# A generating function where the coefficient of $x^r$ is $2^r$

I want to find a generating function for $\displaystyle\sum_{r=0}^\infty 2^rx^r = 1 + 2^2x^2 + 2^3x^3 + ...$

I don't think this is a very difficult question but I'm having an issue with it. All I've done in class are simple questions where the coefficient is either a power of $r$ or $(r+1)(r+2)...(r+n)$, a product of consecutive integers.

As far as I can tell, I need to use the exponential generating function where $e^{2x}=\displaystyle\sum_{r=0}^\infty 2^r\frac{x^r}{r!}$. But I have no idea how to deal with the $r!$ part. Is this the kind of approach I want to be taking and if it is, where can I go from here?

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It’s not clear just what your question is. $e^{2x}$ is the exponential generating function for the sequence $\langle 2^n:n\in\Bbb N\rangle$; what do you want to do with it? If you want the ordinary generating function, as @anon says, it’s $\frac1{1-2x}$, since $\sum_{n\ge 0}2^nx^n$ is just a geometric series with ratio $2x$. – Brian M. Scott Mar 7 '13 at 2:12
The title seems to be about $\sum 2^rx^r=\frac{1}{1-2x}$, but there is no actual question stated in the body of your post. – anon Mar 7 '13 at 2:13
I added what I'm trying to find at the start of the post, sorry about that – MangoPirate Mar 7 '13 at 2:17

If you want the ordinary generating function, you want $\sum_0^\infty 2^n t^n=\sum_0^\infty (2t)^n$. This is the expansion of $\dfrac{1}{1-2t}$. For $1+(2t)+(2t)^2+(2t)^3+\cdots$ is the infinite geometric series with first term $1$ and common ratio $2t$. Note that we have convergence only for $|t|\lt 1/2$.
Remark: For the exponential generating function, we want $\sum_0^\infty \frac{(2t)^n}{n!}$. We recognize this as the Maclaurin series for $e^{2t}$.