Find the middle terms in the expansions of

It is known that in the expansion of $(a+b)^{n}$, if $n$ is odd, then there are two middle terms, namely, $\left(\frac{\mathrm{n}+\mathrm{l}}{2}\right)^{\text {th }}$ term and $\left(\frac{\mathrm{n}+1}{2}+1\right)^{\text {th }}$ term.

Therefore, the middle terms in the expansion of $\left(3-\frac{x^{3}}{6}\right)^{7}$ are $\left(\frac{7+1}{2}\right)^{\text {th }}=4^{\text {th }}$ term and $\left(\frac{7+1}{2}+1\right)^{\text {th }}=5^{\text {th }}$ term

$\mathrm{T}_{4}=\mathrm{T}_{3+1}={ }^{7} \mathrm{C}_{3}(3)^{7-3}\left(-\frac{\mathrm{x}^{3}}{6}\right)^{3}=(-1)^{3} \frac{7 !}{3 ! 4 !} \cdot 3^{4} \cdot \frac{\mathrm{x}^{9}}{6^{3}}$

$=-\frac{7 \cdot 6 \cdot 5.4 !}{3 \cdot 2.4 !} \cdot 3^{4} \cdot \frac{1}{2^{3} \cdot 3^{3}} \cdot x^{9}=-\frac{105}{8} x^{9}$

$\mathrm{T}_{5}=\mathrm{T}_{4+1}={ }^{7} \mathrm{C}_{4}(3)^{7-4}\left(-\frac{\mathrm{x}^{3}}{6}\right)^{4}=(-1)^{+} \frac{7 !}{4 ! 3 !}(3)^{3} \cdot \frac{\mathrm{x}^{12}}{6^{4}}$

$=\frac{7 \cdot 6 \cdot 5.4 !}{4 ! \cdot 3 \cdot 2} \cdot \frac{3^{3}}{2^{4} \cdot 3^{4}} \cdot x^{12}=\frac{35}{48} x^{12}$

Thus, the middle terms in the expansion of $\left(3-\frac{x^{3}}{6}\right)^{7}$ are $-\frac{105}{8} x^{9}$ and $\frac{35}{48} x^{12}$.