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? 
 A: We have
$(1+x+x^2+\cdots)(x^2+x^4+\cdots)(1+x^3+x^5)$
Then we want The coeficient of $x^{80}$, note that $x^nx^m=x^{n+m}$ then searching The ways of write $80$ using the sets $\{0,1,\cdots\},\{2,4,\cdots\},\{0,3,5\}$.
$$\begin{align}
80&=0+80+0\\
&=2+78+0\\
&=4+76+0\\
&\vdots\\
&=78+2+0\\
&=1+76+3\\
&=3+74+3\\
&=5+72+3\\
&\vdots\\
&=75+2+3\\
&=1+74+5\\
&=3+72+5\\
&\vdots\\
&=73+2+5
\end{align}$$
Then The coeficiente will be
$$\begin{align}
C_{80}&=\stackrel{2n}{(40-1+1)}+\stackrel{2n+1}{(37-0+1)}+\stackrel{2n+1}{(36-0+1)}\\
&=40+38+37\\
&=115
\end{align}$$
Then if nothing as wrong The answer must be $115$
A: 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
$$
\frac{1}{1-x}=\sum_{k\ge0}x^k
$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
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 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\bracks{x^{80}}\bracks{\pars{1 + x + x^{2} + x^{3} + x^{4} +\cdots}
\pars{x^{2} + x^{4} + x^{6} + x^{8} + \cdots}
     \pars{1 + x^{3} + x^{5}}}:\ {\large ?}}$

$$
{\rm Lets}\quad
\dsc{\fermi\pars{x}}\equiv
\dsc{\pars{1 + x + x^{2} + x^{3} + x^{4} +\cdots}
\pars{x^{2} + x^{4} + x^{6} + x^{8} + \cdots}}
$$

Then,
$$
\bracks{x^{80}}\bracks{\fermi\pars{x}\pars{1 + x^{3} + x^{5}}}
=\bracks{x^{80}}\fermi\pars{x} + \bracks{x^{77}}\fermi\pars{x} + \bracks{x^{75}}\fermi\pars{x}
$$

and
  \begin{align}
\fermi\pars{x}&=\sum_{i\ =\ 0}^{\infty}x^{i}\sum_{j\ =\ 0}^{\infty}x^{2j + 2}
=\sum_{i\ =\ 0}^{\infty}\sum_{j\ =\ 0}^{\infty}x^{i + 2j + 2}
=\sum_{i\ =\ 0}^{\infty}\sum_{j\ =\ 0}^{\infty}
\sum_{n\ =\ 2}^{\infty}x^{n}\delta_{n,i + 2j + 2}
\\[5mm]&=\sum_{n\ =\ 2}^{\infty}x^{n}\pars{%
\sum_{j\ =\ 0}^{\infty}\sum_{i\ =\ 0}^{\infty}\delta_{i,n - 2j - 2}}
=\sum_{n\ =\ 2}^{\infty}x^{n}
\pars{\left.\sum_{j\ =\ 0}^{\infty}1\right\vert_{n - 2j - 2\ \geq\ 0}}
\\[5mm]&=\sum_{n\ =\ 2}^{\infty}x^{n}
\pars{\left.\sum_{j\ =\ 0}^{\infty}1\right\vert_{j\ \leq\ n/2 - 1}}
=\sum_{n\ =\ 2}^{\infty}x^{n}\pars{\sum_{j\ =\ 0}^{\floor{n/2 - 1}}1}
=\sum_{n\ =\ 2}^{\infty}\pars{\floor{{n \over 2} - 1} + 1}x^{n}
\end{align}

$$
\bracks{x^{n}}\fermi\pars{x} = \floor{n  - 2 \over 2} + 1\,,\qquad n \geq 2 
$$

The $\color{#66f}{\large\mbox{final result}}$ is given by:
  $$
\pars{\floor{78 \over 2} + 1} + \pars{\floor{75 \over 2} + 1}
+\pars{\floor{73 \over 2} + 1}=40 + 38 + 37=\color{#66f}{\Large 115}
$$

