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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Find the term independent of $x,x\neq 0$ in the expansion of $\big(\large\frac{3x^2}{2}-\frac{1}{3x})^{15}$

$\begin{array}{1 1}(A)\;14C_{10}(\large\frac{1}{6})^5\\(B)\;15C_{10}(\large\frac{1}{6})^5\\(C)\;5C_4(\large\frac{1}{6})^4\\(D)\;14C_{10}(\large\frac{1}{4})^5\end{array} $

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1 Answer

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Toolbox:
  • $T_{r+1}=nC_ra^{n-r}b^r$
Given :
$\big(\large\frac{3x^2}{2}-\frac{1}{3x})^{15}$
$T_{r+1}=15C_r(\large\frac{3x^2}{2})^{15-r}(\large\frac{1}{3x})^r$
$\Rightarrow 15C_r(\large\frac{3}{2})^{15-r}(x)^{30-2r}.(\large\frac{1}{3})^r.(\large\frac{1}{x})^r$
$\Rightarrow 15C_r(\large\frac{3}{2})^{15-r}.(\large\frac{1}{3})^r$$.x^{30-2r-r}$
Since the term is independent of $x$ we have
$30-2r-r=0$
$30-3r=0$
$3r=30$
$r=10$
Hence $11^{th}$ term is independent of x
$T_{11}=15C_{10}(\large\frac{3}{2})^{5-10}.(\large\frac{1}{3})^{10}$
$\Rightarrow 15C_{10}(\large\frac{3}{2})^{5}.(\large\frac{1}{3})^{10}$
$\Rightarrow 15C_{10}(\large\frac{3^5}{2^5}).(\large\frac{1}{3^{10}})$
$\Rightarrow 15C_{10}(\large\frac{1}{2^5}).(\large\frac{1}{3^{5}})$
$\Rightarrow 15C_{10}(\large\frac{1}{6^5})$
Hence (B) is the correct answer.
answered Jun 24, 2014 by sreemathi.v
 

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