Question

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

Answer 1

Here expansion given as :

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

Now, to find the term undependent of $x$ we have

${(\frac{3x^2}{2}-\frac{1}{3x})}^9=\sum _{i=0}^{r=0}{}^9{C}_r{\left(\frac{3x^2}{2}\right)}^{9-r}{\left(\frac{-1}{3x}\right)}^r$

$=\sum _{i=0}^{i=9}{}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}{x}^{\left(2\right)(9-r)}{\left(\frac{-1}{3}\right)}^r(x{)}^{-r}$

$=\sum _{i=0}^{i=9}{}^9{C}_r\times 2^{r-9}{x}^{18-3r}$ .......... $\left(1\right)$

$\therefore \left(1\right)$ [term independent of $x$ in (1)]$+\left(1\right)$ [Term containing $x^{-1}$]$+2$ [Term containing x^{-3}$]

$=\left(1\right)\left({}^9{C}_r\right(-1{)}^6{3}^{9-12}\times 2^{-3})+\left(1\right)\left(0\right)+\left[2\left({}^9{C}_7\right)\right(-1\left)\right]$

$=\left(1\right)\frac{9!}{6!3!}\times \frac{1}{27}\times \frac{1}{8}-\frac{2\times 9!}{7!2!}\times \frac{1}{35}\times \frac{1}{4}$

$=\frac{7\times 8\times 9}{6}\times \frac{1}{27}\times \frac{1}{8}-\frac{2\times 8\times 9}{2}\times \frac{1}{35}\times \frac{1}{4}$

$=\frac{7}{8}-\frac{2}{27}$

$=\frac{21-4}{54}=\frac{17}{54}$

Term independent of $x$ is $\frac{17}{54}$