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

Question:

Find the term independent of x in the expansion of :

$\left(2 x+\frac{1}{3 x^{2}}\right)^{9}$

Solution:

To Find : term independent of x, i.e. x0

For $\left(2 x+\frac{1}{3 x^{2}}\right)^{9}$

$a=2 x, b=\frac{1}{3 x^{2}}$ and $n=9$

We have a formula,

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

$=\left(\begin{array}{l}9 \\ r\end{array}\right)(2 x)^{9-r}\left(\frac{1}{3 x^{2}}\right)^{r}$

$=\left(\begin{array}{l}9 \\ r\end{array}\right)(x)^{9-r}(2)^{9-r}\left(\frac{1}{3}\right)^{r}\left(\frac{1}{x^{2}}\right)^{r}$

$=\left(\begin{array}{l}9 \\ r\end{array}\right)(x)^{9-r} \frac{(2)^{9-r}}{(3)^{r}}(x)^{-2 r}$

$=\left(\begin{array}{l}9 \\ r\end{array}\right) \frac{(2)^{9-r}}{(3)^{r}}(x)^{9-r-2 r}$

$=\left(\begin{array}{l}9 \\ r\end{array}\right) \frac{(2)^{9-r}}{(3)^{r}}(x)^{9-3 r}$

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

$(x)^{9-3 r}=x^{0}$

- $9-3 r=0$

- $3 r=9$

- $r=3$

Therefore, coefficient of $x^{0}=\left(\begin{array}{l}9 \\ 3\end{array}\right) \frac{(2)^{9-3}}{(3)^{3}}$

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

$=\frac{1792}{3}$

Conclusion : coefficient of $x^{0}=\frac{1792}{3}$