# Find the term independent of x in the expansion of :

Question:

Find the term independent of x in the expansion of :

$\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{6}$

Solution:

To Find : term independent of $x$, i.e. $x^{0}$

For $\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{6}$

$\mathrm{a}=\frac{3 \mathrm{x}^{2}}{2} \mathrm{~b}=-\frac{1}{3 \mathrm{x}}$ and $\mathrm{n}=6$

We have a formula,

$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

$=\left(\begin{array}{l}6 \\ r\end{array}\right)\left(\frac{3 x^{2}}{2}\right)^{6-r}\left(-\frac{1}{3 x}\right)^{r}$

$=\left(\begin{array}{l}6 \\ r\end{array}\right)\left(\frac{3}{2}\right)^{6-r}\left(x^{2}\right)^{6-r}\left(\frac{-1}{3}\right)^{r}\left(\frac{1}{x}\right)^{r}$

$=\left(\begin{array}{l}6 \\ r\end{array}\right)\left(\frac{3}{2}\right)^{6-r}\left(\frac{-1}{3}\right)^{r}(x)^{12-2 r}(x)^{-r}$

$=\left(\begin{array}{l}6 \\ r\end{array}\right)\left(\frac{3}{2}\right)^{6-r}\left(\frac{-1}{3}\right)^{r}(x)^{12-2 r-r}$

$=\left(\begin{array}{l}6 \\ r\end{array}\right)\left(\frac{3}{2}\right)^{6-r}\left(\frac{-1}{3}\right)^{r}(x)^{12-3 r}$

Now, to get coefficient of term independent of $x$ that is coefficient of $x^{0}$ we must have,

$(x)^{12-3 r}=x^{0}$

- $12-3 r=0$

- $3 r=12$

- $r=4$

Therefore, coefficient of $x^{0}=\left(\begin{array}{l}6 \\ 4\end{array}\right)\left(\frac{3}{2}\right)^{6-4}\left(\frac{-1}{3}\right)^{4}$

$=\left(\begin{array}{l}6 \\ 2\end{array}\right)\left(\frac{3}{2}\right)^{2} \frac{1}{81} \ldots \ldots \ldots\left[\because\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)\right]$

$=\frac{6 \times 5}{2 \times 1} \cdot \frac{9}{4} \cdot \frac{1}{81}$

$=\frac{15}{36}$

$\underline{\text { Conclusion }}$ : coefficient of $x^{0}=\frac{15}{36}$