Question

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

Answer 1

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

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

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

Hence ,the general term in expansion by

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

${}^9{C}_r\left(\frac{3}{2}{)}^{q-r}\right(\frac{-1}{3}{)}^r{x}^{18-3r}{+}^9{C}_r\left(\frac{3}{2}{)}^{9-r}\right(\frac{-1}{3}{)}^r{x}^{19-3r}{+}^{2.9}{C}_r(\frac{3}{2}{)}^{q-r}+(\frac{-1}{3}{)}^r{x}^{21-3r}$

For term independent of x, put

18-3r=0 and 21-3r=0

r=6 and r=7

$\therefore 2^{nd}$ term is not independent of x. (ie.r=6)

$\therefore$ term independent of x is

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

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

$=\frac{84}{8}.\frac{1}{3^3}-\frac{36}{4}-\frac{2}{3^5}=\frac{21-4}{54}=\frac{17}{54}$