# Find the term independent of $x$ in the expansion of $(1+x+2x^3)(\large\frac{3}{2}$$x^2-\large\frac{1}{3x})^9 \begin{array}{1 1}(A)\;25\\(B)\;28\\(C)\;\large\frac{15}{17}\\(D)\;\large\frac{17}{54}\end{array} ## 1 Answer Toolbox: • General term T_{r+1}=nC_r a^{n-r}b^r (1+x+2x^3)(\large\frac{3}{2}$$x^2-\large\frac{1}{3x})^9$
$\Rightarrow (1+x+2x^3)\big[(3/2)x^2]^9-9C_1 [(3/2)x^2]^8.\large\frac{1}{3}$$x+......+9C_6[(3/2)x^2]^3(\large\frac{1}{3}$$x)^6-9C_7[(3/2)x^2]^2(\large\frac{1}{3}$$x)^7......\big] \Rightarrow (1+x+2x^3)\big[[(3/2)x^2]^9-9C_1(3^7/2^8)x^{15}+.....+9C_6(1\times \large\frac{1}{2^3}$$\times 3^3)-9C_7\large\frac{1}{(2^2\times 3^5)}$$1/x^3+......\big] Term independent of x 9C_6\times \large\frac{1}{2^3\times 3^3}$$-9C_7\large\frac{2}{2^2\times 3^5}$
$\Rightarrow \large\frac{9!}{6!\times 3!}.\frac{1}{8\times 27}-\frac{9!}{7!\times 2!}\times \large\frac{1}{2\times 243}$
$\Rightarrow \large\frac{9\times 8\times 7\times 6!}{6!\times 3\times 2}.\frac{1}{8\times 27}-\frac{9\times 8\times 7!}{7!\times 2}.\frac{1}{2\times 243}$
$\Rightarrow \large\frac{7}{18}-\frac{2}{27}$
$\Rightarrow \large\frac{17}{54}$
Hence (C) is the correct answer.