<?xml version="1.0" encoding="UTF-8"?>

<record version="12" id="247">
 <title>binomial theorem</title>
 <name>BinomialTheorem</name>
 <created>2001-10-16 08:50:18</created>
 <modified>2005-02-22 07:54:13</modified>
 <type>Theorem</type>
 <creator id="5" name="KimJ"/>
 <author id="5" name="KimJ"/>
 <classification>
	<category scheme="msc" code="11B65"/>
 </classification>
 <related>
	<object name="BinomialFormula"/>
	<object name="BinomialCoefficient"/>
	<object name="BernoulliDistribution2"/>
	<object name="UsingThePrimitiveElementOfBiquadraticField"/>
 </related>
 <keywords>
	<term>number theory combinatorics</term>
 </keywords>
 <preamble>\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{xypic}</preamble>
 <content>The binomial theorem is a formula for the expansion of $(a+b)^n$, for $n$ a positive integer and $a$ and $b$ any two real (or complex) numbers, into a sum of powers of $a$ and $b$. More precisely,
$$(a+b)^n  = a^n + \binom{n}{1} a^{n-1}b + \binom{n}{2} a^{n-2}b^2 + \cdots + b^n .
$$
For example, if $n$ is 3 or 4, we have:
\begin{eqnarray*}
(a+b)^3 &amp;= a^3 + 3 a^2 b + 3 a b^2 + b^3 \\
(a+b)^4 &amp;= a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4 .
\end{eqnarray*}

This result actually holds more generally if $a$ and $b$ belong to a commutative \PMlinkname{rig}{Rig}.</content>
</record>
