Difference between revisions of "Binomial Theorem"
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==The Theorem== | ==The Theorem== | ||
| − | First discovered 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 discovered by [[Newton]], the '''Binomial Theorem''' states that for [[real number | real]] or [[complex number |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 | + | This may be easily shown for the [[integer]]s:<br> |
<math>\displaystyle (a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}</math>. | <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>. |
==Usage== | ==Usage== | ||
| − | Many factorizations involve complicated | + | 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. |
==See also== | ==See also== | ||
*[[Combinatorics]] | *[[Combinatorics]] | ||
*[[Multinomial Theorem]] | *[[Multinomial Theorem]] | ||
Revision as of 15:08, 18 July 2006
The Theorem
First discovered by Newton, the Binomial Theorem states that for real or complex a,b,
.
This may be easily shown for the integers:
.
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 
. Extending this to all possible values of from 
to , we see that .
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.
