Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Binomial Theorem"
(→‎Usage)
Line 15: Line 15:
 
==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.
 +
 +
In addition, the expansion of a polynomial such as <math>(x+y)^n</math> will have coefficients corresponding to the <math>nth</math> row of [[Pascal's Triangle]]. For example, <math>(x+1)^5</math> = <math>x^5+5x^4+10x^3+10x^2+5x+1</math>, and the integers <math>1</math>, <math>5</math>, <math>10</math>, <math>10</math>, <math>5</math>, and <math>1</math> make up the 5th row of Pascal's Triangle.
  
 
==See also==
 
==See also==

Revision as of 00:07, 11 December 2009

The Binomial Theorem states that for real or complex $a$, $b$, and non-negative integer $n$,

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

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is a binomial coefficient. This result has a nice combinatorial proof: $(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}$. Repeatedly using the distributive property, we see that for a term $a^m b^{n-m}$, 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 $a^m b^{n-m}$ is $\binom{m}{n}$. Extending this to all possible values of $m$ from $0$ to $n$, we see that $(a+b)^n = \sum_{k=0}^{n}{\binom{n}{k}}\cdot a^k\cdot b^{n-k}$.

Generalizations

The Binomial Theorem was generalized by Isaac Newton, who used an infinite series to allow for complex exponents: For any real or complex $a$, $b$, and $r$,

$(a+b)^r = \sum_{k=0}^{\infty}\binom{r}{k}a^{r-k}b^k$.

Proof

Consider the function $f(b)=(a+b)^r$ for constants $a,r$. It is easy to see that $\frac{d^k}{db^k}f=r(r-1)\cdots(r-k+1)(a+b)^{r-k}$. Then, we have $\frac{d^k}{db^k}f(0)=r(r-1)\cdots(r-k+1)a^{r-k}$. So, the Taylor series for $f(b)$ centered at $0$ is

\[(a+b)^k=\sum_{k=0}^\infty \frac{r(r-1)\cdots(r-k+1)a^{r-k}b^k}{k!}=\sum_{k=0}^\infty \binom{r}{k}a^{r-k}b^k.\]

Usage

Many factorizations involve complicated polynomials with binomial coefficients. For example, if a contest problem involved the polynomial $x^5+4x^4+6x^3+4x^2+x$, one could 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, including knowing the first few binomial coefficients.

In addition, the expansion of a polynomial such as $(x+y)^n$ will have coefficients corresponding to the $nth$ row of Pascal's Triangle. For example, $(x+1)^5$ = $x^5+5x^4+10x^3+10x^2+5x+1$, and the integers $1$, $5$, $10$, $10$, $5$, and $1$ make up the 5th row of Pascal's Triangle.

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Binomial_Theorem&oldid=33015"