If the term independent of x in the expansion of

Question:

If the term independent of $x$ in the expansion of

$\left(\frac{3}{2} \mathrm{x}^{2}-\frac{1}{3 \mathrm{x}}\right)^{9}$ is $\mathrm{k}$, then $18 \mathrm{k}$ is equal to :

  1. 9

  2. 11

  3. 5

  4. 7


Correct Option: , 4

Solution:

$\mathrm{T}_{\mathrm{r}+1}={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\frac{3}{2} \mathrm{x}^{2}\right)^{9-\mathrm{r}}\left(-\frac{1}{3 \mathrm{x}}\right)^{\mathrm{r}}$

$\mathrm{T}_{\mathrm{r}+1}={ }^{9} \mathrm{C}_{\mathrm{r}}\left(\frac{3}{2}\right)^{9-\mathrm{r}}\left(-\frac{1}{3}\right)^{\mathrm{r}} \mathrm{x}^{18-3 \mathrm{r}}$

For independent of $x$

$18-3 r=0, r=6$

$\therefore \quad \mathrm{T}_{7}={ }^{9} \mathrm{C}_{6}\left(\frac{3}{2}\right)^{3}\left(-\frac{1}{3}\right)^{6}=\frac{21}{54}=\mathrm{k}$

$\therefore \quad 18 \mathrm{k}=\frac{21}{54} \times 18=7$

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