Binomial theorem

In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, the power ⁠${\displaystyle \textstyle (x+y)^{n}}$⁠ expands into a polynomial with terms of the form ⁠${\displaystyle \textstyle ax^{k}y^{m}}$⁠, where the exponents ⁠${\displaystyle k}$⁠ and ⁠${\displaystyle m}$⁠ are nonnegative integers satisfying ⁠${\displaystyle k+m=n}$⁠ and the coefficient ⁠${\displaystyle a}$⁠ of each term is a specific positive integer depending on ⁠${\displaystyle n}$⁠ and ⁠${\displaystyle k}$⁠. For example, for ⁠${\displaystyle n=4}$⁠, $${\displaystyle (x+y)^{4}=x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4}.}$$

The coefficient ⁠${\displaystyle a}$⁠ in each term ⁠${\displaystyle \textstyle ax^{k}y^{m}}$⁠ is known as the binomial coefficient ⁠${\displaystyle {\tbinom {n}{k}}}$⁠ or ⁠${\displaystyle {\tbinom {n}{m}}}$⁠ (the two have the same value). These coefficients for varying ⁠${\displaystyle n}$⁠ and ⁠${\displaystyle k}$⁠ can be arranged to form Pascal's triangle. These numbers also occur in combinatorics, where ⁠${\displaystyle {\tbinom {n}{k}}}$⁠ gives the number of different combinations (i.e. subsets) of ⁠${\displaystyle k}$⁠ elements that can be chosen from an ⁠${\displaystyle n}$⁠-element set. Therefore ⁠${\displaystyle {\tbinom {n}{k}}}$⁠ is usually pronounced as "⁠${\displaystyle n}$⁠ choose ⁠${\displaystyle k}$⁠".

According to the theorem, the expansion of any nonnegative integer power n of the binomial x + y is a sum of the form $${\displaystyle (x+y)^{n}={\binom {n}{0}}x^{n}y^{0}+{\binom {n}{1}}x^{n-1}y^{1}+{\binom {n}{2}}x^{n-2}y^{2}+\cdots +{\binom {n}{n}}x^{0}y^{n},}$$ where each ${\displaystyle {\tbinom {n}{k}}}$ is a positive integer known as a binomial coefficient, defined as

$${\displaystyle {\binom {n}{k}}={\frac {n!}{k!\,(n-k)!}}={\frac {n(n-1)(n-2)\cdots (n-k+1)}{k(k-1)(k-2)\cdots 2\cdot 1}}.}$$

This formula is also referred to as the binomial formula or the binomial identity. Using summation notation, it can be written more concisely as $${\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}x^{n-k}y^{k}=\sum _{k=0}^{n}{\binom {n}{k}}x^{k}y^{n-k}.}$$

The final expression follows from the previous one by the symmetry of x and y in the first expression, and by comparison it follows that the sequence of binomial coefficients in the formula is symmetric, ${\textstyle {\binom {n}{k}}={\binom {n}{n-k}}.}$

A simple variant of the binomial formula is obtained by substituting 1 for y, so that it involves only a single variable. In this form, the formula reads $${\displaystyle {\begin{aligned}(x+1)^{n}&={\binom {n}{0}}x^{0}+{\binom {n}{1}}x^{1}+{\binom {n}{2}}x^{2}+\cdots +{\binom {n}{n}}x^{n}\\[4mu]&=\sum _{k=0}^{n}{\binom {n}{k}}x^{k}.{\vphantom {\Bigg )}}\end{aligned}}}$$

The first few cases of the binomial theorem are: $${\displaystyle {\begin{aligned}(x+y)^{0}&=1,\\[8pt](x+y)^{1}&=x+y,\\[8pt](x+y)^{2}&=x^{2}+2xy+y^{2},\\[8pt](x+y)^{3}&=x^{3}+3x^{2}y+3xy^{2}+y^{3},\\[8pt](x+y)^{4}&=x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4},\end{aligned}}}$$ In general, for the expansion of (x + y)ⁿ on the right side in the nth row (numbered so that the top row is the 0th row):