Question

The coefficient of $t^{12}$ in the expansion of ${\left(\frac{1-t^9}{1-t^3}\right)}^6$ is

Answer 1

${\left(\frac{1-t^9}{1-t^3}\right)}^6$
$={\left(\frac{(1-t^3)(1+t^3+t^6)}{1-t^3}\right)}^6$
$={[1+t^3(1+t^3\left)\right]}^6$
$={}^6{C}_0+{}^6{C}_1{t}^3(1+t^3)+{}^6{C}_2{t}^6(1+t^3{)}^2+{}^6{C}_3{t}^9(1+t^3{)}^3+{}^6{C}_4{t}^{12}(1+t^3{)}^4+\cdots$

So, for the coefficient of $t^{12}$, consider the terms
${}^6{C}_2{t}^6(1+t^3{)}^2+{}^6{C}_3{t}^9(1+t^3{)}^3+{}^6{C}_4{t}^{12}(1+t^3{)}^4$
$=15t^6(1+t^6+2t^3)+20t^9(1+t^9+3t^3+3t^6)+15t^{12}(1+t^3{)}^4$
Checking only for $t^{12}$ terms and leaving other terms
$15t^6\cdot t^6+20t^9\cdot 3t^3+15t^3\cdot 1=90t^{12}$
Hence, the required coefficient is $90$.

$\underset{–}{\text{Alternate:}}$
${\left(\frac{1-t^9}{1-t^3}\right)}^6$
$=(1-t^9{)}^6\cdot (1-t^3{)}^{-6}$
$=(1-6t^9+\cdots )\cdot (1-t^3{)}^{-6}$

To obtain $t^{12}$, there are only two possibilities,
either $1\cdot (t^3{)}^4$ or $t^9\cdot t^3$
Coefficient of $t^9$ in the expansion of $(1-t^9{)}^6$ is $-6$
Coefficient of $t^3$ and $t^{12}$ in expansion of $(1-t^3{)}^{-6}$ are $6$ and $\frac{6\cdot (6+1)\cdot (6+2)\cdot (6+3)}{4!}$ respectively.
i.e., $6$ and $126$

$\therefore$ Coefficient of $t^{12}$ in the expansion of ${\left(\frac{1-t^9}{1-t^3}\right)}^6$ is
$1\cdot 126-6\cdot 6=90$