Question

Find the term independent of $x$ in the expansion of $(1+x+2x^3){(\frac{3x^2}{2}-\frac{1}{3x})}^9$

Answer 1

Given expansion is $(1+x+2x^3){(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$

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

$T_{r+1}{=}^9{C}_r{\left(\frac{3}{2}{x}^2\right)}^{9.r}{(-\frac{1}{3x})}^r{=}^9{C}_r{\left(\frac{3}{2}\right)}^{9-x}{(-\frac{1}{3})}^r{x}^{100-3r}$

Hence, the general term in the expansion of $(1+x+2x^3){(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$ is

${}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}$ ${\left(\frac{-1}{3}\right)}^r{x}^{18-3r}{+}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}{\left(\frac{-1}{3}\right)}^r{x}^{19-3r}+2{}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}{(-\frac{1}{3})}^r{x}^{21-3r}$

For term independent of x, putting $18-3r=0$, $19-3r=0$ & $21-3r=0$, we get

$r=6,7$

Hence, ground term is not independent of x, therefore, term independent of x is?

${}^9{C}_6{\left(\frac{3}{2}\right)}^{9-6}$ ${(-\frac{1}{3})}^6+2.{}^9{C}_7{\left(\frac{3}{2}\right)}^{9-7}{(-\frac{1}{3})}^7$

$=\frac{9\times 8\times 7\times 6!}{6!\times 3\times 2}\times \frac{1}{2^3{.3}^3}-2\cdot \frac{9\times 8\times 7!}{7!\times 2\times 1}\times \frac{3^2}{2^2}\times \frac{1}{3^7}$

$=dfrac848\times \frac{1}{33}-\frac{36}{4}\times \frac{2}{35}=\frac{21-4}{54}=\frac{17}{54}$.