Combinatorics: Simplify the generating function $x^0+x^2+x^4+\cdots$ This is probably going to have a simple solution, and I'm going to kick myself when I see it, but I have a bit of mental block on this topic in general that I'm trying to clear before the next year of my degree starts.
In trying to solve a combinatoric problem, I've ended up with the following generating function:
$$
(x^{1}+x^{3}+x^{5}+x^{7}+\cdots)^{3} = x^3 (x^0+x^2+x^4+x^6+\cdots)^3
$$
For the life of me, I can't think of how to reduce the second part in parentheses. I'm trying to get it down to $(x^0+x^1+x^2+\cdots)^3$ so that I can use standard solutions that we have covered in the last year, but it's not coming to me.
The original problem is 'How many ways are there to put $12$ pennies into $3$ piles such that each pile has an odd number of pennies?' I'm trying to do it via generating functions for practice, but general advice re this problem would also be appreciated.
I appreciate any help you can give a struggling undergraduate!
 A: Just noticed that it's impossible, and so must be a misprint. The sum of three odd numbers must be odd ($(2p+1)+(2q+1)+(2r+1)=2(p+q+r+1)+1, \forall p,q,r \in Z$) so no three odd bins can add to 12. I see Travis's suggestion of substitution would work in general.
A: Let $a=1+x^2+x^4+x^6+x^8+\cdots$
If we multiply this by $x^2$,
$$ax^2=x^2(1+x^2+x^4+x^6+x^8+\cdots)=x^2+x^4+x^6+x^8+x^{10}+\cdots$$
Notice that this is $1$ less than our original value of $a$.
$$ax^2=a-1$$
Simplifying....
$$ax^2-a=-1$$
$$a(x^2-1)=-1$$
$$a=\frac{-1}{x^2-1}=\frac{1}{1-x^2}$$
This is often not the most common method, but it is still quite effective and simple to understand.
A: Hint $$1 + u + u^2 + u^3 + \cdots = \frac{1}{1 - u}.$$
A: The number of ways, n, to put 12 pennies into 3 bins so that each bin has an odd number of pennies is the coefficient of x^n in the expansion of
 ( x/(1-x) )^3.  Which is 0.
nn = 15; CoefficientList[Series[(x/(1 - x^2))^3, {x, 0, nn}], x]
{0, 0, 0, 1, 0, 3, 0, 6, 0, 10, 0, 15, 0, 21, 0, 28}.
