Binomial Theorem

Integral Index
Let $X$ be one of the set of numbers $\N, \Z, \Q, \R, \C$.

Let $x, y \in X$.

Then:
 * $\displaystyle \forall n \in \Z_+: \left({x+y}\right)^n = \sum_{k=0}^n {n\choose k}x^{n-k}y^k$

Ring Theory
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:


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

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

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


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

Base Case
For $n = 0$ we have:


 * $\displaystyle \left({x+y}\right)^0 = 1 = {0\choose 0}x^{0-0}y^0 = \sum_{k=0}^0 {0\choose k}x^{0-k}y^k$

Therefore the base case holds.

Inductive Hypothesis
This is our inductive hypothesis:


 * $\displaystyle \left({x+y}\right)^j = \sum_{k=0}^j {j\choose k}x^{j-k}y^k$ for all $j \ge 1$

Inductive Step
This is our inductive step:

And so we are done by the Principle of Mathematical Induction.

Proof for Ring Theory
The proof for the Ring Theory version follows the same strategy.

Also see

 * Pascal's Triangle
 * Pascal's Rule