Find the coefficient of $x^{80}$ in the power series expansion $\dfrac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}$ Find the coefficient of $x^{80}$ in the power series expansion $$\dfrac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}.$$

I don't know how to find coefficients in power series, solutions are greatly appreciated!
 A: Write it as $$\frac{x^2(1+x^2+x^5)}{(1-x)(1-x^2)}=x^2(1+x^2+x^5)\left(\sum_{h=0}^\infty x^h\right)\left(\sum_{s=0}^\infty x^{2s}\right)=\\=(x^2+x^4+x^7)\sum_{k=0}^\infty x^k\#\left\{(h,s)\,:\,0\le h,s\wedge h+2s=k\right\}=$$
if you call $a_k:=\#\left\{(h,s)\,:\,0\le h,s\wedge h+2s=k\right\}$, it continues as
$$=\sum_{k=0}^\infty (a_{k-7}+a_{k-4}+a_{k-2})x^k$$
So, you just have to compute $a_{73}+a_{76}+a_{78}$. You can easily see that a pair $(h,s)$ is uniquely determined by the choice of $s$ such that $0\le 2s\le k$. Hence $a_k=\begin{cases}0&\text{if }k<0\\ 1+\left\lfloor\frac k2\right\rfloor&\text{if }k\ge0\end{cases}$.
A: HINT
You can decompose it into partial fractions and get $$x^4+x^3+2x^2+3x+4-\frac{23}{4}(1-x)^{-1}+\frac 14(1+x)^{-1}+\frac 32(1-x)^{-2}$$
Then apply the generalised binomial theorem
A: $$\frac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}=x^2(1+x^2+x^5)\frac{1}{4}\left(\frac{2}{1-x^2}+\frac{1}{(1-x)^2}\right)$$
then use the following
$${\frac {1}{1-x^2}}=\sum _{n=0}^{\infty }x^{2n}$$
$${\frac {1}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n-1}$$
