Difference between revisions of "Binomial Theorem"

Revision as of 14:53, 17 August 2013

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. In other words, the coefficients when $(a + b)^n$ is expanded and like terms are collected are the same as the entries in the $n$ th row of Pascal's Triangle.

For example, $(a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5$ , with coefficients $1 = \binom{5}{0}$ , $5 = \binom{5}{1}$ , $10 = \binom{5}{2}$ , 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 $(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}$ . 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 the number of ways to choose $m$ objects from a set of size $n$ , or $\binom{n}{m}$ . Extending this to all possible values of $m$ from $0$ to $n$ , we see that $(a+b)^n = \sum_{m=0}^{n}{\binom{n}{m}}\cdot a^m\cdot b^{n-m}$ , 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 $a$ , $b$ , 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.

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 ==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.
+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==

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