Question

Find the two middle terms of ${(3a-\frac{a^3}{6})}^9$.

Answer 1

There will be $10$ terms.
So, we have two middle terms $T_5$ and $T_6$

General term $T_{r+1}$ in the expansion of $(x-y{)}^n$ is given by

$T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-r}{y}^r$

$\therefore T_5={}^9{C}_4{\left(3a\right)}^5{(-\frac{a^3}{6})}^4$
$=\frac{189}{8}{a}^{17}$
$T_6={}^9{C}_5{\left(3a\right)}^4{(-\frac{a^3}{6})}^5$
$=-\frac{21}{16}{a}^{19}$