$\begin{aligned}
& \left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \
& \text { General term } m\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \
& ={ }^9 C_r \cdot \frac{3^{9-2 r}}{2^{9-r}}(-1)^r \cdot x^{18-3 r}
\end{aligned}Putr=6togetcoeff.ofx^0={ }^9 C_6 \cdot \frac{1}{6^3} \cdot x^0=\frac{7}{18} x^0Put\mathrm{r}=7togetcoeff.of\mathrm{x}^{-3}={ }^9 \mathrm{C}_{\mathrm{r}} \cdot \frac{3^{-5}}{2^2}(-1)^7 \cdot \mathrm{x}^{-3}\begin{aligned}
& =-{ }^9 C_7 \cdot \frac{1}{3^5 \cdot 2^2} \cdot x^{-3}=\frac{-1}{27} x^{-3} \
& \left(1+2 x-3 x^3\right)\left(\frac{7}{18} x^0-\frac{1}{27} x^{-3}\right) \
& \frac{7}{18}+\frac{3}{27}=\frac{7}{18}+\frac{1}{9}=\frac{7+2}{18}=\frac{9}{18}=\frac{1}{2} \
& \therefore 108 \cdot \frac{1}{2}=54
\end{aligned}$
If the constant term in the expansion of (1+2 x-3 x^3 ) ( 3 2 x^2- 1 3 x )^9 is , then 108 is equal to
JEE Main 2024 — Mathematics Algebra
If the constant term in the expansion of (1+2x−3x3)(23x2−3x1)9 is p, then 108p is equal to
Held on 5 Apr 2024 · Verified 6 Jul 2026.
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