Generating function, determining coefficient

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $n=97$