If the term independent of x in the expansion

Question:

If the term independent of $x$ in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ is $k$, then $18 k$ is equal to :

  1. (1) 5

  2. (2) 9

  3. (3) 7

  4. (4) 11


Correct Option: , 3

Solution:

General term $=T_{r+1}={ }^{9} C_{r}\left(\frac{3 x^{2}}{2}\right)^{9-r}\left(-\frac{1}{3 x}\right)^{r}$

$={ }^{9} C_{r}\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^{r} x^{18-3 r}$

The term is independent of $x$, then

$18-3 r=0 \Rightarrow r=6$

$\therefore T_{7}={ }^{9} C_{6}\left(\frac{3}{2}\right)^{3}\left(-\frac{1}{3}\right)^{6}={ }^{9} C_{3}\left(\frac{1}{6}\right)^{3}$

$=\frac{9 \times 8 \times 7}{3 \times 2 \times 1}\left(\frac{1}{6}\right)^{3}=\left(\frac{7}{18}\right)$

$\therefore 18 k=18 \times \frac{7}{18}=7$.

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