Difference between revisions of "Binomial Theorem"
Chebyshev128 (talk | contribs) (→Proof) |
Chebyshev128 (talk | contribs) (→Proof via Induction (Long)) |
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<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> | <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> | ||
− | ===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}b^{ | + | 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}) | ||
− | + (\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> | + | + (\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+a} /</math> for <math>m\leq n</math>) | ||
==Usage== | ==Usage== |
Revision as of 01:21, 12 August 2019
The Binomial Theorem states that for real or complex ,
, and non-negative integer
,
![$(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k$](http://latex.artofproblemsolving.com/4/9/4/494769588adb2ac3700e25b086fe2b7d41bba70a.png)
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
,
![$(a+b)^r = \sum_{k=0}^{\infty}\binom{r}{k}a^{r-k}b^k$](http://latex.artofproblemsolving.com/0/8/9/0896503fb81e2e64fc5d22e03210b9fc46b7ce32.png)
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.