If the constant term in the expansion of $\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3…
If the constant term in the expansion of $\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$ is $\mathrm{p}$, then $108 \mathrm{p}$ is equal to
Solution
$\begin{aligned}
& \left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \\
& \text { General term } m\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \\
& ={ }^9 C_r \cdot \frac{3^{9-2 r}}{2^{9-r}}(-1)^r \cdot x^{18-3 r}
\end{aligned}$
Put $r=6$ to get coeff. of $x^0={ }^9 C_6 \cdot \frac{1}{6^3} \cdot x^0=\frac{7}{18} x^0$
Put $\mathrm{r}=7$ to get coeff. of $\mathrm{x}^{-3}={ }^9 \mathrm{C}_{\mathrm{r}} \cdot \frac{3^{-5}}{2^2}(-1)^7 \cdot \mathrm{x}^{-3}$
$\begin{aligned}
& =-{ }^9 C_7 \cdot \frac{1}{3^5 \cdot 2^2} \cdot x^{-3}=\frac{-1}{27} x^{-3} \\
& \left(1+2 x-3 x^3\right)\left(\frac{7}{18} x^0-\frac{1}{27} x^{-3}\right) \\
& \frac{7}{18}+\frac{3}{27}=\frac{7}{18}+\frac{1}{9}=\frac{7+2}{18}=\frac{9}{18}=\frac{1}{2} \\
& \therefore 108 \cdot \frac{1}{2}=54
\end{aligned}$