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Difference between revisions of "Binomial Theorem"

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==Usage==
 
==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.
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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 13:55, 26 June 2006

The Theorem

First discovered 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

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