Binomial Theorem

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

Let $$x, y \in X$$.

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

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:

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

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

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

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

General Binomial Theorem
Let $$\alpha \in \R$$ be a real number.

Let $$x \in \R$$ be a real number such that $$\left|{x}\right| < 1$$.

Then $$\left({1 + x}\right)^\alpha = \sum_{n=0}^\infty \frac {\prod_{k+0}^{n-1}\left({\alpha - k}\right)} {n!} x^n$$.

That is: $$\left({1 + x}\right)^\alpha = 1 + \alpha x + \frac {\alpha \left({\alpha - 1}\right)} {2!} x^2 + \frac {\alpha \left({\alpha - 1}\right) \left({\alpha - 2}\right)} {3!} x^3 + \cdots$$.

Base Case
$$n = 0$$

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

Therefore the base case holds.

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

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.

Proof of General Binomial Theorem
Let $$R$$ be the radius of convergence of the power series $$f \left({x}\right) = \sum_{n=0}^\infty \frac {\prod_{k+0}^{n-1}\left({\alpha - k}\right)} {n!} x^n$$.

Then by Radius of Convergence from Limit of Sequence, we have:

$$\frac 1 R = \lim_{n \to \infty} \frac {\left|{\alpha \left({\alpha - 1}\right) \cdots \left({\alpha - n}\right)}\right|} {\left({n+1}\right)!} \frac {n!} {\left|{\alpha \left({\alpha - 1}\right) \cdots \left({\alpha - n + 1}\right)}\right|}$$

$$ $$ $$

Thus for $$\left|{x}\right| < 1$$ we can use Power Series Differentiable on Interval of Convergence and get:

$$D_x f \left({x}\right) = \sum_{n=1}^\infty \frac {\prod_{k+0}^{n-1}\left({\alpha - k}\right)} {n!} n x^{n-1}$$

This leads us to:

$$ $$ $$ $$ $$

Gathering up what we've got, that is: $$\left({1 + x}\right) D_x f \left({x}\right) = \alpha f \left({x}\right)$$.

Thus we get $$D_x \left({\frac {f \left({x}\right)} {\left({1 + x}\right)^\alpha}}\right) = \alpha \left({1 + x}\right)^{-\alpha - 1} f \left({x}\right) + \left({1 + x}\right)^{-\alpha} D_x f \left({x}\right) = 0$$

So $$f \left({x}\right) = c \left({1 + x}\right)^\alpha$$ when $$\left|{x}\right| < 1$$ for some constant $$c$$.

But $$f \left({0}\right) = 1$$ and hence $$c = 1$$.

Also see

 * Pascal's Triangle
 * Pascal's Rule