The coefficient of the term independent of x | vedclass

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(C) The general term in the expansion of $\left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$ is $T_{r+1} = \binom{9}{r} \left( \frac{3}{2}x^2 \right)^{9-

Vedclass · Accepted Answer

$\frac{17}{54}$

Vedclass · Answer

(C) The general term in the expansion of $\left( \frac{3}{2}x^2 - \frac{1}{3x} ight)^9$ is $T_{r+1} = \binom{9}{r} \left( \frac{3}{2}x^2 ight)^{9-r} \left( -\frac{1}{3x} ight)^r = \binom{9}{r} \left( \frac{3}{2} ight)^{9-r} \left( -\frac{1}{3} ight)^r x^{18-3r}$.We need the coefficient of the term independent of $x$ in $(1 + x + 2x^3) \left( \frac{3}{2}x^2 - \frac{1}{3x} ight)^9$.This is equal to the sum of the coefficients of $x^0$,$x^{-1}$,and $x^{-3}$ in the expansion of $\left( \frac{3}{2}x^2 - \frac{1}{3x} ight)^9$.$1$. For $x^0$: $18 - 3r = 0 \Rightarrow r = 6$. The coefficient is $\binom{9}{6} \left( \frac{3}{2} ight)^3 \left( -\frac{1}{3} ight)^6 = 84 	imes \frac{27}{8} 	imes \frac{1}{729} = \frac{84 	imes 27}{8 	imes 729} = \frac{7}{18}$.$2$. For $x^{-1}$: $18 - 3r = -1 \Rightarrow 3r = 19$,which has no integer solution for $r$.$3$. For $x^{-3}$: $18 - 3r = -3$ $\Rightarrow 3r = 21$ $\Rightarrow r = 7$. The coefficient is $2 	imes \binom{9}{7} \left( \frac{3}{2} ight)^2 \left( -\frac{1}{3} ight)^7 = 2 	imes 36 	imes \frac{9}{4} 	imes \left( -\frac{1}{2187} ight) = 162 	imes \left( -\frac{1}{2187} ight) = -\frac{2}{27}$.Summing these,the coefficient is $\frac{7}{18} - \frac{2}{27} = \frac{21 - 4}{54} = \frac{17}{54}$.

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

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