Difference between revisions of "Binomial Theorem"
Aimesolver (talk | contribs) (→Usage) |
|||
Line 2: | Line 2: | ||
<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]]. |
+ | |||
+ | 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. | ||
+ | |||
+ | ==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== | ||
Line 15: | Line 20: | ||
==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. | ||
− | |||
− | |||
==See also== | ==See also== |
Revision as of 22:08, 17 December 2009
The Binomial Theorem states that for real or complex ,
, and non-negative integer
,
![$(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$](http://latex.artofproblemsolving.com/4/9/4/494769588adb2ac3700e25b086fe2b7d41bba70a.png)
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
,
![$(a+b)^r = \sum_{k=0}^{\infty}\binom{r}{k}a^{r-k}b^k$](http://latex.artofproblemsolving.com/0/8/9/0896503fb81e2e64fc5d22e03210b9fc46b7ce32.png)
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.