Determining coefficients of an infinite series Here is my question:
Let $s=\sum_{i\ge 0} a_ix^i = (1-4x^2)^{-20}$
$t=\sum_{j\ge 0} b_jx^j = (1+x^5)^{-17}$
Determine $a_i$ and $b_j$ for all $i,j \ge 0$. (The answer will be divided into cases. Ex $a_i$ will depend on whether $i\equiv0\pmod 2$)
Then, determine the coefficient of $x^{16}$ in $3x^4st$

I believe I have the correct first steps but I'm not sure how to proceed further:
$(1-4x^2)^{-20} = \sum_{k\ge0} (_{k}^{19+k})(4x^2)^k = \sum_{k\ge0} (_{k}^{19+k})4^kx^{2k}$
$(1+x^5)^{-17} =(1-(-x^5))^{-17}= \sum_{l\ge0} (_{l}^{16+l})(-x^5)^l= \sum_{l\ge0} (_{l}^{16+l})(-1)^l(x^{5l})$
For the second part, it should be the same as calculating the coefficient on $x^{12}$ in $3st$ which I believe is straight forward once I know what all the coefficients in $s$ and $t$.
 A: The only way for degree 12 terms to appear in $st$ is if you take a degree $n$ term from $s$ and a degree $m$ term from $t$, such that $n+m = 12$. $s$ has terms of degree $0, 2, 4, 6, 8, 10, 12, ...$ and $t$ has terms of degree $0, 5, 10, 15, ...$. So we have only these ways of making $12$: $2+10, 12 + 0$.
Calculate the 0-th and 10-th coefficient of $t$: $\left(1+x^5\right)^{-17} = 1+...+153 x^{10}+...$
Calculate the 2-th and 12-th coefficient of $s$: $\left(1-x^2\right)^{-20} = ...+80 x^2+...+725401600 x^{12}...$
So $1*725401600 + 153*80 = 725413840$ is the $12$-th degree coefficient of $st$. So $3st$ has $3*725413840= 2176241520$ as its $12$-th degree coefficient which is the answer.
A: It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.

In order to determine the coefficient of $3x^{16}$ in $s(x)t(x)$ we obtain
  \begin{align*}
\color{blue}{[x^{16}]3x^4 s(x)t(x)}
&=3[x^{12}]\left(\sum_{k\geq 0}\binom{19+k}{k}4^kx^{2k}\right)
\left(\sum_{l\geq 0}\binom{16+l}{l}(-1)^lx^{5l}\right)\tag{1}\\
&=3\sum_{k=0}^6\binom{19+k}{k}4^k[x^{12-2k}]\sum_{l\geq 0}\binom{16+l}{l}(-1)^lx^{5l}\tag{2}\\
&=3\left(\binom{20}{1}4[x^{10}]+\binom{25}{6}4^k[x^0]\right)\sum_{l\geq 0}\binom{16+l}{l}(-1)^lx^{5l}\tag{3}\\
&=12\binom{20}{1}\binom{18}{2}+3\cdot 4^6\binom{25}{6}\binom{16}{0}\tag{4}\\
&\color{blue}{=2176241520}
\end{align*}

Comment:


*

*In (1) we use the linearity of the coefficient of operator and apply the rule
$[x^{p-q}]A(x)=[x^p]x^qA(x)$.

*In (2) we apply the same rule as in (1) to the left series. We restrict the upper limit of the left-hand series with $6$ since the exponent of $x^{12-2k}$ is non-negative.

*In (3) we skip terms $[x^p]$ which are not multiples of $5$ in the exponent.

*In (4) we select the coefficients of $x^{10}$ and $x^0$ accordingly.
