# Difference between revisions of "Binomial Theorem"

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===Proof via Induction=== | ===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> | + | 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}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}) | <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}) |

## Latest revision as of 15:08, 6 July 2020

The **Binomial Theorem** states that for real or complex , , and non-negative integer ,

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.

## 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 . 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.

Similarly, the coefficients of will be the entries of the row of Pascal's Triangle. This is explained further in the Counting and Probability textbook [AoPS].

### Proof via Induction

Given the constants are all natural numbers, it's clear to see that . Assuming that , Therefore, if the theorem holds under , it must be valid. (Note that for )

## 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 ,

### 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.