JEE MAIN - Mathematics (2023 - 13th April Evening Shift - No. 7)
The coefficient of $$x^{5}$$ in the expansion of $$\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}$$ is :
$$\frac{26}{3}$$
$$\frac{80}{9}$$
9
8
Explanation
Given, $$\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}$$
General term,
$$ \begin{aligned} & T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r} \\\\ & \therefore 15-5 \mathrm{r}=5 \\\\ & \therefore \mathrm{r}=2 \\\\ & T_3=10\left(\frac{8}{9}\right) x^5 \end{aligned} $$
So, coefficient is $\frac{80}{9}$.
General term,
$$ \begin{aligned} & T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r} \\\\ & \therefore 15-5 \mathrm{r}=5 \\\\ & \therefore \mathrm{r}=2 \\\\ & T_3=10\left(\frac{8}{9}\right) x^5 \end{aligned} $$
So, coefficient is $\frac{80}{9}$.
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