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 ________.
Answer (integer)
54
Solution
<p>$\text { General term of }\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$</p>
<p>$$T_{r+1}={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r={ }^9 C_r(-1)^r 3^{9-2 r} 2^{r-9} x^{18-35}$$</p>
<p>Constant term in expansion of $\left(1+2 x-3 x^3\right)$</p>
<p>$$\begin{aligned}
& \left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \\
& =T_7-3 T_8={ }^9 C_6 3^{-3} \cdot 2^{-3}+3{ }^9 C_7 \cdot 3^{-5} \cdot 2^{-2} \\
& =\frac{3 \times 4 \times 7}{3^3 \cdot 2^3}+\frac{3 \times 9 \times 4}{3^5 \times 2^2}= \\
& p=\frac{42+12}{108}=\frac{54}{108} \\
& 108 p=54
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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