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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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The power of x occuring in the $7^{th}$ term in the expansion of $\big(\large\frac{4x}{5}-\frac{8}{5x}\big)^9$ is

$\begin{array}{1 1}(A)\;3\\(B)\;-3\\(C)\;\large\frac{160}{27}\\(D)\;\large\frac{80}{3}\end{array} $

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1 Answer

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Toolbox:
  • $T_{r+1}=nC_r (a)^{n-r} b^r$
$T_{n+r}=6C_r (2x)^{6-r} (\large\frac{1}{3x})^r$
$\qquad=6C_r \large\frac{2^{6-r}}{3^r}$$x^{6-2r}$
Let $T_{r+1}$ be independent of $x$
$\therefore 6-2r=0$
$r=3$
$\therefore T_{r+1}=T_{3+1}=6C_3 \large\frac{2^{6-3}}{3^3}$$\times x^{6-2(3)}$
$\Rightarrow \large\frac{20\times 8}{27}$
$\Rightarrow \large\frac{160}{27}$
Hence (C) is the correct answer.
answered Jun 25, 2014 by sreemathi.v
 

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