In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial.
According to the theorem, it is possible to expand the polynomial  into a sum involving terms of the form , where the exponents  and  are nonnegative integers with , and the coefficient  of each term is a specific positive integer depending on  and .
For example, for ,
(x+y)^4 = x^4 + 4 x^3y + 6 x^2 y^2 + 4 x y^3 + y^4.
The coefficient  in the term of  is known as the binomial coefficient \tbinom{n}{b} or \tbinom{n}{c} (the two have the same value).
These coefficients for varying  and  can be arranged to form Pascal's triangle.
These numbers also arise in combinatorics, where \tbinom{n}{b} gives the number of different combinations of  elements that can be chosen from an -element set.
Therefore \tbinom{n}{b} is often pronounced as " choose ".
History
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 .
There is evidence that the binomial theorem for cubes was known by the 6th century AD in India.
Binomial coefficients, as combinatorial quantities expressing the number of ways of selecting  objects out of  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.
The commentator Halayudha from the 10th century AD explains this method using what is now known as Pascal's triangle.
By the 6th century AD, the Indian mathematicians probably knew how to express this as a quotient \frac{n!}{(n-k)!k!}, and a clear statement of this rule can be found in the 12th century text Lilavati by Bhaskara.
The first formulation of the binomial theorem and the table of binomial coefficients, to our knowledge, can be found in a work by Al-Karaji, quoted by Al-Samaw'al in his "al-Bahir".
Al-Karaji described the triangular pattern of the binomial coefficients and also provided a mathematical proof of both the binomial theorem and Pascal's triangle, using an early form of mathematical induction.
The Persian poet and mathematician Omar Khayyam was probably familiar with the formula to higher orders, although many of his mathematical works are lost.
The binomial expansions of small degrees were known in the 13th century mathematical works of Yang Hui and also Chu Shih-Chieh.
Yang Hui attributes the method to a much earlier 11th century text of Jia Xian, although those writings are now also lost.
In 1544, Michael Stifel introduced the term "binomial coefficient" and showed how to use them to express (1+a)^n in terms of (1+a)^{n-1}, via "Pascal's triangle".
Blaise Pascal studied the eponymous triangle comprehensively in his Traité du triangle arithmétique.
However, the pattern of numbers was already known to the European mathematicians of the late Renaissance,  including Stifel, Niccolò Fontana Tartaglia, and Simon Stevin.
Isaac Newton is generally credited with the generalized binomial theorem, valid for any rational exponent.
Statement
According to the theorem, it is possible to expand any nonnegative power of  into a sum of the form
(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 n \geq 0 is an integer and each  \tbinom nk  is a positive integer known as a binomial coefficient.
(When an exponent is zero, the corresponding power expression is taken to be 1 and this multiplicative factor is often omitted from the term.
Hence one often sees the right hand side written as \binom{n}{0} x^n + \ldots.)
This formula is also referred to as the binomial formula or the binomial identity.
Using summation notation, it can be written as
(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  and  in the first expression, and by comparison it follows that the sequence of binomial coefficients in the formula is symmetrical.
A simple variant of the binomial formula is obtained by substituting  for , so that it involves only a single variable.
In this form, the formula reads
(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,
or equivalently
(1+x)^n = \sum_{k=0}^n {n \choose k}x^k.
or more explicitly
(1+x)^n = 1 + n x + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots.
Examples
Here are the first few cases of the binomial theorem:
\begin{align} (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^2y + 3xy^2 + y^3, \\[8pt] (x+y)^4 & = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4, \\[8pt] (x+y)^5 & = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5, \\[8pt] (x+y)^6 & = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6, \\[8pt] (x+y)^7 & = x^7 + 7x^6y + 21x^5y^2 + 35x^4y^3 + 35x^3y^4 + 21x^2y^5 + 7xy^6 + y^7, \\[8pt] (x+y)^8 & = x^8 + 8x^7y + 28x^6y^2 + 56x^5y^3 + 70x^4y^4 + 56x^3y^5 + 28x^2y^6 + 8xy^7 + y^8.
\end{align}
In general, for the expansion of  on the right side in the th row (numbered so that the top row is the 0th row):
the exponents of  in the terms are  (the last term implicitly contains );
the exponents of  in the terms are  (the first term implicitly contains );
the coefficients form the th row of Pascal's triangle;
before combining like terms, there are  terms  in the expansion (not shown);
after combining like terms, there are  terms, and their coefficients sum to .
An example illustrating the last two points:
\begin{align} (x+y)^3 & = xxx + xxy + xyx + xyy + yxx + yxy + yyx + yyy & (2^3 \; \mathrm{ terms}) \\         & = x^3 + 3x^2y + 3xy^2 + y^3 & (3 + 1 \; \mathrm{ terms}) \end{align}
with 1 + 3 + 3 + 1 = 2^3.
A simple example with a specific positive value of :
\begin{align} (x+2)^3 &= x^3 + 3x^2(2) + 3x(2)^2 + 2^3 \\ &= x^3 + 6x^2 + 12x + 8.
\end{align}
A simple example with a specific negative value of :
\begin{align} (x-2)^3 &= x^3 - 3x^2(2) + 3x(2)^2 - 2^3 \\ &= x^3 - 6x^2 + 12x - 8.
\end{align}
Geometric explanation
thumb|300px|Visualisation of binomial expansion up to the 4th power For positive values of  and , the binomial theorem with  is the geometrically evident fact that a square of side  can be cut into a square of side , a square of side , and two rectangles with sides  and .
With , the theorem states that a cube of side  can be cut into a cube of side , a cube of side , three  rectangular boxes, and three  rectangular boxes.
In calculus, this picture also gives a geometric proof of the derivative (x^n)'=nx^{n-1}: if one sets a=x and b=\Delta x, interpreting  as an infinitesimal change in , then this picture shows the infinitesimal change in the volume of an -dimensional hypercube, (x+\Delta x)^n, where the coefficient of the linear term (in \Delta x) is nx^{n-1}, the area of the  faces, each of dimension :
(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, (\Delta x)^2 and higher, become negligible, and yields the formula (x^n)'=nx^{n-1}, interpreted as
"the infinitesimal rate of change in volume of an -cube as side length varies is the area of  of its -dimensional faces".
If one integrates this picture, which corresponds to applying the fundamental theorem of calculus, one obtains Cavalieri's quadrature formula, the integral \textstyle{\int x^{n-1}\,dx = \tfrac{1}{n} x^n} – see proof of Cavalieri's quadrature formula for details.
Binomial coefficients
The coefficients that appear in the binomial expansion are called binomial coefficients.
These are usually written \tbinom{n}{k}, and pronounced " choose ".
Formulae
The coefficient of  is given by the formula
\binom{n}{k} = \frac{n!}{k! \; (n-k)!},
which is defined in terms of the factorial function .
Equivalently, this formula can be written
\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  factors in both the numerator and denominator of the fraction.
Although this formula involves a fraction, the binomial coefficient \tbinom{n}{k} is actually an integer.
Combinatorial interpretation
The binomial coefficient  \tbinom nk  can be interpreted as the number of ways to choose  elements from an -element set.
This is related to binomials for the following reason: if we write  as a product
(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  or  from each of the binomials of the product.
For example, there will only be one term , corresponding to choosing  from each binomial.
However, there will be several terms of the form , one for each way of choosing exactly two binomials to contribute a .
Therefore, after combining like terms, the coefficient of  will be equal to the number of ways to choose exactly  elements from an -element set.
Proofs
Combinatorial proof
Example
The coefficient of  in
\begin{align}    (x+y)^3 &= (x+y)(x+y)(x+y) \\    &= xxx + xxy + xyx + \underline{xyy} + yxx + \underline{yxy} + \underline{yyx} + yyy \\    &= x^3 + 3x^2y + \underline{3xy^2} + y^3  \end{align}
equals \tbinom{3}{2}=3 because there are three  strings of length 3 with exactly two s, namely,
xyy, \; yxy, \; yyx,
corresponding to the three 2-element subsets of , namely,
\{2,3\},\;\{1,3\},\;\{1,2\},
where each subset specifies the positions of the  in a corresponding string.
General case
Expanding  yields the sum of the  products of the form  where each  is  or .
Rearranging factors shows that each product equals  for some  between  and .
For a given , the following are proved equal in succession:
the number of copies of  in the expansion
the number of -character  strings having  in exactly  positions
the number of -element subsets of
\tbinom{n}{k}, either by definition, or by a short combinatorial argument if one is defining \tbinom{n}{k} as \tfrac{n!}{k! (n-k)!}.
This proves the binomial theorem.
Inductive proof
Induction yields another proof of the binomial theorem.
When , both sides equal , since  and \tbinom{0}{0}=1.
Now suppose that the equality holds for a given ; we will prove it for .
For , let  denote the coefficient of  in the polynomial .
By the inductive hypothesis,  is a polynomial in  and  such that  is \tbinom{n}{k} if , and  otherwise.
The identity
(x+y)^{n+1} = x(x+y)^n + y(x+y)^n
shows that  is also a polynomial in  and , and
[(x+y)^{n+1}]_{j,k} = [(x+y)^n]_{j-1,k} + [(x+y)^n]_{j,k-1},
since if , then  and .
Now, the right hand side is
\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k},
by Pascal's identity.Binomial theorem – inductive proofs  On the other hand, if , then  and , so we get .
Thus
(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  substituted for  and so completes the inductive step.
Generalizations
Newton's generalized binomial theorem
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 , one can define
{r \choose k}=\frac{r(r-1) \cdots (r-k+1)}{k!} =\frac{(r)_k}{k!},
where (\cdot)_k is the Pochhammer symbol, here standing for a falling factorial.
This agrees with the usual definitions when  is a nonnegative integer.
Then, if  and  are real numbers with ,This is to guarantee convergence.
Depending on , the series may also converge sometimes when .
and  is any complex number, one has
\begin{align}    (x+y)^r & =\sum_{k=0}^\infty {r \choose k} x^{r-k} y^k \\    &= x^r + r x^{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{align}
When  is a nonnegative integer, the binomial coefficients for  are zero, so this equation reduces to the usual binomial theorem, and there are at most  nonzero terms.
For other values of , the series typically has infinitely many nonzero terms.
For example,  gives the following series for the square root:
\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 , the generalized binomial series gives the geometric series formula, valid for :
(1+x)^{-1} = \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \cdots
More generally, with :
\frac{1}{(1-x)^s} = \sum_{k=0}^\infty {s+k-1 \choose k} x^k.
So, for instance, when ,
\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
Further generalizations
The generalized binomial theorem can be extended to the case where  and  are complex numbers.
For this version, one should again assume  and define the powers of  and  using a holomorphic branch of log defined on an open disk of radius  centered at .
The generalized binomial theorem is valid also for elements  and  of a Banach algebra as long as , and  is invertible, and .
A version of the binomial theorem is valid for the following Pochhammer symbol-like family of polynomials: for a given real constant , define  x^{(0)} = 1  and
x^{(n)} = \prod_{k=1}^{n}[x+(k-1)c]
for  n > 0.
Then
(a + b)^{(n)} = \sum_{k=0}^{n}\binom{n}{k}a^{(n-k)}b^{(k)}.
The case  recovers the usual binomial theorem.
More generally, a sequence \{p_n\}_{n=0}^\infty of polynomials is said to be binomial if
\deg p_n = n  for all n,
p_0(0) = 1 , and
p_n(x+y) = \sum_{k=0}^n \binom{n}{k} p_k(x) p_{n-k}(y)  for all x, y, and n.
An operator Q on the space of polynomials is said to be the basis operator of the sequence \{p_n\}_{n=0}^\infty if Qp_0 = 0 and  Q p_n = n p_{n-1}  for all  n \geqslant 1 .
A sequence \{p_n\}_{n=0}^\infty is binomial if and only if its basis operator is a Delta operator.
Writing  E^a  for the shift by  a  operator, the Delta operators corresponding to the above "Pochhammer" families of polynomials are the backward difference  I - E^{-c}  for  c>0 , the ordinary derivative for  c=0 , and the forward difference  E^{-c} - I  for  c<0 .
Multinomial theorem
The binomial theorem can be generalized to include powers of sums with more than two terms.
The general version is
(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots +k_m = n} \binom{n}{k_1, k_2, \ldots, k_m}  x_1^{k_1} x_2^{k_2} \cdots x_m^{k_m},
where the summation is taken over all sequences of nonnegative integer indices  through  such that the sum of all  is .
(For each term in the expansion, the exponents must add up to ).
The coefficients  \tbinom{n}{k_1,\cdots,k_m}  are known as multinomial coefficients, and can be computed by the formula
\binom{n}{k_1, k_2, \ldots, k_m} = \frac{n!}{k_1! \cdot k_2! \cdots k_m!}.
Combinatorially, the multinomial coefficient \tbinom{n}{k_1,\cdots,k_m} counts the number of different ways to partition an -element set into disjoint subsets of sizes .
{{anchor|multi-binomial}} Multi-binomial theorem
When working in more dimensions, it is often useful to deal with products of binomial expressions.
By the binomial theorem this is equal to
(x_1+y_1)^{n_1}\dotsm(x_d+y_d)^{n_d} = \sum_{k_1=0}^{n_1}\dotsm\sum_{k_d=0}^{n_d} \binom{n_1}{k_1} x_1^{k_1}y_1^{n_1-k_1} \dotsc \binom{n_d}{k_d} x_d^{k_d}y_d^{n_d-k_d}.
This may be written more concisely, by multi-index notation, as
(x+y)^\alpha = \sum_{\nu \le \alpha} \binom{\alpha}{\nu} x^\nu y^{\alpha - \nu}.
General Leibniz rule
The general Leibniz rule gives the th derivative of a product of two functions in a form similar to that of the binomial theorem:
(fg)^{(n)}(x) = \sum_{k=0}^n \binom{n}{k} f^{(n-k)}(x) g^{(k)}(x).
Here, the superscript  indicates the th derivative of a function.
If one sets  and , and then cancels the common factor of  from both sides of the result, the ordinary binomial theorem is recovered.
Applications
Multiple-angle identities
For the complex numbers the binomial theorem can be combined with de Moivre's formula to yield multiple-angle formulas for the sine and cosine.
According to De Moivre's formula,
\cos\left(nx\right)+i\sin\left(nx\right) = \left(\cos x+i\sin x\right)^n.
Using the binomial theorem, the expression on the right can be expanded, and then the real and imaginary parts can be taken to yield formulas for  and .
For example, since
\left(\cos x+i\sin x\right)^2 = \cos^2 x + 2i \cos x \sin x - \sin^2 x,
De Moivre's formula tells us that
\cos(2x) = \cos^2 x - \sin^2 x \quad\text{and}\quad\sin(2x) = 2 \cos x \sin x,
which are the usual double-angle identities.
Similarly, since
\left(\cos x+i\sin x\right)^3 = \cos^3 x + 3i \cos^2 x \sin x - 3 \cos x \sin^2 x - i \sin^3 x,
De Moivre's formula yields
\cos(3x) = \cos^3 x - 3 \cos x \sin^2 x \quad\text{and}\quad \sin(3x) = 3\cos^2 x \sin x - \sin^3 x.
In general,
\cos(nx) = \sum_{k\text{ even}} (-1)^{k/2} {n \choose k}\cos^{n-k} x \sin^k x
and
\sin(nx) = \sum_{k\text{ odd}} (-1)^{(k-1)/2} {n \choose k}\cos^{n-k} x \sin^k x.
Series for ''e''
The number  is often defined by the formula
e = \lim_{n\to\infty} \left(1 + \frac{1}{n}\right)^n.
Applying the binomial theorem to this expression yields the usual infinite series for .
In particular:
\left(1 + \frac{1}{n}\right)^n = 1 + {n \choose 1}\frac{1}{n} + {n \choose 2}\frac{1}{n^2} + {n \choose 3}\frac{1}{n^3} + \cdots + {n \choose n}\frac{1}{n^n}.
The th term of this sum is
{n \choose k}\frac{1}{n^k} = \frac{1}{k!}\cdot\frac{n(n-1)(n-2)\cdots (n-k+1)}{n^k}
As , the rational expression on the right approaches , and therefore
\lim_{n\to\infty} {n \choose k}\frac{1}{n^k} = \frac{1}{k!}.
This indicates that  can be written as a series:
e=\sum_{k=0}^\infty\frac{1}{k!}=\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots.
Indeed, since each term of the binomial expansion is an increasing function of , it follows from the monotone convergence theorem for series that the sum of this infinite series is equal to .
Probability
The binomial theorem is closely related to the probability mass function of the negative binomial distribution.
The probability of a (countable) collection of independent Bernoulli trials \{X_t\}_{t\in S} with probability of success p\in [0,1] all not happening is
P\left(\bigcap_{t\in S} X_t^C\right) = (1-p)^{|S|} = \sum_{n=0}^{|S|} {|S| \choose n} (-p)^n.
A useful upper bound for this quantity is  e^{-p|S|}.
In abstract algebra
The binomial theorem is valid more generally for two elements  and  in a ring, or even a semiring, provided that .
For example, it holds for two  matrices, provided that those matrices commute; this is useful in computing powers of a matrix.Artin, Algebra, 2nd edition, Pearson, 2018, equation (4.7.11).
The binomial theorem can be stated by saying that the polynomial sequence   is of binomial type.
In popular culture
The binomial theorem is mentioned in the Major-General's Song in the comic opera The Pirates of Penzance.
Professor Moriarty is described by Sherlock Holmes as having written a treatise on the binomial theorem.
The Portuguese poet Fernando Pessoa, using the heteronym Álvaro de Campos, wrote that "Newton's Binomial is as beautiful as the Venus de Milo.
The truth is that few people notice it."
In the 2014 film The Imitation Game, Alan Turing makes reference to Isaac Newton's work on the binomial theorem during his first meeting with Commander Denniston at Bletchley Park.
See also
Binomial approximation
Binomial distribution
Binomial inverse theorem
Stirling's approximation
Tannery's theorem
Notes
References
Further reading
External links
Binomial Theorem by Stephen Wolfram, and "Binomial Theorem (Step-by-Step)" by Bruce Colletti and Jeff Bryant, Wolfram Demonstrations Project, 2007.
