Question

The middle term in the expansion of ${(x-\frac{1}{2x})}^{10}$ is equal to

• A. $\frac{105}{32x^2}$
• B. $\frac{63}{8}$
• C. $\frac{-105}{32x^2}$
• D. $\frac{-63}{8}$

Answer 1

The correct option is D $\frac{-63}{8}$

General term

$T_{r+1}={}^{10}{C}_r{x}^{10-r}{(-\frac{1}{2x})}^r$

For middle term, $r=5$

$T_6={}^{10}{C}_5{x}^5\left(\frac{-1}{2^5{x}^5}\right)$

$=-\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2\times 2^5}$

$=-\frac{3\times 2\times 7\times 3\times 2}{2\times 2\times 2\times 2\times 2}$

$T_6=\frac{-63}{8}$.