Question

The term independent of $x$ in $(1+x+2x^3){(\frac{{3x}^2}{2}-\frac{1}{3x})}^9$ is

• A. $\frac{25}{54}$
• B. $\frac{217}{72}$
• C. $\frac{250}{54}$
• D. $17$

Answer 1

The correct option is B $\frac{217}{72}$

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

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

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

$18-3r=0$

$r=6$

$2(9-r)-r=1$

$18-3r=1$

$12=3r$

$r=17/3$

$2(9-r)-r=3$

$18-3r=3$

$15=3r$

$r=\frac{15}{3}=5$

$r=5$

${}^9{C}_6{\left(\frac{3}{2}\right)}^3{\left(\frac{1}{3}\right)}^6$ ${}^9{C}_5{\left(\frac{3}{2}\right)}^4{\left(\frac{1}{3}\right)}^5$

$\Rightarrow \frac{987}{3\times 2}\left(\frac{1}{8}\right)\left(\frac{1}{27}\right)$ $\frac{9876}{432}\frac{1}{16}\times 3$

$\Rightarrow \frac{7}{18}$ $\frac{21}{8}$

$\Rightarrow \frac{7}{18}+\frac{21}{8}=\frac{28+189}{172}$

$=\frac{217}{72}$