# Binomial theorem

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In elementary algebra, the binomial theorem or the binomial expansion is a mechanism by which expressions of the form ${\displaystyle (x+y)^{n}}$ can be expanded. It is the identity that states that for any non-negative integer n,

${\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k},}$

where

${\displaystyle {n \choose k}={\frac {n!}{k!(n-k)!}}}$

is a binomial coefficient. Another useful way of stating it is the following:

${\displaystyle (x+y)^{n}={n \choose 0}x^{n}+{n \choose 1}x^{n-1}y+{n \choose 2}x^{n-2}y^{2}+\ldots +{n \choose n}y^{n}}$

### Pascal's triangle

An alternate way to find the binomial coefficients is by using Pascal's triange. The triangle is built from apex down, starting with the number one alone on a row. Each number is equal to the sum of the two numbers directly above it.

n=0         1
n=1        1 1
n=2       1 2 1
n=3      1 3 3 1
n=4     1 4 6 4 1
n=5   1 5 10 10 5 1


Thus, the binomial coefficients for the expression ${\displaystyle (x+y)^{4}}$ are 1, 3, 6, 4, and 1.

## Proof

One way to prove this identity is by mathematical induction.

Base case: n = 0

${\displaystyle (x+y)^{0}=\sum _{k=0}^{0}{0 \choose k}x^{0-k}y^{k}=1}$

Induction case: Now suppose that it is true for n : ${\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k},}$ and prove it for n + 1.

${\displaystyle (x+y)^{n+1}=(x+y)(x+y)^{n}\,}$
${\displaystyle =(x+y)\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}\,}$
${\displaystyle =\sum _{k=0}^{n}{n \choose k}x^{n+1-k}y^{k}+\sum _{j=0}^{n}{n \choose j}x^{n-j}y^{j+1}\,}$
${\displaystyle =\sum _{k=0}^{n}{n \choose k}x^{n+1-k}y^{k}+\sum _{j=0}^{n}{n \choose {(j+1)-1}}x^{n-j}y^{j+1}\,}$
${\displaystyle =\sum _{k=0}^{n}{n \choose k}x^{n+1-k}y^{k}+\sum _{k=1}^{n+1}{n \choose {k-1}}x^{n+1-k}y^{k}\,}$
${\displaystyle =\sum _{k=0}^{n+1}{n \choose k}x^{n+1-k}y^{k}-{n \choose {n+1}}x^{0}y^{n+1}+\sum _{k=0}^{n+1}{n \choose {k-1}}x^{n+1-k}y^{k}-{n \choose {-1}}x^{n+1}y^{0}\,}$
${\displaystyle =\sum _{k=0}^{n+1}\left[{n \choose k}+{n \choose {k-1}}\right]x^{n+1-k}y^{k}\,}$
${\displaystyle =\sum _{k=0}^{n+1}{{n+1} \choose k}x^{n+1-k}y^{k},}$

and the proof is complete.

## Examples

These are the expansions from 0 to 6.

{\displaystyle {\begin{aligned}(x+y)^{0}&=1\\(x+y)^{1}&=x+y\\(x+y)^{2}&=x^{2}+2xy+y^{2}\\(x+y)^{3}&=x^{3}+3x^{2}y+3xy^{2}+y^{3}\\(x+y)^{4}&=x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4}\\(x+y)^{5}&=x^{5}+5x^{4}y+10x^{3}y^{2}+6x^{2}y^{3}+y^{5}\\(x+y)^{6}&=x^{6}+6x^{5}y+15x^{4}y^{2}+20x^{3}y^{3}+15x^{2}y^{4}+6xy^{5}+y^{6}\end{aligned}}}

## Newton's binomial theorem

There is also Newton's binomial theorem, proved by Isaac Newton, that goes beyond elementary algebra into mathematical analysis, which expands the same sum (x + y)n as an infinite series when n is not an integer or is not positive.