Browse Questions

# Find the middle term in the expansion of $(3x-\large\frac{x^3}{6})^9$

$\begin{array}{1 1}(A)\;\large\frac{189}{8}\normalsize x^{17};-\large\frac{21}{16}\normalsize x^{19}\\(B)\;\large\frac{189}{8}\normalsize x^{17};-\large\frac{22}{16}\normalsize x^{15}\\(C)\;\large\frac{8}{189}\normalsize x^{17};-\large\frac{21}{16}\normalsize x^{19}\\(D)\;\large\frac{8}{189}\normalsize x^{17};-\large\frac{25}{4}\normalsize x^{13}\end{array}$

Toolbox:
• If $n+1$ is even term
• First middle term =$\large\frac{n+1}{2}^{th}$ term
• Second middle term =$(\large\frac{n+1}{2}$$+1)^{th} term Number of terms in the expansion is 9+1=10 There are two middle terms T_5 and T_6 Hence we are to find T_5 and T_6 (3x-\large\frac{x^6}{6})^9=$$[3x+(-\large\frac{x^6}{6}]^9$
$T_{r+1}=C(9,r) (3x)^{9-r}.(\large\frac{x^6}{6})^r$
$T_{r+1}=T_5$
$r+1=5$
$r=4$
Putting $r=4$ we have
$T_{4+1}=C(9,4)(3x)^{9-4}(\large\frac{x^3}{6})^4$
$\Rightarrow C(9,4)(3x)^{5}(\large\frac{x^{12}}{6^4})$
$\Rightarrow \large\frac{9!}{4!5!}$$(3^5x^{5})(\large\frac{x^{12}}{6^4}) \Rightarrow \large\frac{9!}{4!5!}\frac{3^5x^{17}}{6^4} \Rightarrow \large\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times5!}\frac{3^5x^{17}}{6^4} \Rightarrow \large\frac{126}{1}\frac{3^5x^{17}}{6^4} \Rightarrow \large\frac{189}{8}$$x^{17}$
Second term :-
$T_6=C(9,5).(3x)^{9-5}(\large\frac{-x^3}{6})^5$
$\Rightarrow C(9,5).(3x)^{4}\large\frac{-x^{15}}{6^5}$
$\Rightarrow \large\frac{9!}{5!4!}$$3^4.x^4.\large\frac{-x^{15}}{6^5} \Rightarrow \large\frac{9\times 8\times 7\times 6\times 5!}{5!4\times 3\times 2}$$3^4.\large\frac{-x^{19}}{6^5}$
$\Rightarrow \large\frac{-126.3^4}{6^5}$$x^{19} \Rightarrow \large\frac{-21}{16}$$x^{19}$
Hence (A) is the correct answer.