^{12}$ is
(a) 495
(b) −495
(c) −7920
(d) 7920
Solution:
(d) 7920
Suppose the $(\mathrm{r}+1)$ th term in the given expansion is independent of $x$.
Then, we have :
$T_{r+1}={ }^{12} C_{r}(2 x)^{12-r}\left(\frac{-1}{2 x^{2}}\right)^{r}$
$=(-1)^{r}{ }^{12} C_{r} \quad 2^{12-2 r} \quad x^{12-r-2 r}$
For this term to be independent of $x$, we must have:
$12-3 r=0$
$\Rightarrow r=4$
$\therefore$ Required term :
$(-1)^{4}{ }^{12} C_{4} 2^{12-8}$
$=\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2} \times 16$
$=7920$