The coefficient of $x^{5}$ in the expansion of $\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}$ is :
Solution
Given, $\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}$
<br/><br/>General term,
<br/><br/>$$
\begin{aligned}
& T_{r+1}={ }^5 C_r\left(2 x^3\right)^{5-r}\left(\frac{-1}{3 x^2}\right)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r} \\\\
& \therefore 15-5 \mathrm{r}=5 \\\\
& \therefore \mathrm{r}=2 \\\\
& T_3=10\left(\frac{8}{9}\right) x^5
\end{aligned}
$$
<br/><br/>So, coefficient is $\frac{80}{9}$.
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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