Finding coefficient of $x^n$ in this series I was doing an assignment question, I came across this:

I understand everything (I know the theorem used in the answer), but I don't get how solution switched from first "i.e." to second "i.e.". I mean how did it happen? And what is happening after that Geometric Progression formula.
Thank you!

They are using the fact that $(1-x^6)^4=1-4x^6+{4\choose2}x^{12}-\cdots$ and noting that since you're only looking for the coefficient of $x^8$, you can drop all those higher-order terms, including the ${4\choose2}x^{12}$.
As for the rest, it's a matter of observing that
$${1\over(1-x)^4}={1\over3!}\left(1\over1-x\right)'''={1\over3!}(1+x+x^2+x^3+\cdots)'''={1\over3!}\left({3!\over0!}+{4!\over1!}x+{5!\over2!}x^2+\cdots \right)\\={3\choose3}+{4\choose3}x+{5\choose3}x^2+\cdots+{11\choose3}x^8+\cdots$$
When you multiply by $1-4x^6$ and pick out the resulting coefficient for $x^8$. you get the $11\choose3$ from the multiplication by the $1$ and $-4{5\choose3}$ from the multiplication by the $-4x^6$.
