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

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Question

$\left(2 x^3-\frac{1}{3 x^2}\right)^5$

$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}

Vedclass · Accepted Answer

$\frac{80}{9}$

Vedclass · Answer

$\left(2 x^3-\frac{1}{3 x^2}ight)^5$

$T_{r+1}={ }^5 C_r\left(2 x^3ight)^{5-r}\left(\frac{-1}{3 x^2}ight)^r={ }^5 C_r \frac{(2)^{5-r}}{(-3)^r}(x)^{15-5 r}$

$	herefore 15-5 r =5$

$	herefore r =2$

$T_3=10\left(\frac{8}{9}ight) x^5$

So, coefficient is $\frac{80}{9}$

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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