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Binomial theorem

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Binomial theorem

The binomial coefficients appear as the entries of Pascal's triangle where each entry is the sum of the two above it.

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 power (x + y)n into a sum involving terms of the form a xbyc, 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,

(x+y)^4 \;=\; x^4 \,+\, 4 x^3y \,+\, 6 x^2 y^2 \,+\, 4 x y^3 \,+\, y^4.

The coefficient a in the term of a xbyc is known as the binomial coefficient \tbinom nb or \tbinom nc (the two have the same value). These coefficients for varying n and b can be arranged to form Pascal's triangle. These numbers also arise in combinatorics, where \tbinom nb gives the number of different combinations of b elements that can be chosen from an n-element set.


  • History 1
  • Statement of the theorem 2
  • Examples 3
    • Geometric explanation 3.1
  • The binomial coefficients 4
    • Formulae 4.1
    • Combinatorial interpretation 4.2
  • Proofs 5
    • Combinatorial proof 5.1
      • Example 5.1.1
      • General case 5.1.2
    • Inductive proof 5.2
  • Generalisations 6
    • Newton's generalised binomial theorem 6.1
      • Generalisations 6.1.1
    • The multinomial theorem 6.2
    • The multi-binomial theorem 6.3
  • Applications 7
    • Multiple-angle identities 7.1
    • Series for e 7.2
    • Derivative of the power function 7.3
    • Nth derivative of a product 7.4
  • The binomial theorem in abstract algebra 8
  • In popular culture 9
  • See also 10
  • Notes 11
  • References 12
  • Further reading 13
  • External links 14


Special cases of the binomial theorem were known from ancient times. The 4th century B.C. Greek mathematician Euclid mentioned the special case of the binomial theorem for exponent 2.[1][2] There is evidence that the binomial theorem for cubes was known by the 6th century in India.[1][2]

Binomial coefficients, as combinatorial quantities expressing the number of ways of selecting k objects out of n without replacement, were of interest to the ancient Hindus. The earliest known reference to this combinatorial problem is the Chandaḥśāstra by the Hindu lyricist Pingala (c. 200 B.C.), which contains a method for its solution.[3]:230 The commentator Halayudha from the 10th century A.D. explains this method using what is now known as Pascal's triangle.[3] By the 6th century A.D., the Hindu mathematicians probably knew how to express this as a quotient \frac{n!}{(n-k)!k!},[4] and a clear statement of this rule can be found in the 12th century text Lilavati by Bhaskara.[4]

The binomial theorem as such can be found in the work of 11th-century Arabian mathematician Al-Karaji, who described the triangular pattern of the binomial coefficients.[5] He also provided a mathematical proof of both the binomial theorem and Pascal's triangle, using a primitive form of mathematical induction.[5] The Persian poet and mathematician Omar Khayyam was probably familiar with the formula to higher orders, although many of his mathematical works are lost.[2] The binomial expansions of small degrees were known in the 13th century mathematical works of Yang Hui[6] and also Chu Shih-Chieh.[2] Yang Hui attributes the method to a much earlier 11th century text of Jia Xian, although those writings are now also lost.[3]:142

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".[7] Blaise Pascal studied the eponymous triangle comprehensively in the treatise Traité du triangle arithmétique (1653). However, the pattern of numbers was already known to the European mathematicians of the late Renaissance, including Stifel, Niccolò Fontana Tartaglia, and Simon Stevin.[7]

Isaac Newton is generally credited with the generalised binomial theorem, valid for any rational exponent.[8][7]

Statement of the theorem

According to the theorem, it is possible to expand any power of x + y 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 each \tbinom nk is a specific 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 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 x and y 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 1 for y, 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.


Pascal's triangle

The most basic example of the binomial theorem is the formula for the square of x + y:

(x + y)^2 = x^2 + 2xy + y^2.\!

The binomial coefficients 1, 2, 1 appearing in this expansion correspond to the second row of Pascal's triangle (Note that the top is row 0). The coefficients of higher powers of x + y correspond to later rows of the triangle:

\begin{align} \\[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. \end{align}

Notice that

  1. the powers of x go down until it reaches 0 (x^0=1), starting value is n (the n in (x+y)^n.)
  2. the powers of y go up from 0 (y^0=1) until it reaches n (also the n in (x+y)^n.)
  3. the nth row of the Pascal's Triangle will be the coefficients of the expanded binomial.
  4. for each line, the number of products (i.e. the sum of the coefficients) is equal to 2^n.
  5. for each line, the number of product groups is equal to n+1.

The binomial theorem can be applied to the powers of any binomial. For example,

\begin{align} (x+2)^3 &= x^3 + 3x^2(2) + 3x(2)^2 + 2^3 \\ &= x^3 + 6x^2 + 12x + 8.\end{align}

For a binomial involving subtraction, the theorem can be applied as long as the opposite of the second term is used. This has the effect of changing the sign of every other term in the expansion:

(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3.\!

Another useful example is that of the expansion of the following square roots:

\sqrt{1+x} = \textstyle 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
\frac{1}{\sqrt{1+x}} = \textstyle 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

Sometimes it may be useful to expand negative exponents when |x| < 1:

(1+x)^{-1} = \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \cdots

Geometric explanation

Visualisation of binomial expansion up to the 4th power

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.

In calculus, this picture also gives a geometric proof of the derivative (x^n)'=nx^{n-1}:[9] if one sets a=x and 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, (x+\Delta x)^n, where the coefficient of the linear term (in \Delta x) is nx^{n-1}, the area of the n faces, each of dimension (n-1):

(x+\Delta x)^n = x^n + nx^{n-1}\Delta x + \tbinom{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 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 \textstyle{\int x^{n-1}\,dx = \tfrac{1}{n} x^n} – see proof of Cavalieri's quadrature formula for details.[9]

The binomial coefficients

The coefficients that appear in the binomial expansion are called binomial coefficients. These are usually written \tbinom nk , and pronounced “n choose k”.


The coefficient of xnkyk is given by the formula

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

which is defined in terms of the factorial function n!. Equivalently, this formula can be written

{n \choose 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. Note that, although this formula involves a fraction, the binomial coefficient \tbinom nk is actually an integer.

Combinatorial interpretation

The binomial coefficient \tbinom nk 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


then, according to the distributive law, there will be one term in the expansion for each choice of either x or y 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.


Combinatorial proof


The coefficient of xy2 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 x,y strings of length 3 with exactly two y's, namely,

xyy, \; yxy, \; yyx,

corresponding to the three 2-element subsets of { 1, 2, 3 }, namely,


where each subset specifies the positions of the y in a corresponding string.

General case

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

  • the number of copies of xn − kyk 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}
  • {n \choose k} (this is either by definition, or by a short combinatorial argument if one is defining {n \choose k} as \frac{n!}{k!\,(n-k)!}).

This proves the binomial theorem.

Inductive proof

Induction yields another proof of the binomial theorem. When n = 0, both sides equal 1, since x0 = 1 and \tbinom{0}{0}=1. Now suppose that the equality holds for a given n; we will prove it for n + 1. For jk ≥ 0, let [ƒ(xy)] j,k denote the coefficient of xjyk in the polynomial ƒ(xy). By the inductive hypothesis, (x + y)n is a polynomial in x and y such that [(x + y)n] j,k is \tbinom{n}{k} if j + k = n, and 0 otherwise. The identity

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

shows that (x + y)n+1 also is a polynomial in x and y, and

[(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

\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k},

by Pascal's identity.[10] 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

(x+y)^{n+1} = \sum_{k=0}^{n+1} \tbinom{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.


Newton's generalised binomial theorem

Around 1665, Isaac Newton generalised the formula to allow real exponents other than nonnegative integers. In addition, the formula can be generalised to complex exponents. In this generalisation, 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 above formula with factorials; however factoring out (n − k)! from numerator and denominator in that formula, and replacing n by r which now stands 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. Then, if x and y are real numbers with |x| > |y|,[Notes 1] and r is any complex number, one has

\begin{align} (x+y)^r & =\sum_{k=0}^\infty {r \choose k} x^{r-k} y^k \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(2) \\ & = 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 r is a nonnegative integer, the binomial coefficients for k > r are zero, so (2) specializes to (1), and there are at most r + 1 nonzero terms. For other values of r, the series (2) has infinitely many nonzero terms, at least if x and y are nonzero.

This is important when one is working with infinite series and would like to represent them in terms of generalised hypergeometric functions.

Taking r = −s leads to a useful formula:

\frac{1}{(1-x)^s} = \sum_{k=0}^\infty {s+k-1 \choose k} x^k \equiv \sum_{k=0}^\infty {s+k-1 \choose s-1} x^k.

Further specializing to s = 1 yields the geometric series formula.


Formula (2) can be generalised to the case where x and y are complex numbers. For this version, one should assume |x| > |y|[Notes 1] and define the powers of x + y and x using a holomorphic branch of log defined on an open disk of radius |x| centered at x.

Formula (2) is valid also for elements x and y of a Banach algebra as long as xy = yx, x is invertible, and ||y/x|| < 1.

The multinomial theorem

The binomial theorem can be generalised 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} {n \choose 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 k1 through km such that the sum of all ki is n. (For each term in the expansion, the exponents must add up to n). The coefficients \tbinom n{k_1,\cdots,k_m} are known as multinomial coefficients, and can be computed by the formula

{n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, 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 n-element set into disjoint subsets of sizes k1, ..., km.

The multi-binomial theorem

It is often useful when working in more dimensions, 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}.


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 cos(nx) and sin(nx). 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


\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 e 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 e. 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 kth 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 n → ∞, the rational expression on the right approaches one, and therefore

\lim_{n\to\infty} {n \choose k}\frac{1}{n^k} = \frac{1}{k!}.

This indicates that e can be written as a series:

e = \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 n, it follows from the monotone convergence theorem for series that the sum of this infinite series is equal to e.

Derivative of the power function

In finding the derivative of the power function f(x) = xn for integer n using the definition of derivative, one must expand the binomial (x + h)n.

Nth derivative of a product

To indicate the formula for the derivative of order n of the product of two functions, the formula of the binomial theorem is used symbolically.[11]

The binomial theorem in abstract algebra

Formula (1) is valid more generally for any elements x and y of a semiring satisfying xy = yx. The theorem is true even more generally: alternativity suffices in place of associativity.

The binomial theorem can be stated by saying that the polynomial sequence { 1, xx2x3, ... } is of binomial type.

In popular culture

See also


  1. ^ a b This is to guarantee convergence. Depending on r, the series may also converge sometimes when |x| = |y|.


  1. ^ a b
  2. ^ a b c d
  3. ^ a b c
  4. ^ a b
  5. ^ a b .
  6. ^
  7. ^ a b c
  8. ^
  9. ^ a b
  10. ^ Binomial theorem - inductive proofs Archived February 24, 2015 at the Wayback Machine
  11. ^
  12. ^

Further reading

External links

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