If the constant term in the expansion of (1+2x-3x^3)((3)/(2)x^2-(1)/(3x))^9 is p, then 108…

Question

If the constant term in the expansion of $(1+2x-3{x}^{3}){(\frac{3}{2}{x}^{2}-\frac{1}{3x})}^{9}$ is $p$, then $108p$ is equal to \_\_\_\_\_\_\_\_.

Answer: 54

Accepted Answer

${(\frac{3}{2}{x}^{2}-\frac{1}{3x})}^{9}$

${T}_{r+1}={}^{9}{C}_{r}{(\frac{3}{2}{x}^{2})}^{9-r}{(-\frac{1}{3x})}^{r}={}^{9}{C}_{r}(-1{)}^{r}{3}^{9-2r}{2}^{r-9}{x}^{18-35}$

Constant term in expansion of $(1+2x-3{x}^{3})$

$\begin{gathered} {(\frac{3}{2}{x}^{2}-\frac{1}{3x})}^{9} \ ={T}_{7}-3{T}_{8}={}^{9}{C}_{6}{3}^{-3}\cdot {2}^{-3}+3{}^{9}{C}_{7}\cdot {3}^{-5}\cdot {2}^{-2} \ =\frac{3	imes 4	imes 7}{{3}^{3}\cdot {2}^{3}}+\frac{3	imes 9	imes 4}{{3}^{5}	imes {2}^{2}}= \ p=\frac{42+12}{108}=\frac{54}{108} \ 108p=54 \end{gathered}$

Solutions · 1