Question

In the expansion of ${(x^2-\frac{1}{3x})}^9$, the term without x is equal to

• A. $\frac{28}{81}$
• B. $\frac{-28}{243}$
• C. $\frac{28}{243}$
• D. None of these

Answer 1

The correct option is C

$\frac{28}{243}$

$\frac{28}{243}$

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

Then, we have:

$T_{r=1}={}^9{C}_r(x^2{)}^{9-r}{\left(\frac{-1}{3x}\right)}^r$

$=(-1{)}^r{}^9{C}_r\frac{1}{3^r}{x}^{18-2r-r}$

$=(-1{)}^r{}^9{C}_r\frac{1}{3^r}{x}^{18-2r-r}$

For this term to be independent of x, we must have: 18-3r=0

$\Rightarrow r=6$

$\therefore$ Required term $=(-1{)}^6{}^9{C}_6\frac{1}{3^6}$

$=\frac{9\times 8\times 7}{3\times 2}\times \frac{1}{3^6}=\frac{28}{243}$