The coefficient of x^5 in the expansion of le | vedclass

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(C) The general term of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is given by $T_{r+1} = { }^5 C_r (2 x^3)^{5-r} (-\frac{1}{3 x^2})^r$. Simplifying the e

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$\frac{80}{9}$

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(C) The general term of $\left(2 x^3-\frac{1}{3 x^2}ight)^5$ is given by $T_{r+1} = { }^5 C_r (2 x^3)^{5-r} (-\frac{1}{3 x^2})^r$. Simplifying the expression: $T_{r+1} = { }^5 C_r (2)^{5-r} (x^3)^{5-r} (-\frac{1}{3})^r (x^{-2})^r = { }^5 C_r (2)^{5-r} (-\frac{1}{3})^r x^{15-3r-2r} = { }^5 C_r (2)^{5-r} (-\frac{1}{3})^r x^{15-5r}$. To find the coefficient of $x^5$,we set the exponent of $x$ equal to $5$: $15 - 5r = 5 \implies 5r = 10 \implies r = 2$. Substituting $r = 2$ into the coefficient part: ${ }^5 C_2 (2)^{5-2} (-\frac{1}{3})^2 = 10 	imes 2^3 	imes \frac{1}{9} = 10 	imes 8 	imes \frac{1}{9} = \frac{80}{9}$.

The coefficient of $x^5$ in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is

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