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The $5^{th}$ term from the end in the expansion of $\bigg(\large\frac{x^3}{2}-\frac{2}{x}\bigg)^{12}$ is ?

$\begin{array}{1 1} =7920x^4 \\ =-7920x^4 \\ =\frac{7920}{x^4} \\ =\frac{-7920}{x^4} \end{array} $

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  • General term $T_{r+1}=^nC_r x^{n-r}.y^r$ in the expansion $(x+y)^n$
$r^{th} $ term from end = $(n-r+2)^{th} $ term from begining.
in the expansion of $(x+y)^n$
$\therefore \:5^{th}$ term from end = $(12-5+2)^{th}$ term from begining.
$=9^{th}$ term = $T_{8+1}$
$=^{12}C_8(\large\frac{x^3}{2})^4.(-\large\frac{2}{x})^8$
$=495\times 16.x^4$
$=7920x^4$
answered Sep 19, 2013 by rvidyagovindarajan_1
 

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