Find the term independent of $x$ in the expansion of
$$\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$$
Given expansion is $\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$
Now, consider $\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$
$$\begin{aligned} T_{r+1} & ={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r \\ & ={ }^9 C_r\left(\frac{3}{2}\right)^{9-r} x^{18-2 r}\left(-\frac{1}{3}\right)^r x^{-r}={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{18-3 r} \end{aligned}$$
Hence, the general term in the expansion of $\left(1+x+2 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$
$$={ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{18-3 r}+{ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{19-3 r}+2 \cdot{ }^9 C_r\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^r x^{21-3 r}$$
For term independent of $x$, putting $18-3 r=0,19-3 r=0$ and $21-3 r=0$, we get
$$r=6, r=19 / 3, r=7$$
Since, the possible value of $r$ are 6 and 7.
Hence, second term is not independent of $x$.
$$\begin{aligned} & =\frac{9 \times 8 \times 7 \times 6!}{6!\times 3 \times 2} \cdot \frac{3^3}{2^3} \cdot \frac{1}{3^6}-2 \cdot \frac{9 \times 8 \times 7!}{7!\times 2 \times 1} \cdot \frac{3^2}{2^2} \cdot \frac{1}{3^7} \\ & =\frac{84}{8} \cdot \frac{1}{3^3}-\frac{36}{4} \cdot \frac{2}{3^5}=\frac{7}{18}-\frac{2}{27}=\frac{21-4}{54}=\frac{17}{54} \end{aligned}$$
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