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

Question:

Find the term independent of x in the expansion of :

$\left(3 x-\frac{2}{x^{2}}\right)^{15}$

Solution:

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

For $\left(3 x-\frac{2}{x^{2}}\right)^{15}$

$a=3 x, \quad b=\frac{-2}{x^{2}}$ and $n=15$

We have a formula

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

$=\left(\begin{array}{c}15 \\ \mathrm{r}\end{array}\right)(3 \mathrm{x})^{15-\mathrm{r}}\left(\frac{-2}{\mathrm{x}^{2}}\right)^{\mathrm{r}}$

$=\left(\begin{array}{c}15 \\ \mathrm{r}\end{array}\right)(3)^{15-\mathrm{r}}(\mathrm{x})^{15-\mathrm{r}}(-2)^{\mathrm{r}}\left(\frac{1}{\mathrm{x}^{2}}\right)^{\mathrm{r}}$

$=\left(\begin{array}{c}15 \\ \mathrm{r}\end{array}\right)(3)^{15-\mathrm{r}}(\mathrm{x})^{15-\mathrm{r}}(-2)^{\mathrm{r}}(\mathrm{x})^{-2 \mathrm{r}}$

$=\left(\begin{array}{c}15 \\ \mathrm{r}\end{array}\right)(3)^{15-\mathrm{r}}(-2)^{\mathrm{r}}(\mathrm{x})^{15-\mathrm{r}-2 \mathrm{r}}$

$=\left(\begin{array}{c}15 \\ \mathrm{r}\end{array}\right)(3)^{15-\mathrm{r}}(-2)^{\mathrm{r}}(\mathrm{x})^{15-3 \mathrm{r}}$

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

$(x)^{15-3 r}=x^{0}$

- $15-3 r=0$

- $3 r=15$

- $r=5$

Therefore, coefficient of $x^{0^{0}}=\left(\begin{array}{c}15 \\ 5\end{array}\right)(3)^{15-5}(-2)^{5}$

$=\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \cdot(3)^{10} \cdot(-32)$

$=-3003 .(3)^{10} .(32)$

$\underline{\text { Conclusion }}$ : coefficient of $x^{0}=-3003 \cdot(3)^{10} \cdot(32)$