Generating function, determining coefficient Here is a question I encountered the other day:
Determine the coefficient of $x^{98}$ in the following generating function:
$$f(x)=\frac{x}{(1-2x)^{21}}$$
I'm thrown off a bit by the large exponent in the denominator and the fact that we see a $1-2x$ instead of $1-x$. Let me just start by asking, does the above function equal
$$x \sum_{n \geq0}{n \choose 20}(2x)^{n-20}$$
?
If so, I think I can work out the rest. If not, I'm rather lost. Thanks.
 A: There is only a small mistake in your expression. It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.

We obtain
  \begin{align*}
[x^{98}]\frac{x}{(1-2x)^{21}}&=[x^{98}]x\sum_{n\geq0}\binom{-21}{n}(-2x)^{n}\tag{1}\\
&=[x^{97}]\sum_{n\geq 0}\binom{n+20}{20}(2x)^n\tag{2}\\
&=\binom{117}{20}2^{97}\tag{3}
\end{align*}

Comment:


*

*In (1) we use the binomial series expansion

*In (2) we use the rule $[x^{p-q}]=[x^p]x^q$ and the identity
\begin{align*}
\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q
\end{align*}

*In (3) we select the coefficient with $n=97$
A: Maybe you should write the following way :
$$\frac{x}{(1-2x)^{21}}=x(\sum_{n\geq 0}2^nx^n)^{21})$$
$$\frac{x}{(1-2x)^{21}}=x\sum_{n_1\geq0}2^{n_1}x^{n_1}\times\cdots \sum_{n_{21}\geq0}2^{n_{21}}x^{n_{21}}$$
$$\frac{x}{(1-2x)^{21}}=x\sum_{n_1,\dots,n_{21}\geq0}2^{n_1+\cdots+n_{21}}x^{n_1+\cdots+n_{21}} $$
$$\frac{x}{(1-2x)^{21}}=x\sum_{n\geq 0}\sum_{n_1,\dots,n_{21}\geq0\text{ and } n_1+\cdots+n_{21}=n}2^{n}x^{n} $$
Now you will be able to find the associated generating function whence the coefficient.
A: It can  be written as $(1-2x)^{-21}$ so in general for non nonegative integer the coefficient of $x^r$ is $$\frac{(-1)^r n(n-1)...(n-(r-1))x^r(2)^r}{r!}$$ so can you do it now with some manipulations. 
