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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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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} $

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1 Answer

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  • 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.
answered Jun 24, 2014 by sreemathi.v
 

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