If k is the coefficient of x^5 in the expansi | vedclass

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

Vedclass · Accepted Answer

-$40$

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(B) The general term $T_{r+1}$ in the expansion of $(2x^2 - \frac{1}{3x^3})^5$ is given by:$T_{r+1} = {}^5C_r (2x^2)^{5-r} (-\frac{1}{3x^3})^r$$T_{r+1} = {}^5C_r (2)^{5-r} (-\frac{1}{3})^r x^{10-2r-3r}$$T_{r+1} = {}^5C_r (2)^{5-r} (-\frac{1}{3})^r x^{10-5r}$For the coefficient of $x^5$,we set the exponent $10-5r = 5$:$5r = 5 \implies r = 1$Substituting $r=1$ to find $k$:$k = {}^5C_1 (2)^{5-1} (-\frac{1}{3})^1 = 5 	imes 16 	imes (-\frac{1}{3}) = -\frac{80}{3}$Finally,calculating $\frac{3k}{2}$:$\frac{3k}{2} = \frac{3}{2} 	imes (-\frac{80}{3}) = -40$

If $k$ is the coefficient of $x^5$ in the expansion of $(2x^2 - \frac{1}{3x^3})^5$,then $\frac{3k}{2} = $

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