Binomial Theorem/Ring Theory

Theorem
Let $\left({R, +, \odot}\right)$ be a ringoid such that $\left({R, \odot}\right)$ is a commutative semigroup.

Let $n \in \Z: n \ge 2$. Then:


 * $\displaystyle \forall x, y \in R: \odot^n \left({x + y}\right) = \odot^n x + \sum_{k \mathop = 1}^{n-1} \binom n k \left({\odot^{n-k} x}\right) \odot \left({\odot^k y}\right) + \odot^n y$

where $\displaystyle \binom n k = \frac {n!} {k!\ \left({n-k}\right)!}$ (see Binomial Coefficient).

If $\left({R, \odot}\right)$ has an identity element $e$, then:


 * $\displaystyle \forall x, y \in R: \odot^n \left({x + y}\right) = \sum_{k \mathop = 0}^n \binom n k \left({\odot^{n-k} x}\right) \odot \left({\odot^k y}\right)$

Proof
First we establish the result for when $\left({R, \odot}\right)$ has an identity element $e$.

For $n = 0$ we have:


 * $\displaystyle \odot^0 \left({x + y}\right) = e = {0 \choose 0} \odot^{0-0} x \odot^0 y = \sum_{k \mathop = 0}^0 {0 \choose k} x^{0-k} y^k$

Base Case
For $n = 0$ we have:


 * $\displaystyle \odot^0 \left({x + y}\right) = e = {0 \choose 0} \odot^{0-0} x \odot^0 y = \sum_{k \mathop = 0}^0 {0 \choose k} x^{0-k} y^k$

For $n = 2$ we have:

Therefore the base case holds.

Induction Hypothesis
This is our inductive hypothesis:


 * $\displaystyle \forall n \ge 2: \odot^n \left({x+y}\right) = \odot^n x + \sum_{k \mathop = 1}^{n-1} {n \choose k} \left({\odot^{n-k} x}\right) \odot \left({\odot^k y}\right) + \odot^n y$

Induction Step
This is the induction step:

The result follows by the Principle of Mathematical Induction.