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# binomial theorem

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:

$\displaystyle(a+b)^{3}$ | $\displaystyle=a^{3}+3a^{2}b+3ab^{2}+b^{3}$ | |||

$\displaystyle(a+b)^{4}$ | $\displaystyle=a^{4}+4a^{3}b+6a^{2}b^{2}+4ab^{3}+b^{4}.$ |

This result actually holds more generally if $a$ and $b$ belong to a commutative rig.

Keywords:

number theory combinatorics

Related:

BinomialFormula, BinomialCoefficient, BernoulliDistribution2, UsingThePrimitiveElementOfBiquadraticField

Type of Math Object:

Theorem

Major Section:

Reference

## Mathematics Subject Classification

11B65*no label found*06F25

*no label found*03E20

*no label found*

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## Attached Articles

## Corrections

0^0 convention by matte ✓

0^0 by matte ✓

This entry is broken in page images mode. by cgibbard ✓

generality by rspuzio ✓

spelling by mps ✘

0^0 by matte ✓

This entry is broken in page images mode. by cgibbard ✓

generality by rspuzio ✓

spelling by mps ✘

## Comments

## generality

It might be worth pointing out that this theorem also holds if a and b belong to an commutative rig (the spelling is right; I really mean "rig", not "ring" here) since we only use some basic algebraic properties of real or complex numbers in the proof.

## Re: generality

I'm sorry. I meant to post this as a correction, but I hit the post button instead and didn't notice what happened until it was too late. I apologize that this correction is going to hang around in the wrong place underneath the article and possibly confuse people in the future.