The independent term in the expansion of (1+x | vedclass

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

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$\frac{7}{18}$

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(B) The general term in the expansion of $(\frac{3x^2}{2}-\frac{1}{3x})^9$ is $T_{r+1} = {}^9C_r (\frac{3x^2}{2})^{9-r} (-\frac{1}{3x})^r = {}^9C_r (\frac{3}{2})^{9-r} (-\frac{1}{3})^r x^{18-3r}$.Expanding $(1+x+2x^2) 	imes {}^9C_r (\frac{3}{2})^{9-r} (-\frac{1}{3})^r x^{18-3r}$,we get terms with $x^{18-3r}$,$x^{19-3r}$,and $x^{20-3r}$.For the independent term,we set the exponents to $0$:$1$) $18-3r = 0 \Rightarrow r = 6$. The term is ${}^9C_6 (\frac{3}{2})^3 (-\frac{1}{3})^6 = 84 	imes \frac{27}{8} 	imes \frac{1}{729} = \frac{84 	imes 27}{8 	imes 729} = \frac{21}{54} = \frac{7}{18}$.$2$) $19-3r = 0 \Rightarrow r = 19/3$ (not an integer).$3$) $20-3r = 0 \Rightarrow r = 20/3$ (not an integer).Thus,the independent term is $\frac{7}{18}$.

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

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