Contents

Binomial theorem

Algebraic expansion of powers of a binomial

${\displaystyle {\begin{array}{c}1\\1\quad 1\\1\quad 2\quad 1\\1\quad 3\quad 3\quad 1\\1\quad 4\quad 6\quad 4\quad 1\\1\quad 5\quad 10\quad 10\quad 5\quad 1\\1\quad 6\quad 15\quad 20\quad 15\quad 6\quad 1\\1\quad 7\quad 21\quad 35\quad 35\quad 21\quad 7\quad 1\end{array}}}$

The binomial coefficient ${\displaystyle {\tbinom {n}{k}}}$ appears as the kth entry in the nth row of Pascal's triangle (where the top is the 0th row ${\displaystyle {\tbinom {0}{0}}}$). Each entry is the sum of the two above it.

Inelementary algebra, the binomial theorem (orbinomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial (x + y)ⁿ into a sum involving terms of the form ax^by^c, where the exponents b and c are nonnegative integers with b + c = n, and the coefficient a of each term is a specific positive integer depending on n and b. For example, for n = 4, $${\displaystyle (x+y)^{4}=x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4}.}$$

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

History[edit]

Special cases of the binomial theorem were known since at least the 4th century BC when Greek mathematician Euclid mentioned the special case of the binomial theorem for exponent ${\displaystyle n=2}$.^[1] Greek mathematician Diophantus cubed various binomials, including ${\displaystyle x-1}$.^[1] Indian mathematician Aryabhata's method for finding cube roots, from around 510 CE, suggests that he knew the binomial formula for exponent ${\displaystyle n=3}$.^[1]

Binomial coefficients, as combinatorial quantities expressing the number of ways of selecting k objects out of n without replacement, were of interest to ancient Indian mathematicians. The earliest known reference to this combinatorial problem is the Chandaḥśāstra by the Indian lyricist Pingala (c. 200 BC), which contains a method for its solution.^[2]: 230 The commentator Halayudha from the 10th century AD explains this method.^[2]: 230 By the 6th century AD, the Indian mathematicians probably knew how to express this as a quotient ${\textstyle {\frac {n!}{(n-k)!k!}}}$,^[3] and a clear statement of this rule can be found in the 12th century text LilavatibyBhaskara.^[3]

The first known formulation of the binomial theorem and the table of binomial coefficients appears in a work by Al-Karaji, quoted by Al-Samaw'al in his "al-Bahir".^[4][5][6] Al-Karaji described the triangular pattern of the binomial coefficients^[7] and also provided a mathematical proof of both the binomial theorem and Pascal's triangle, using an early form of mathematical induction.^[7] The Persian poet and mathematician Omar Khayyam was probably familiar with the formula to higher orders, although many of his mathematical works are lost.^[1] The binomial expansions of small degrees were known in the 13th century mathematical works of Yang Hui^[8] and also Chu Shih-Chieh.^[1] Yang Hui attributes the method to a much earlier 11th century text of Jia Xian, although those writings are now also lost.^[2]: 142

In 1544, Michael Stifel introduced the term "binomial coefficient" and showed how to use them to express ${\displaystyle (1+x)^{n}}$ in terms of ${\displaystyle (1+x)^{n-1}}$, via "Pascal's triangle".^[9] Blaise Pascal studied the eponymous triangle comprehensively in his Traité du triangle arithmétique.^[10] However, the pattern of numbers was already known to the European mathematicians of the late Renaissance, including Stifel, Niccolò Fontana Tartaglia, and Simon Stevin.^[9]

Isaac Newton is generally credited with discovering the generalized binomial theorem, valid for any real exponent, in 1665.^[9][11] It was discovered independently in 1670 by James Gregory.^[12]

Statement[edit]

According to the theorem, the expansion of any nonnegative integer power n of the binomial x + y is a sum of the form $${\displaystyle (x+y)^{n}={n \choose 0}x^{n}y^{0}+{n \choose 1}x^{n-1}y^{1}+{n \choose 2}x^{n-2}y^{2}+\cdots +{n \choose n-1}x^{1}y^{n-1}+{n \choose n}x^{0}y^{n},}$$ 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 $${\displaystyle (x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}.}$$

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 symmetrical, ${\textstyle {\binom {n}{k}}={\binom {n}{n-k}}.}$

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 $${\displaystyle {\begin{aligned}(1+x)^{n}&={n \choose 0}x^{0}+{n \choose 1}x^{1}+{n \choose 2}x^{2}+\cdots +{n \choose {n-1}}x^{n-1}+{n \choose n}x^{n}\\[4mu]&=\sum _{k=0}^{n}{n \choose k}x^{k}.{\vphantom {\Bigg )}}\end{aligned}}}$$

Examples[edit]

Here are the first few cases of the binomial theorem: $${\displaystyle {\begin{aligned}(x+y)^{0}&=1,\\[8pt](x+y)^{1}&=x+y,\\[8pt](x+y)^{2}&=x^{2}+2xy+y^{2},\\[8pt](x+y)^{3}&=x^{3}+3x^{2}y+3xy^{2}+y^{3},\\[8pt](x+y)^{4}&=x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4},\\[8pt](x+y)^{5}&=x^{5}+5x^{4}y+10x^{3}y^{2}+10x^{2}y^{3}+5xy^{4}+y^{5},\\[8pt](x+y)^{6}&=x^{6}+6x^{5}y+15x^{4}y^{2}+20x^{3}y^{3}+15x^{2}y^{4}+6xy^{5}+y^{6},\\[8pt](x+y)^{7}&=x^{7}+7x^{6}y+21x^{5}y^{2}+35x^{4}y^{3}+35x^{3}y^{4}+21x^{2}y^{5}+7xy^{6}+y^{7},\\[8pt](x+y)^{8}&=x^{8}+8x^{7}y+28x^{6}y^{2}+56x^{5}y^{3}+70x^{4}y^{4}+56x^{3}y^{5}+28x^{2}y^{6}+8xy^{7}+y^{8}.\end{aligned}}}$$ 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):

• the exponents of x in the terms are n, n − 1, ..., 2, 1, 0 (the last term implicitly contains x⁰ = 1);
• the exponents of y in the terms are 0, 1, 2, ..., n − 1, n (the first term implicitly contains y⁰ = 1);
• the coefficients form the nth row of Pascal's triangle;
• before combining like terms, there are 2ⁿ terms xⁱy^j in the expansion (not shown);
• after combining like terms, there are n + 1 terms, and their coefficients sum to 2ⁿ.

An example illustrating the last two points: $${\displaystyle {\begin{aligned}(x+y)^{3}&=xxx+xxy+xyx+xyy+yxx+yxy+yyx+yyy&(2^{3}{\text{ terms}})\\&=x^{3}+3x^{2}y+3xy^{2}+y^{3}&(3+1{\text{ terms}})\end{aligned}}}$$ with ${\displaystyle 1+3+3+1=2^{3}}$.

A simple example with a specific positive value of y: $${\displaystyle {\begin{aligned}(x+2)^{3}&=x^{3}+3x^{2}($$2)+3x(2)^{2}+2^{3}\\&=x^{3}+6x^{2}+12x+8.\end{aligned}}}

A simple example with a specific negative value of y: $${\displaystyle {\begin{aligned}(x-2)^{3}&=x^{3}-3x^{2}($$2)+3x(2)^{2}-2^{3}\\&=x^{3}-6x^{2}+12x-8.\end{aligned}}}

Geometric explanation[edit]

For positive values of a and b, the binomial theorem with n = 2 is the geometrically evident fact that a square of side a + b can be cut into a square of side a, a square of side b, and two rectangles with sides a and b. With n = 3, the theorem states that a cube of side a + b can be cut into a cube of side a, a cube of side b, three a × a × b rectangular boxes, and three a × b × b rectangular boxes.

Incalculus, this picture also gives a geometric proof of the derivative ${\displaystyle (x^{n})'=nx^{n-1}:}$^[13] if one sets ${\displaystyle a=x}$ and ${\displaystyle b=\Delta x,}$ interpreting b as an infinitesimal change in a, then this picture shows the infinitesimal change in the volume of an n-dimensional hypercube, ${\displaystyle (x+\Delta x)^{n},}$ where the coefficient of the linear term (in${\displaystyle \Delta x}$) is ${\displaystyle nx^{n-1},}$ the area of the n faces, each of dimension n − 1: $${\displaystyle (x+\Delta x)^{n}=x^{n}+nx^{n-1}\Delta x+{\binom {n}{2}}x^{n-2}(\Delta x)^{2}+\cdots .}$$ Substituting this into the definition of the derivative via a difference quotient and taking limits means that the higher order terms, ${\displaystyle (\Delta x)^{2}}$ and higher, become negligible, and yields the formula ${\displaystyle (x^{n})'=nx^{n-1},}$ interpreted as

"the infinitesimal rate of change in volume of an n-cube as side length varies is the area of n of its (n − 1)-dimensional faces".

If one integrates this picture, which corresponds to applying the fundamental theorem of calculus, one obtains Cavalieri's quadrature formula, the integral ${\displaystyle \textstyle {\int x^{n-1}\,dx={\tfrac {1}{n}}x^{n}}}$ – see proof of Cavalieri's quadrature formula for details.^[13]

Binomial coefficients[edit]

The coefficients that appear in the binomial expansion are called binomial coefficients. These are usually written ${\displaystyle {\tbinom {n}{k}},}$ and pronounced "n choose k".

Formulas[edit]

The coefficient of x^n−ky^k is given by the formula $${\displaystyle {\binom {n}{k}}={\frac {n!}{k!\;(n-k)!}},}$$ which is defined in terms of the factorial function n!. Equivalently, this formula can be written $${\displaystyle {\binom {n}{k}}={\frac {n(n-1)\cdots (n-k+1)}{k(k-1)\cdots 1}}=\prod _{\ell =1}^{k}{\frac {n-\ell +1}{\ell }}=\prod _{\ell =0}^{k-1}{\frac {n-\ell }{k-\ell }}}$$ with k factors in both the numerator and denominator of the fraction. Although this formula involves a fraction, the binomial coefficient ${\displaystyle {\tbinom {n}{k}}}$ is actually an integer.

Combinatorial interpretation[edit]

The binomial coefficient ${\displaystyle {\tbinom {n}{k}}}$ can be interpreted as the number of ways to choose k elements from an n-element set. This is related to binomials for the following reason: if we write (x + y)ⁿ as a product $${\displaystyle (x+y)(x+y)(x+y)\cdots (x+y),}$$ then, according to the distributive law, there will be one term in the expansion for each choice of either xory from each of the binomials of the product. For example, there will only be one term xⁿ, corresponding to choosing x from each binomial. However, there will be several terms of the form xⁿ⁻²y², one for each way of choosing exactly two binomials to contribute a y. Therefore, after combining like terms, the coefficient of xⁿ⁻²y² will be equal to the number of ways to choose exactly 2 elements from an n-element set.

Proofs[edit]

Combinatorial proof[edit]

Expanding (x + y)ⁿ yields the sum of the 2ⁿ products of the form e₁e₂ ... e_n where each e_iisx or y. Rearranging factors shows that each product equals x^n−ky^k for some k between 0 and n. For a given k, the following are proved equal in succession:

• the number of terms equal to x^n−ky^k in the expansion
• the number of n-character x,y strings having y in exactly k positions
• the number of k-element subsets of {1, 2, ..., n}
• ${\displaystyle {\tbinom {n}{k}},}$ either by definition, or by a short combinatorial argument if one is defining ${\displaystyle {\tbinom {n}{k}}}$as${\displaystyle {\tfrac {n!}{k!(n-k)!}}.}$

This proves the binomial theorem.

Example[edit]

The coefficient of xy²in$${\displaystyle {\begin{aligned}(x+y)^{3}&=(x+y)(x+y)(x+y)\\&=xxx+xxy+xyx+{\underline {xyy}}+yxx+{\underline {yxy}}+{\underline {yyx}}+yyy\\&=x^{3}+3x^{2}y+{\underline {3xy^{2}}}+y^{3}\end{aligned}}}$$ equals ${\displaystyle {\tbinom {3}{2}}=3}$ because there are three x,y strings of length 3 with exactly two y's, namely, $${\displaystyle xyy,\;yxy,\;yyx,}$$ corresponding to the three 2-element subsets of {1, 2, 3}, namely, $${\displaystyle \{2,3\},\;\{1,3\},\;\{1,2\},}$$ where each subset specifies the positions of the y in a corresponding string.

Inductive proof[edit]

Induction yields another proof of the binomial theorem. When n = 0, both sides equal 1, since x⁰ = 1 and ${\displaystyle {\tbinom {0}{0}}=1.}$ Now suppose that the equality holds for a given n; we will prove it for n + 1. For j, k ≥ 0, let [f(x, y)]_j,k denote the coefficient of x^jy^k in the polynomial f(x, y). By the inductive hypothesis, (x + y)ⁿ is a polynomial in x and y such that [(x + y)ⁿ]_j,kis${\displaystyle {\tbinom {n}{k}}}$ifj + k = n, and 0 otherwise. The identity $${\displaystyle (x+y)^{n+1}=x(x+y)^{n}+y(x+y)^{n}}$$ shows that (x + y)ⁿ⁺¹ is also a polynomial in x and y, and $${\displaystyle [(x+y)^{n+1}]_{j,k}=[(x+y)^{n}]_{j-1,k}+[(x+y)^{n}]_{j,k-1},}$$ since if j + k = n + 1, then (j − 1) + k = n and j + (k − 1) = n. Now, the right hand side is $${\displaystyle {\binom {n}{k}}+{\binom {n}{k-1}}={\binom {n+1}{k}},}$$byPascal's identity.^[14] On the other hand, if j + k ≠ n + 1, then (j – 1) + k ≠ n and j + (k – 1) ≠ n, so we get 0 + 0 = 0. Thus $${\displaystyle (x+y)^{n+1}=\sum _{k=0}^{n+1}{\binom {n+1}{k}}x^{n+1-k}y^{k},}$$ which is the inductive hypothesis with n + 1 substituted for n and so completes the inductive step.

Generalizations[edit]

Newton's generalized binomial theorem[edit]

Around 1665, Isaac Newton generalized the binomial theorem to allow real exponents other than nonnegative integers. (The same generalization also applies to complex exponents.) In this generalization, the finite sum is replaced by an infinite series. In order to do this, one needs to give meaning to binomial coefficients with an arbitrary upper index, which cannot be done using the usual formula with factorials. However, for an arbitrary number r, one can define $${\displaystyle {r \choose k}={\frac {r(r-1)\cdots (r-k+1)}{k!}}={\frac {($$r)_{k}}{k!}},} where ${\displaystyle (\cdot )_{k}}$ is the Pochhammer symbol, here standing for a falling factorial. This agrees with the usual definitions when r is a nonnegative integer. Then, if x and y are real numbers with |x| > |y|,^{[Note 1]} and r is any complex number, one has $${\displaystyle {\begin{aligned}(x+y)^{r}&=\sum _{k=0}^{\infty }{r \choose k}x^{r-k}y^{k}\\&=x^{r}+rx^{r-1}y+{\frac {r(r-1)}{2!}}x^{r-2}y^{2}+{\frac {r(r-1)(r-2)}{3!}}x^{r-3}y^{3}+\cdots .\end{aligned}}}$$

When r is a nonnegative integer, the binomial coefficients for k > r are zero, so this equation reduces to the usual binomial theorem, and there are at most r + 1 nonzero terms. For other values of r, the series typically has infinitely many nonzero terms.

For example, r = 1/2 gives the following series for the square root: $${\displaystyle {\sqrt {1+x}}=1+{\frac {1}{2}}x-{\frac {1}{8}}x^{2}+{\frac {1}{16}}x^{3}-{\frac {5}{128}}x^{4}+{\frac {7}{256}}x^{5}-\cdots .}$$

Taking r = −1, the generalized binomial series gives the geometric series formula, valid for |x| <1: $${\displaystyle (1+x)^{-1}={\frac {1}{1+x}}=1-x+x^{2}-x^{3}+x^{4}-x^{5}+\cdots .}$$

More generally, with r = −s, we have for |x| <1:^[15] $${\displaystyle {\frac {1}{(1+x)^{s}}}=\sum _{k=0}^{\infty }{-s \choose k}x^{k}=\sum _{k=0}^{\infty }{s+k-1 \choose k}(-1)^{k}x^{k}.}$$

So, for instance, when s = 1/2, $${\displaystyle {\frac {1}{\sqrt {1+x}}}=1-{\frac {1}{2}}x+{\frac {3}{8}}x^{2}-{\frac {5}{16}}x^{3}+{\frac {35}{128}}x^{4}-{\frac {63}{256}}x^{5}+\cdots .}$$