Question

The term independent of $y$ in the expansion of $(y^{1/6}-y^{-1/3}{)}^9$ is

• A. 84
• B. 8.4
• C. -0-84
• D. -84

Answer 1

Answer: (D)

$T_{r+1}={}^9{c}_r{\left(y^{\frac{1}{6}}\right)}^{9-r}{\left(-y^{\frac{1}{3}}\right)}^r$
$={}^9{c}_r{y}^{\frac{9-r}{6}}\cdot {\left(-1\right)}^r{y}^{\frac{-r}{3}}$
$\therefore \frac{9-r}{6}-\frac{r}{3}=0$
$\Rightarrow r=3$
$\therefore$ reqd. term $={}^9{c}_3{\left(-1\right)}^3=-{}^9{c}_3$
$=\frac{9\times 8\times 7}{1\times 2\times 3}$
$=-3\times 4\times 7=-84$