<s>[INST] How to find the coefficient of a power in a power series?? How can I find the coefficient of $x^{80}$ in the power series $$(1+x+x^{2}+x^{3}+x^{4}+\cdots)(x^{2}+x^{4}+x^{6}+x^{8}+\cdots)(1+x^{3}+x^{5})\,?$$
Is there a general method to this? 
 [/INST] We can use the fact that
\begin{align}
1+x+x^2+\dotsb&=\frac{1}{1-x}\\
x^2+x^4+x^6+\dotsb&=\frac{x^2}{1-x^2}
\end{align}
so the expression is
$$
\frac{x^2(1+x^3+x^5)}{(1-x)^2(1+x)}
$$
and we can try doing partial fraction decomposition:
$$
\frac{x^2(1+x^3+x^5)}{(1-x)^2(1+x)}=
x^4+x^3+3x^2+3x+5+\frac{6x^2+2x-5}{(1-x)^2(1+x)}
$$
so we want to decompose the fraction into
$$
\frac{A}{(1-x)^2}+\frac{B}{1-x}+\frac{C}{1+x}
$$
which gives $A=3/2$, $B=-25/4$, $C=-1/4$. Therefore your product can be written as
$$
x^4+x^3+3x^2+3x+5
+\frac{3}{2}\sum_{k\ge0}(k+1)x^k
-\frac{25}{4}\sum_{k\ge0}x^k
-\frac{1}{4}\sum_{k\ge0}(-1)^kx^k
$$
and the coefficient of $x^{80}$ can be read directly as
$$
\frac{3}{2}\cdot 81-\frac{25}{4}-\frac{1}{4}=115
$$
Note that this method gives all coefficients.
I've used the formula
$$
\frac{1}{(1-x)^2}=\sum_{k\ge0}(k+1)x^k
$$
that can be deduced by differentiating
$$
\frac{1}{1-x}=\sum_{k\ge0}x^k
$$
</s>