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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Find the middle term in the expansion of $(3-\large\frac{x^3}{6})^7$

$\begin{array}{1 1}(A)\;\large\frac{35}{48}\\(B)\;\large\frac{48}{35}\\(C)\;\large\frac{35}{48}\normalsize x^{12}\\(D)\;\large\frac{35}{48}\normalsize x^{10}\end{array} $

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1 Answer

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Toolbox:
  • If $n+1$ is even term .
  • $1^{st}$ middle term=$(\large\frac{n+1}{2})^{th}$ term
  • $2^{nd}$ middle term=$(\large\frac{n+1}{2}\normalsize +1)^{th}$ term
Number of terms in the expansion is $7+1=8$
There are two middle terms which are $T_4$ and $T_5$
Hence we are to find $T_4$ and $T_5$ in the given expansion
$(3-\large\frac{x^3}{6})^7=$$[3+(-\large\frac{x^3}{6})]^7$
$T_{r+1}=C(7,r)3^{7-r}(-\large\frac{x^3}{6})$------(1)
$T_{r+1}=T_4$
$r+1=4$
$r=3$
Putting $r=3$ we have
$T_{3+1}=C(7,3)3^{7-3}(-\large\frac{x^3}{6})^3$
$\Rightarrow C(7,3)3^4(-1)^3\large\frac{x^9}{6^3}$
$\Rightarrow \large\frac{-7}{3!4!}\frac{3}{2^3}$$x^9$
$\Rightarrow -\large\frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}.\frac{3}{2^3}$$x^9$
$\Rightarrow -\large\frac{105}{8}$$x^9$
Again $T_{r+1}=T_5$ or $r+1=5$
$r=4$
Putting $r=4$ in (1) we have
$T_{4+1}=T_5=C(7,4)3^{7-4}(-1)^4-\large\frac{x^{12}}{6^4}$
$\Rightarrow \large\frac{7!}{3!4!}\frac{3^3x^{12}}{3^42^4}$
$\Rightarrow \large\frac{35}{48}$$x^{12}$
Hence (C) is the correct answer.
answered Jun 23, 2014 by sreemathi.v
 
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