Difference between revisions of "Binomial Theorem"
Chebyshev128 (talk | contribs) (→Proof via Induction (Long)) |
Chebyshev128 (talk | contribs) (→Proof via Induction) |
||
Line 31: | Line 31: | ||
<cmath>=\sum_{k=0}^{n+1}\binom{n+1}{k}a^{(n+1)-k}b^{k}</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. | 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+ | + | (Note that <math>\binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1} </math> for <math>m\leq n</math>) |
==Usage== | ==Usage== |
Revision as of 00:23, 12 August 2019
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].
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
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 )
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.