Question

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

Answer 1

It is known that in the expansion of $(a+b{)}^n$ if n is odd, then there are two middle terms namely ${\left(\frac{n+1}{2}\right)}^{th}$ term and ${(\frac{n+1}{2}+1)}^{th}$ term.
Therefore, the middle terms in the expansion of ${(3-\frac{x^3}{6})}^7$ are
$\left(\frac{7+1}{2}\right)=4^{th}$ term and $(\frac{7+1}{2}+1)=5^{th}$ term.
$T_4=T_{3+1}{=}^7{C}_3(3{)}^{7-3}{(-\frac{x^3}{6})}^3=(-1{)}^3\frac{7!}{3!4!}{.3}^4.\frac{x^9}{6^3}$
$=-\frac{\mathrm{7.6.5.4}!}{\mathrm{3.2.4}!}{.3}^4.\frac{1}{2^3{.3}^3}.x^9=-\frac{105}{8}{x}^9$
$T_5=T_{4+1}{=}^7{C}_4(3{)}^{7-4}{(-\frac{x^3}{6})}^4=(-1{)}^4\frac{7!}{4!3!}.(3{)}^3.\frac{x^{12}}{6^4}$
$=\frac{\mathrm{7.6.5.4}!}{4!3.2}\frac{3^3}{2^4{.3}^4}.x^{12}=\frac{35}{48}{x}^{12}$
Thus, the middle terms in the expansion of ${(3-\frac{x^3}{6})}^7$ are
$-\frac{105}{8}{x}^9$ and $\frac{35}{48}{x}^{12}$