Say I have a Ring with set $G$ and binary operations $+$ and $\times$. If $G$ has order 2 under addition (meaning $A+A=0,\forall A\in G$, where $0$ is the additive identity), how can I reproduce the binomial theorem for this ring? What I mean is, if for regular numbers we have: $$(x+y)^n = \sum_{i=0}^n={n \choose i}x^{n-i}y^i$$ Can we have a similar expression for $(A+B)^n,\forall A,B\in G$ that can cancel out the $A+A$ terms?
Much appreciated.
${n \choose i}$ is going to be 1 in your ring if it is odd, and otherwise 0, so you just need to know when ${n \choose i}$ is even. For that, see http://en.wikipedia.org/wiki/Lucas%27_theorem.