# Binomial Theorem/Integral Index

## Theorem

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_{\ge 0}: \ \$ $\displaystyle \paren {x + y}^n$ $=$ $\displaystyle \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$ $\displaystyle$ $=$ $\displaystyle x^n + \binom n 1 x^{n - 1} y + \binom n 2 x^{n - 2} y^2 + \binom n 3 x^{n - 3} y^3 + \cdots$ $\displaystyle$ $=$ $\displaystyle x^n + n x^{n - 1} y + \frac {n \paren {n - 1} } {2!} x^{n - 2} y^2 + \frac {n \paren {n - 1} \paren {n - 3} } {3!} x^{n - 3} y^3 + \cdots$

where $\dbinom n k$ is $n$ choose $k$.

## Proof

### Basis for the Induction

For $n = 0$ we have:

$\displaystyle \paren {x + y}^0 = 1 = \binom 0 0 x^{0 - 0} y^0 = \sum_{k \mathop = 0}^0 \binom 0 k x^{0 - k} y^k$

This is the basis for the induction.

### Induction Hypothesis

This is our induction hypothesis:

$\displaystyle \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$

### Induction Step

This is our induction step:

 $\displaystyle \paren {x + y}^{n + 1}$ $=$ $\displaystyle \paren {x + y} \paren {x + y}^n$ $\displaystyle$ $=$ $\displaystyle x \sum_{k \mathop = 0}^n \binom n k x^{n - k}y^k + y \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$ Inductive Hypothesis $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^n \binom n k x^{n + 1 - k} y^k + \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^{k + 1}$ $\displaystyle$ $=$ $\displaystyle \binom n 0 x^{n + 1} + \sum_{k \mathop = 1}^n \binom n k x^{n + 1 - k} y^k + \binom n n y^{n + 1} + \sum_{k \mathop = 0}^{n - 1} \binom n k x^{n - k} y^{k + 1}$ $\displaystyle$ $=$ $\displaystyle x^{n + 1} + y^{n + 1} + \sum_{k \mathop = 1}^n \binom n k x^{n + 1 - k} y^k + \sum_{k \mathop = 0}^{n - 1} \binom n k x^{n - k} y^{k + 1}$ $\displaystyle$ $=$ $\displaystyle \binom {n + 1} 0 x^{n + 1} + \binom {n + 1} {n + 1} y^{n + 1} + \sum_{k \mathop = 1}^n \binom n k x^{n + 1 - k} y^k + \sum_{k \mathop = 1}^n \binom n {k - 1} x^{n + 1 - k} y^k$ $\displaystyle$ $=$ $\displaystyle \binom {n + 1} 0 x^{n + 1} + \binom {n + 1} {n + 1} y^{n + 1} + \sum_{k \mathop = 1}^n \paren {\binom n k + \binom n {k - 1} } x^{n + 1 - k} y^k$ $\displaystyle$ $=$ $\displaystyle \binom {n + 1} 0 x^{n + 1} + \binom {n + 1} {n + 1} y^{n + 1} + \sum_{k \mathop = 1}^n \binom {n + 1} k x^{n + 1 - k} y^k$ Pascal's Rule $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^{n + 1} \binom {n + 1} k x^{n + 1 - k} y^k$

The result follows by the Principle of Mathematical Induction.

$\blacksquare$