Binomial theorem

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

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

or, equivalently,

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

where

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

is a binomial coefficient.

One way to prove this identity is by mathematical induction.

Proof:

Base case: n = 0

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

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

$(x+y)^{n+1}=(x+y)(x+y)^{n}\,$

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

$=\sum _{k=0}^{n}{n \choose k}x^{n+1-k}y^{k}+\sum _{j=0}^{n}{n \choose j}x^{n-j}y^{j+1}\,$

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

$=\sum _{k=0}^{n}{n \choose k}x^{n+1-k}y^{k}+\sum _{k=1}^{n+1}{n \choose {k-1}}x^{n+1-k}y^{k}\,$

$=\sum _{k=0}^{n+1}{n \choose k}x^{n+1-k}y^{k}-{n \choose {n+1}}x^{0}y^{n+1}+\sum _{k=0}^{n+1}{n \choose {k-1}}x^{n+1-k}y^{k}-{n \choose {-1}}x^{n+1}y^{0}\,$

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

$=\sum _{k=0}^{n+1}{{n+1} \choose k}x^{n+1-k}y^{k},$

and the proof is complete.

The first several cases

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

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.