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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem

Find the term independent of $x$ in the expansion of $(3x-\large\frac{2}{x^2})^{15}$

$\begin{array}{1 1}(A)\;-3003(3^{10})(2^5)\\(B)\;3003(3^{10})(2^5)\\(C)\;300(3^{11})(2^5)\\(D)\;-300(3^{15})(2^{10})\end{array} $

1 Answer

Toolbox:
  • $T_{r+1}=nC_ra^{n-r}b^r$
$(3x-\large\frac{2}{x^2})^{15}$$=(-1)^r 15C_r(3x)^{n-r}(\large\frac{2}{x^2})^r$
$T_{r+1}=(-1)^r 15C_r(3)^{15-r}(x)^{n-r}\large\frac{2^r}{x^{2r}}$
$\Rightarrow (-1)^r 15C_r(3)^{15-r}(x)^{15-r}\large\frac{2^r}{x^{2r}}$
Since the term is independent of $x$ in the expansion we have
$15-r-2r=0$
$15-3r=0$
$r=5$
Hence $6^{th}$ term is independent of $x$
$T_6=(-1)^515C_5(3)^{15-5}.2^5$
$T_6=(-1)^515C_5 3^{10}.2^5$
$\Rightarrow \large\frac{15!}{5!10!}$$ 3^{10}.2^5$
$\Rightarrow \large\frac{-15\times 14\times 13\times 12\times 11\times 10!}{5\times 4\times 3\times 2\times10!}$$ 3^{10}.2^5$
$\Rightarrow -3003(3^{10})(2^5)$
Hence (A) is the correct answer.
answered Jun 24, 2014 by sreemathi.v
 

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