Question

Numerically greatest term in the expansion of $(x-\frac{1}{x}{)}^{11}$ when x=2 is given by

• A. 0
• B. 2¹¹
• C. 2¹¹.¹¹C₅
• D. None

Answer 1

The correct option is D

None

Let $T_{r+1}$ be the greatest term, then
$\frac{r}{n-r+1}\le x\le \frac{r+1}{n-r}\frac{r}{12-r}\le \frac{1}{4}\le \frac{r+1}{11-r}$

$\Rightarrow \frac{7}{5}\le r\le \frac{12}{5}\Rightarrow r=2$

$\therefore$ Greatest term in the expansion
=${}^{11}{C}_2(\frac{1}{4}{)}^2{.2}^{11}=55\times 2^7$