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(D) Consider the expansion $(1-3x+2x^3)(\frac{3x^2}{2}-\frac{1}{3x})^9$. First,find the general term $T_{r+1}$ in the expansion of $(\frac{3x^2}{2}-\f

Vedclass · Accepted Answer

$\frac{17}{54}$

Vedclass · Answer

(D) Consider the expansion $(1-3x+2x^3)(\frac{3x^2}{2}-\frac{1}{3x})^9$. First,find the general term $T_{r+1}$ in the expansion of $(\frac{3x^2}{2}-\frac{1}{3x})^9$: $T_{r+1} = {}^9C_r (\frac{3x^2}{2})^{9-r} (-\frac{1}{3x})^r = {}^9C_r (\frac{3}{2})^{9-r} (-\frac{1}{3})^r x^{18-2r-r} = {}^9C_r (\frac{3}{2})^{9-r} (-\frac{1}{3})^r x^{18-3r}$. To find the term independent of $x$ in the product,we look for terms in the expansion that result in $x^0$,$x^1$,and $x^{-3}$ from the second bracket to multiply with $1$,$-3x$,and $2x^3$ respectively. $1$. For $x^0$: $18-3r=0 \Rightarrow r=6$. Term $= 1 	imes {}^9C_6 (\frac{3}{2})^3 (-\frac{1}{3})^6 = 84 	imes \frac{27}{8} 	imes \frac{1}{729} = \frac{84 	imes 27}{8 	imes 729} = \frac{21}{2 	imes 27} = \frac{21}{54} = \frac{7}{18}$. $2$. For $x^1$: $18-3r=1 \Rightarrow 3r=17$ (no integer solution). $3$. For $x^{-3}$: $18-3r=-3$ $\Rightarrow 3r=21$ $\Rightarrow r=7$. Term $= 2x^3 	imes {}^9C_7 (\frac{3}{2})^2 (-\frac{1}{3})^7 = 2 	imes 36 	imes \frac{9}{4} 	imes (-\frac{1}{2187}) = 72 	imes \frac{9}{4} 	imes (-\frac{1}{2187}) = 162 	imes (-\frac{1}{2187}) = -\frac{162}{2187} = -\frac{2}{27}$. Total constant term $= \frac{7}{18} - \frac{2}{27} = \frac{21-4}{54} = \frac{17}{54}$.

The term independent of $x$ in the expansion of $(1-3x+2x^3)(\frac{3x^2}{2}-\frac{1}{3x})^9$ is

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