Difference between revisions of "Binomial Theorem"

Latest revision as of 12:11, 20 February 2024

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.

Proof

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.

Similarly, the coefficients of $(x+y)^n$ will be the entries of the $n^\text{th}$ row of Pascal's Triangle. This is explained further in the Counting and Probability textbook [AoPS].

Proof via Induction

Given the constants $a,b,n$ are all natural numbers, it's clear to see that $(a+b)^{1} = a+b$ . Assuming that $(a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$ , $(a+b)^{n+1} = (\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k})(a+b)$ $=(\binom{n}{0}a^{n}b^{0} + \binom{n}{1}a^{n-1}b^{1} + \binom{n}{2}a^{n-2}b^{2}+\cdots+\binom{n}{n}a^{0}b^{n})(a+b)$ $=(\binom{n}{0}a^{n+1}b^{0} + \binom{n}{1}a^{n}b^{1} + \binom{n}{2}a^{n-1}b^{2}+\cdots+\binom{n}{n}a^{1}b^{n}) + (\binom{n}{0}a^{n}b^{1} + \binom{n}{1}a^{n-1}b^{2} + \binom{n}{2}a^{n-2}b^{3}+\cdots+\binom{n}{n}a^{0}b^{n+1})$ $=(\binom{n}{0}a^{n+1}b^{0} + (\binom{n}{0}+\binom{n}{1})(a^{n}b^{1}) + (\binom{n}{1}+\binom{n}{2})(a^{n-1}b^{2})+\cdots+(\binom{n}{n-1}+\binom{n}{n})(a^{1}b^{n})+\binom{n}{n}a^{0}b^{n+1})$ $=\binom{n+1}{0}a^{n+1}b^{0} + \binom{n+1}{1}a^{n}b^{1} + \binom{n+1}{2}a^{n-1}b^{2}+\cdots+\binom{n+1}{n}a^{1}b^{n} + \binom{n+1}{n+1}a^{0}b^{n+1}$ $=\sum_{k=0}^{n+1}\binom{n+1}{k}a^{(n+1)-k}b^{k}$ Therefore, if the theorem holds under $n+1$ , it must be valid. (Note that $\binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1}$ for $m\leq n$ )

Proof using calculus

The Taylor series for $e^x$ is $\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots$ for all $x$ .

Since $e^{a+b} = e^ae^b$ , and power series for the same function are termwise equal, the series at $x = a + b$ is the convolution of the series at $x = a$ and $x = b$ . Examining the degree- $n$ term of each, $\frac{(a+b)^n}{n!} = \sum_{k=0}^{n} \left( \frac{a^k}{k!} \right) \left( \frac{b^{n-k}}{(n-k)!} \right),$ which simplifies to $(a+b)^n = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!}a^kb^{n-k}$ for all natural numbers $n$ .

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

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

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.

@@ Line 2: / Line 2: @@
 <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]].  This result has a nice [[combinatorial proof]]: <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 <math>\binom{m}{n}</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}{\binom{n}{k}}\cdot a^k\cdot b^{n-k}</math>.
+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.
+==Proof==
+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.
+Similarly, the coefficients of <math>(x+y)^n</math> will be the entries of the <math>n^\text{th}</math> row of [[Pascal's Triangle]]. This is explained further in the Counting and Probability textbook [AoPS].
+===Proof via Induction===
+Given the constants <math>a,b,n</math> are all natural numbers, it's clear to see that <math>(a+b)^{1} = a+b</math>. Assuming that <math>(a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}</math>, <cmath>(a+b)^{n+1} = (\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k})(a+b)</cmath>
+<cmath>=(\binom{n}{0}a^{n}b^{0} + \binom{n}{1}a^{n-1}b^{1} + \binom{n}{2}a^{n-2}b^{2}+\cdots+\binom{n}{n}a^{0}b^{n})(a+b)</cmath>
+<cmath>=(\binom{n}{0}a^{n+1}b^{0} + \binom{n}{1}a^{n}b^{1} + \binom{n}{2}a^{n-1}b^{2}+\cdots+\binom{n}{n}a^{1}b^{n})
++ (\binom{n}{0}a^{n}b^{1} + \binom{n}{1}a^{n-1}b^{2} + \binom{n}{2}a^{n-2}b^{3}+\cdots+\binom{n}{n}a^{0}b^{n+1})</cmath>
+<cmath>=(\binom{n}{0}a^{n+1}b^{0} + (\binom{n}{0}+\binom{n}{1})(a^{n}b^{1}) + (\binom{n}{1}+\binom{n}{2})(a^{n-1}b^{2})+\cdots+(\binom{n}{n-1}+\binom{n}{n})(a^{1}b^{n})+\binom{n}{n}a^{0}b^{n+1})</cmath>
+<cmath>=\binom{n+1}{0}a^{n+1}b^{0} + \binom{n+1}{1}a^{n}b^{1} + \binom{n+1}{2}a^{n-1}b^{2}+\cdots+\binom{n+1}{n}a^{1}b^{n} + \binom{n+1}{n+1}a^{0}b^{n+1}</cmath>
+<cmath>=\sum_{k=0}^{n+1}\binom{n+1}{k}a^{(n+1)-k}b^{k}</cmath>
+Therefore, if the theorem holds under <math>n+1</math>, it must be valid.
+(Note that <math>\binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1} </math> for <math>m\leq n</math>)
+===Proof using calculus===
+The [[Taylor series]] for <math>e^x</math> is <cmath>\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots</cmath> for all <math>x</math>.
+Since <math>e^{a+b} = e^ae^b</math>, and power series for the same function are termwise equal, the series at <math>x = a + b</math> is the [[Generating function#Convolutions|convolution]] of the series at <math>x = a</math> and <math>x = b</math>. Examining the degree-<math>n</math> term of each, <cmath>\frac{(a+b)^n}{n!} = \sum_{k=0}^{n} \left( \frac{a^k}{k!} \right) \left( \frac{b^{n-k}}{(n-k)!} \right),</cmath> which simplifies to <cmath>(a+b)^n = \sum_{k=0}^{n} \frac{n!}{k!(n-k)!}a^kb^{n-k}</cmath> for all [[Natural number|natural numbers]] <math>n</math>.
 ==Generalizations==
@@ Line 11: / Line 36: @@
 Consider the function <math>f(b)=(a+b)^r</math> for constants <math>a,r</math>.  It is easy to see that <math>\frac{d^k}{db^k}f=r(r-1)\cdots(r-k+1)(a+b)^{r-k}</math>.  Then, we have <math>\frac{d^k}{db^k}f(0)=r(r-1)\cdots(r-k+1)a^{r-k}</math>.  So, the [[Taylor series]] for <math>f(b)</math> centered at <math>0</math> is
-<cmath>(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.</cmath>
+<cmath>(a+b)^r=\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.</cmath>
 ==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.
-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==

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Difference between revisions of "Binomial Theorem"

Latest revision as of 12:11, 20 February 2024

Contents

Proof

Proof via Induction

Proof using calculus

Generalizations

Proof

Usage

See also