Difference between revisions of "Binomial Theorem"

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==The Theorem==
 
First invented by [[Newton]], the Binomial Theorem states that for real or complex ''a'',''b'',<br><math>(a+b)^n = \sum_{k=0}^{n}{n \choose k}\cdot a^k\cdot b^{n-k}</math>  
 
First invented by [[Newton]], the Binomial Theorem states that for real or complex ''a'',''b'',<br><math>(a+b)^n = \sum_{k=0}^{n}{n \choose k}\cdot a^k\cdot b^{n-k}</math>  
  
 
This may be shown for the integers easily:<br>
 
This may be shown for the integers easily:<br>
<center><math>\displaystyle (a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}</math></center>
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<math>\displaystyle (a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}</math>
 
<br>Repeatedly using the distributive property, we see that for a term <math>\displaystyle 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>\displaystyle a^m b^{n-m}</math> is <math>\displaystyle n \choose 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_{k=0}^{n}{n \choose k}\cdot a^k\cdot b^{n-k}</math>.
 
<br>Repeatedly using the distributive property, we see that for a term <math>\displaystyle 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>\displaystyle a^m b^{n-m}</math> is <math>\displaystyle n \choose 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_{k=0}^{n}{n \choose k}\cdot a^k\cdot b^{n-k}</math>.
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==Usage==
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Many factorizations involve complicated polynomials with binomial coefficients. For example, If a contest problem involved the polynomial <math>x^5+4x^4+6x^3+4x^2+x</math>, I would 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, and knowing the first few coefficients would also be beneficial.
  
 
==See also==
 
==See also==
 
*[[Combinatorics]]
 
*[[Combinatorics]]

Revision as of 10:17, 23 June 2006

The Theorem

First invented by Newton, the Binomial Theorem states that for real or complex a,b,
$(a+b)^n = \sum_{k=0}^{n}{n \choose k}\cdot a^k\cdot b^{n-k}$

This may be shown for the integers easily:
$\displaystyle (a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}$
Repeatedly using the distributive property, we see that for a term $\displaystyle a^m b^{n-m}$, we must choose $m$ of the $n$ terms to contribute an $a$ to the term, and then each of the other $n-m$ terms of the product must contribute a $b$. Thus the coefficient of $\displaystyle a^m b^{n-m}$ is $\displaystyle n \choose m$. Extending this to all possible values of $m$ from $0$ to $n$, we see that $(a+b)^n = \sum_{k=0}^{n}{n \choose k}\cdot a^k\cdot b^{n-k}$.

Usage

Many factorizations involve complicated polynomials with binomial coefficients. For example, If a contest problem involved the polynomial $x^5+4x^4+6x^3+4x^2+x$, I would factor it as such: $x(x^4+4x^3+6x^2+4x+1)=x(x+1)^{4}$. It is a good idea to be familiar with binomial expansions, and knowing the first few coefficients would also be beneficial.

See also