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Difference between revisions of "Binomial Theorem"

(Clarification. Seemed a bit incognoscible previously.)
m (Fixed \binom{}{} to {} \choose {})
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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>
 
<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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<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 \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_{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>.

Revision as of 08:20, 22 June 2006

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 $(a+b)^n = \sum_{k=0}^{n}{n \choose k}\cdot a^k\cdot b^{n-k}$.

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