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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} $

1 Answer

  • 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
Putting $r=3$ we have
$\Rightarrow C(7,3)3^4(-1)^3\large\frac{x^9}{6^3}$
$\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}$$x^9$
Again $T_{r+1}=T_5$ or $r+1=5$
Putting $r=4$ in (1) we have
$\Rightarrow \large\frac{7!}{3!4!}\frac{3^3x^{12}}{3^42^4}$
$\Rightarrow \large\frac{35}{48}$$x^{12}$
Hence (C) is the correct answer.
answered Jun 23, 2014 by sreemathi.v