Generating Functions of Partitions?

Question

Show that $2(1-x)^{-3} [(1-x)^{-3} + (1+x)^{-3}]$ is the generating function for the number of ways to toss $r$ identical dice and obtain an even sum.

Workings:

I'm not too sure on this problem.

With the $-3$ I'm guessing that the dice are 6 sided.

I suppose that for the sum to be even there could be an even number of odd numbers on the die, an even number of even numbers shown on the die or an odd number of even numbers shown on the die.

But I'm not sure how to use this info into getting a generating function.

Any help will be appreciated.

Answer 1

Note: The formulation of the problem seems to be incorrect. Nevertheless here's a hint for a 6-sided die.

The generating function for tossing one die once is $x^1+x^2+\ldots+x^6$, the exponent of $x$ indicating the result of the throw. Let's denote with $A(x)$ the generating function for tossing $r$ dice.

We observe, applying the formula for finite geometric series:

\begin{align*} A(x)&=(x^1+x^2+\ldots+x^6)^r\\ &=x^r(1+x+\ldots+x^5)^r\\ &=x^r\left(\frac{1-x^6}{1-x}\right)^r \end{align*}

Since we are interested in even results only, we are looking for a generating function which contains only the even powers of $x$ in $A(x)$.

This job is done with

\begin{align*}\frac{1}{2}\left(A(x)+A(-x)\right)\tag{1}\end{align*}

Note: (1) is valid for all generating functions $A(x)=\sum_{n=0}^{\infty}a_nx^n$, since \begin{align*} \frac{1}{2}\left(A(x)+A(-x)\right)& =\frac{1}{2}\left(\sum_{n=0}^{\infty}a_nx^n+\sum_{n=0}^{\infty}a_n(-x)^n\right)\\ &=\sum_{n=0}^{\infty}a_n\frac{x^n+(-x)^n}{2}\\ &=\sum_{{n=0}\atop{n\equiv 0(2)}}^{\infty}a_nx^n\\ &=\sum_{n=0}^{\infty}a_{2n}x^{2n} \end{align*}