# Difference between revisions of "Binomial Theorem"

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<center><math>(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k</math></center> | <center><math>(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k</math></center> | ||

− | where <math>\binom{n}{k} = \frac{n!}{k!(n-k)!}</math> is a [[binomial coefficient]]. | + | where <math>\binom{n}{k} = \frac{n!}{k!(n-k)!}</math> is a [[binomial coefficient]]. In other words, the coefficients when <math>(a + b)^n</math> is expanded and like terms are collected are the same as the entries in the <math>n</math>th row of [[Pascal's Triangle]]. |

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+ | For example, <math>(a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5</math>, with coefficients <math>1 = \binom{5}{0}</math>, <math>5 = \binom{5}{1}</math>, <math>10 = \binom{5}{2}</math>, etc. | ||

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+ | ==Proofs== | ||

+ | There are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of [[mathematical induction]]. The Binomial Theorem also has a nice [[combinatorial proof]]: we can write <math>(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}</math>. Repeatedly using the [[distributive property]], we see that for a term <math>a^m b^{n-m}</math>, we must choose <math>m</math> of the <math>n</math> terms to contribute an <math>a</math> to the term, and then each of the other <math>n-m</math> terms of the product must contribute a <math>b</math>. Thus, the coefficient of <math>a^m b^{n-m}</math> is the number of ways to choose <math>m</math> objects from a set of size <math>n</math>, or <math>\binom{n}{m}</math>. Extending this to all possible values of <math>m</math> from <math>0</math> to <math>n</math>, we see that <math>(a+b)^n = \sum_{m=0}^{n}{\binom{n}{m}}\cdot a^m\cdot b^{n-m}</math>, as claimed. | ||

==Generalizations== | ==Generalizations== | ||

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==Usage== | ==Usage== | ||

Many [[factoring | factorizations]] involve complicated [[polynomial]]s with [[binomial coefficient]]s. For example, if a contest problem involved the polynomial <math>x^5+4x^4+6x^3+4x^2+x</math>, one could factor it as such: <math> x(x^4+4x^3+6x^2+4x+1)=x(x+1)^{4}</math>. It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients. | Many [[factoring | factorizations]] involve complicated [[polynomial]]s with [[binomial coefficient]]s. For example, if a contest problem involved the polynomial <math>x^5+4x^4+6x^3+4x^2+x</math>, one could factor it as such: <math> x(x^4+4x^3+6x^2+4x+1)=x(x+1)^{4}</math>. It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients. | ||

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==See also== | ==See also== |

## Revision as of 21:08, 17 December 2009

The **Binomial Theorem** states that for real or complex , , and non-negative integer ,

where is a binomial coefficient. In other words, the coefficients when is expanded and like terms are collected are the same as the entries in the th row of Pascal's Triangle.

For example, , with coefficients , , , etc.

## Contents

## Proofs

There are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of mathematical induction. The Binomial Theorem also has a nice combinatorial proof: we can write . Repeatedly using the distributive property, we see that for a term , we must choose of the terms to contribute an to the term, and then each of the other terms of the product must contribute a . Thus, the coefficient of is the number of ways to choose objects from a set of size , or . Extending this to all possible values of from to , we see that , as claimed.

## Generalizations

The Binomial Theorem was generalized by Isaac Newton, who used an infinite series to allow for complex exponents: For any real or complex , , and ,

### Proof

Consider the function for constants . It is easy to see that . Then, we have . So, the Taylor series for centered at is

## Usage

Many factorizations involve complicated polynomials with binomial coefficients. For example, if a contest problem involved the polynomial , one could factor it as such: . It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients.