Browse Questions

# Find the middle term in the expansion of $(3-\large\frac{x^3}{6})^7$

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

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
$T_{4+1}=T_5=C(7,4)3^{7-4}(-1)^4-\large\frac{x^{12}}{6^4}$
$\Rightarrow \large\frac{7!}{3!4!}\frac{3^3x^{12}}{3^42^4}$