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

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

Vedclass · Accepted Answer

$\frac{80}{9}$

Vedclass · Answer

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

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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