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:


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

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

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

Base Case
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.

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

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

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

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

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

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

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

Historical Note
The General Binomial Theorem was announced by Isaac Newton in 1676. However, he had no real proof. Euler made an incomplete attempt in 1774, but the full proof had to wait for Gauss to provide it in 1812.