Question

The term without x in the expansion of $(2x-\frac{1}{2x^2})$ is

• A. 495
• B. -495
• C. -7920
• D. 7920

Answer 1

The correct option is D

7920

Suppose the (r+1)th term in the given expansion is independent of x.

Then, we have:

$T_{n+1}={}^{12}{C}_r(2x{)}^{12-r}{\left(\frac{-1}{2x^2}\right)}^r$

$=(-1{)}^r{}^{12}{C}_r{2}^{12-2r}{x}^{12-r-2r}$

For this term to be independent of x, we must have:

12-3r =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