Question

Find the middle term(s) in the expansion of:

${(3a-\frac{a^3}{6})}^9$

Answer 1

${(3a-\frac{a^3}{6})}^9$

Here, $n$ is odd therefore, middle terms are $\left(\frac{9+1}{2}\right)th$ and $(\frac{9+1}{2}+1)th$.
It means $T_5$ and $T_6$ are middle terms
$T_5={}^9{C}_4(3a{)}^{9-4}{(-\frac{a^3}{6})}^4=\frac{189}{8}{a}^{17}$
$T_6={}^9{C}_5(3a{)}^{9-5}{(-\frac{a^3}{6})}^5=-\frac{21}{16}{a}^{19}$.