The Binomial Theorem: Formulas - MywallpapersMobi

# 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 polynomial (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 (for n = 4),

$\displaystyle (x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4.$

(
x
+
y

)

4

=

x

4

+
4

x

3

y
+
6

x

2

y

2

+
4
x

y

3

+

y

4

.

\displaystyle (x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4.

The coefficient a in the term of a xbyc is known as the binomial coefficient

$\displaystyle \tbinom nb$

(

n
b

)

\displaystyle \tbinom nb

or

$\displaystyle \tbinom nc$

(

n
c

)

\displaystyle \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

$\displaystyle \tbinom nb$

(

n
b

)

\displaystyle \tbinom nb

gives the number of different combinations of b elements that can be chosen from an n-element set .

## Contents

• 1 History
• 2 Theorem statement
• 3 Examples
• 3.1 Geometric explanation
• 4 Binomial coefficients
• 4.1 Formulae
• 4.2 Combinatorial interpretation
• 5 Proofs
• 5.1 Combinatorial proof
• 5.1.1 Example
• 5.1.2 General case
• 5.2 Inductive proof
• 6 Generalizations
• 6.1 Newton’s generalized binomial theorem
• 6.2 Further generalizations
• 6.3 Multinomial theorem
• 6.4 Multi-binomial theorem
• 6.5 General Leibniz rule
• 7 Applications
• 7.1 Multiple-angle identities
• 7.2 Series for e
• 7.3 Probability
• 8 The binomial theorem in abstract algebra
• 9 In popular culture
• 11 Notes
• 12 References

## History[ edit ]

Special cases of the binomial theorem were known since at least the 4th century B.C. when 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 ancient Indian mathematicians. The earliest known reference to this combinatorial problem is the Chandaḥśāstra by the Indian 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 Indian mathematicians probably knew how to express this as a quotient

$\displaystyle \frac n!(n-k)!k!$

n
!

(
n

k
)
!
k
!

\displaystyle \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 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”. [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. [2] The binomial expansions of small degrees were known in the 13th century mathematical works of Yang Hui [8] 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

$\displaystyle (1+a)^n$

(
1
+
a

)

n

\displaystyle (1+a)^n

in terms of

$\displaystyle (1+a)^n-1$

(
1
+
a

)

n

1

\displaystyle (1+a)^n-1

, via “Pascal’s triangle”. [9] 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 . [9]

Isaac Newton is generally credited with the generalized binomial theorem, valid for any rational exponent. [9] [10]

## Theorem statement[ edit ]

According to the theorem, it is possible
to expand any power of x + y into a sum of the form

$\displaystyle (x+y)^n=n \choose 0x^ny^0+n \choose 1x^n-1y^1+n \choose 2x^n-2y^2+\cdots +n \choose n-1x^1y^n-1+n \choose nx^0y^n,$

(
x
+
y

)

n

=

(

n
0

)

x

n

y

0

+

(

n
1

)

x

n

1

y

1

+

(

n
2

)

x

n

2

y

2

+

+

(

n

n

1

)

x

1

y

n

1

+

(

n
n

)

x

0

y

n

,

\displaystyle (x+y)^n=n \choose 0x^ny^0+n \choose 1x^n-1y^1+n \choose 2x^n-2y^2+\cdots +n \choose n-1x^1y^n-1+n \choose nx^0y^n,

where each

$\displaystyle \tbinom nk$

(

n
k

)

\displaystyle \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

$\displaystyle \binom n0x^n+\ldots$

(

n
0

)

x

n

+

\displaystyle \binom n0x^n+\ldots

.) This formula is also referred to as the binomial formula or the binomial identity. Using summation notation , it can be written as

$\displaystyle (x+y)^n=\sum _k=0^nn \choose kx^n-ky^k=\sum _k=0^nn \choose kx^ky^n-k.$

(
x
+
y

)

n

=

k
=
0

n

(

n
k

)

x

n

k

y

k

=

k
=
0

n

(

n
k

)

x

k

y

n

k

.

\displaystyle (x+y)^n=\sum _k=0^nn \choose kx^n-ky^k=\sum _k=0^nn \choose kx^ky^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

$\displaystyle (1+x)^n=n \choose 0x^0+n \choose 1x^1+n \choose 2x^2+\cdots +n \choose n-1x^n-1+n \choose nx^n,$

(
1
+
x

)

n

=

(

n
0

)

x

0

+

(

n
1

)

x

1

+

(

n
2

)

x

2

+

+

(

n

n

1

)

x

n

1

+

(

n
n

)

x

n

,

\displaystyle (1+x)^n=n \choose 0x^0+n \choose 1x^1+n \choose 2x^2+\cdots +n \choose n-1x^n-1+n \choose nx^n,

or equivalently

$\displaystyle (1+x)^n=\sum _k=0^nn \choose kx^k.$

(
1
+
x

)

n

=

k
=
0

n

(

n
k

)

x

k

.

\displaystyle (1+x)^n=\sum _k=0^nn \choose kx^k.

## Examples[ edit ]

Pascal’s triangle

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

$\displaystyle (x+y)^2=x^2+2xy+y^2.$

(
x
+
y

)

2

=

x

2

+
2
x
y
+

y

2

.

\displaystyle (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 “1” of the triangle is considered to be row 0, by convention.) The coefficients of higher powers of x + y correspond to lower rows of the triangle:

\displaystyle \beginaligned(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.\endaligned

(
x
+
y

)

3

=

x

3

+
3

x

2

y
+
3
x

y

2

+

y

3

,

(
x
+
y

)

4

=

x

4

+
4

x

3

y
+
6

x

2

y

2

+
4
x

y

3

+

y

4

,

(
x
+
y

)

5

=

x

5

+
5

x

4

y
+
10

x

3

y

2

+
10

x

2

y

3

+
5
x

y

4

+

y

5

,

(
x
+
y

)

6

=

x

6

+
6

x

5

y
+
15

x

4

y

2

+
20

x

3

y

3

+
15

x

2

y

4

+
6
x

y

5

+

y

6

,

(
x
+
y

)

7

=

x

7

+
7

x

6

y
+
21

x

5

y

2

+
35

x

4

y

3

+
35

x

3

y

4

+
21

x

2

y

5

+
7
x

y

6

+

y

7

.

\displaystyle \beginaligned(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.\endaligned

Several patterns can be observed from these examples. In general, for the expansion (x + y)n:

1. the powers of x start at n and decrease by 1 in each term until they reach 0 (with x0 = 1, often unwritten);
2. the powers of y start at 0 and increase by 1 until they reach n;
3. the nth row of Pascal’s Triangle will be the coefficients of the expanded binomial when the terms are arranged in this way;
4. the number of terms in the expansion before like terms are combined is the sum of the coefficients and is equal to 2n; and
5. there will be n + 1 terms in the expression after combining like terms in the expansion.

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

\displaystyle \beginaligned(x+2)^3&=x^3+3x^2(2)+3x(2)^2+2^3\\&=x^3+6x^2+12x+8.\endaligned

(
x
+
2

)

3

=

x

3

+
3

x

2

(
2
)
+
3
x
(
2

)

2

+

2

3

=

x

3

+
6

x

2

+
12
x
+
8.

\displaystyle \beginaligned(x+2)^3&=x^3+3x^2(2)+3x(2)^2+2^3\\&=x^3+6x^2+12x+8.\endaligned

For a binomial involving subtraction, the theorem can be applied by using the form (xy)n = (x + (−y))n. This has the effect of changing the sign of every other term in the expansion:

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

(
x

y

)

3

=
(
x
+
(

y
)

)

3

=

x

3

+
3

x

2

(

y
)
+
3
x
(

y

)

2

+
(

y

)

3

=

x

3

3

x

2

y
+
3
x

y

2

y

3

.

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

### Geometric explanation[ edit ]

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

$\displaystyle (x^n)'=nx^n-1:$

(

x

n

)

=
n

x

n

1

:

\displaystyle (x^n)’=nx^n-1:

[11] if one sets

$\displaystyle a=x$

a
=
x

\displaystyle a=x

and

$\displaystyle b=\Delta x,$

b
=
Δ
x
,

\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 ,

$\displaystyle (x+\Delta x)^n,$

(
x
+
Δ
x

)

n

,

\displaystyle (x+\Delta x)^n,

where the coefficient of the linear term (in

$\displaystyle \Delta x$

Δ
x

\displaystyle \Delta x

) is

$\displaystyle nx^n-1,$

n

x

n

1

,

\displaystyle nx^n-1,

the area of the n faces, each of dimension

$\displaystyle (n-1):$

(
n

1
)
:

\displaystyle (n-1):

$\displaystyle (x+\Delta x)^n=x^n+nx^n-1\Delta x+\tbinom n2x^n-2(\Delta x)^2+\cdots .$

(
x
+
Δ
x

)

n

=

x

n

+
n

x

n

1

Δ
x
+

(

n
2

)

x

n

2

(
Δ
x

)

2

+

.

\displaystyle (x+\Delta x)^n=x^n+nx^n-1\Delta x+\tbinom n2x^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,

$\displaystyle (\Delta x)^2$

(
Δ
x

)

2

\displaystyle (\Delta x)^2

and higher, become negligible, and yields the formula

$\displaystyle (x^n)'=nx^n-1,$

(

x

n

)

=
n

x

n

1

,

\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

$\displaystyle (n-1)$

(
n

1
)

\displaystyle (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

$\displaystyle \textstyle \int x^n-1\,dx=\tfrac 1nx^n$

x

n

1

d
x
=

1
n

x

n

\displaystyle \textstyle \int x^n-1\,dx=\tfrac 1nx^n

– see proof of Cavalieri’s quadrature formula for details. [11]

## Binomial coefficients[ edit ]

Main article: Binomial coefficient

The coefficients that appear in the binomial expansion are called binomial coefficients. These are usually written

$\displaystyle \tbinom nk$

(

n
k

)

\displaystyle \tbinom nk

, and pronounced “n choose k”.

### Formulae[ edit ]

The coefficient of xnkyk is given by the formula

$\displaystyle n \choose k=\frac n!k!(n-k)!$

(

n
k

)

=

n
!

k
!
(
n

k
)
!

\displaystyle n \choose k=\frac n!k!(n-k)!

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

$\displaystyle n \choose k=\frac n(n-1)\cdots (n-k+1)k(k-1)\cdots 1=\prod _\ell =1^k\frac n-\ell +1\ell =\prod _\ell =0^k-1\frac n-\ell k-\ell$

(

n
k

)

=

n
(
n

1
)

(
n

k
+
1
)

k
(
k

1
)

1

=

=
1

k

n

+
1

=

=
0

k

1

n

k

\displaystyle 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

$\displaystyle \tbinom nk$

(

n
k

)

\displaystyle \tbinom nk

is actually an integer .

### Combinatorial interpretation[ edit ]

The binomial coefficient

$\displaystyle \tbinom nk$

(

n
k

)

\displaystyle \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

$\displaystyle (x+y)(x+y)(x+y)\cdots (x+y),$

(
x
+
y
)
(
x
+
y
)
(
x
+
y
)

(
x
+
y
)
,

\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 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.

## Proofs[ edit ]

### Combinatorial proof[ edit ]

#### Example[ edit ]

The coefficient of xy2 in

\displaystyle \beginaligned(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.\endaligned

(
x
+
y

)

3

=
(
x
+
y
)
(
x
+
y
)
(
x
+
y
)

=
x
x
x
+
x
x
y
+
x
y
x
+

x
y
y

_

+
y
x
x
+

y
x
y

_

+

y
y
x

_

+
y
y
y

=

x

3

+
3

x

2

y
+

3
x

y

2

_

+

y

3

.

\displaystyle \beginaligned(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.\endaligned

equals

$\displaystyle \tbinom 32=3$

(

3
2

)

=
3

\displaystyle \tbinom 32=3

because there are three x,y strings of length 3 with exactly two y’s, namely,

$\displaystyle xyy,\;yxy,\;yyx,$

x
y
y
,

y
x
y
,

y
y
x
,

\displaystyle xyy,\;yxy,\;yyx,

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

$\displaystyle \2,3\,\;\1,3\,\;\1,2\,$

2
,
3

,

1
,
3

,

1
,
2

,

\displaystyle \2,3\,\;\1,3\,\;\1,2\,

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

#### General case[ edit ]

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

• $\displaystyle n \choose k$

(

n
k

)

\displaystyle n \choose k

(this is either by definition, or by a short combinatorial argument if one is defining

$\displaystyle n \choose k$

(

n
k

)

\displaystyle n \choose k

as

$\displaystyle \frac n!k!(n-k)!$

n
!

k
!
(
n

k
)
!

\displaystyle \frac n!k!(n-k)!

).

This proves the binomial theorem.

### Inductive proof[ edit ]

Induction yields another proof of the binomial theorem. When n = 0, both sides equal 1, since x0 = 1 and

$\displaystyle \tbinom 00=1$

(

0
0

)

=
1

\displaystyle \tbinom 00=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

$\displaystyle \tbinom nk$

(

n
k

)

\displaystyle \tbinom nk

if j + k = n, and 0 otherwise.
The identity

$\displaystyle (x+y)^n+1=x(x+y)^n+y(x+y)^n$

(
x
+
y

)

n
+
1

=
x
(
x
+
y

)

n

+
y
(
x
+
y

)

n

\displaystyle (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

$\displaystyle [(x+y)^n+1]_j,k=[(x+y)^n]_j-1,k+[(x+y)^n]_j,k-1,$

[
(
x
+
y

)

n
+
1

]

j
,
k

=
[
(
x
+
y

)

n

]

j

1
,
k

+
[
(
x
+
y

)

n

]

j
,
k

1

,

\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

$\displaystyle \binom nk+\binom nk-1=\binom n+1k,$

(

n
k

)

+

(

n

k

1

)

=

(

n
+
1

k

)

,

\displaystyle \binom nk+\binom nk-1=\binom n+1k,

by Pascal’s identity . [12] 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

$\displaystyle (x+y)^n+1=\sum _k=0^n+1\tbinom n+1kx^n+1-ky^k,$

(
x
+
y

)

n
+
1

=

k
=
0

n
+
1

(

n
+
1

k

)

x

n
+
1

k

y

k

,

\displaystyle (x+y)^n+1=\sum _k=0^n+1\tbinom n+1kx^n+1-ky^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 ]

Main article: Binomial series

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

$\displaystyle r \choose k=\frac r\,(r-1)\cdots (r-k+1)k!=\frac (r)_kk!,$

(

r
k

)

=

r

(
r

1
)

(
r

k
+
1
)

k
!

=

(
r

)

k

k
!

,

\displaystyle r \choose k=\frac r\,(r-1)\cdots (r-k+1)k!=\frac (r)_kk!,

where

$\displaystyle (\cdot )_k$

(

)

k

\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 \beginaligned(x+y)^r&=\sum _k=0^\infty r \choose kx^r-ky^k\\&=x^r+rx^r-1y+\frac r(r-1)2!x^r-2y^2+\frac r(r-1)(r-2)3!x^r-3y^3+\cdots .\endaligned

(
x
+
y

)

r

=

k
=
0

(

r
k

)

x

r

k

y

k

=

x

r

+
r

x

r

1

y
+

r
(
r

1
)

2
!

x

r

2

y

2

+

r
(
r

1
)
(
r

2
)

3
!

x

r

3

y

3

+

.

\displaystyle \beginaligned(x+y)^r&=\sum _k=0^\infty r \choose kx^r-ky^k\\&=x^r+rx^r-1y+\frac r(r-1)2!x^r-2y^2+\frac r(r-1)(r-2)3!x^r-3y^3+\cdots .\endaligned

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:

$\displaystyle \sqrt 1+x=\textstyle 1+\frac 12x-\frac 18x^2+\frac 116x^3-\frac 5128x^4+\frac 7256x^5-\cdots$

1
+
x

=

1
+

1
2

x

1
8

x

2

+

1
16

x

3

5
128

x

4

+

7
256

x

5

\displaystyle \sqrt 1+x=\textstyle 1+\frac 12x-\frac 18x^2+\frac 116x^3-\frac 5128x^4+\frac 7256x^5-\cdots

Taking

$\displaystyle r=-1$

r
=

1

\displaystyle r=-1

, the generalized binomial series gives the geometric series formula , valid for

$\displaystyle$

|

x

|

<
1

x

:

$\displaystyle (1+x)^-1=\frac 11+x=1-x+x^2-x^3+x^4-x^5+\cdots$

(
1
+
x

)

1

=

1

1
+
x

=
1

x
+

x

2

x

3

+

x

4

x

5

+

\displaystyle (1+x)^-1=\frac 11+x=1-x+x^2-x^3+x^4-x^5+\cdots

More generally, with r = −s:

$\displaystyle \frac 1(1-x)^s=\sum _k=0^\infty s+k-1 \choose kx^k\equiv \sum _k=0^\infty s+k-1 \choose s-1x^k.$

1

(
1

x

)

s

=

k
=
0

(

s
+
k

1

k

)

x

k

k
=
0

(

s
+
k

1

s

1

)

x

k

.

\displaystyle \frac 1(1-x)^s=\sum _k=0^\infty s+k-1 \choose kx^k\equiv \sum _k=0^\infty s+k-1 \choose s-1x^k.

So, for instance, when

$\displaystyle s=1/2$

s
=
1

/

2

\displaystyle s=1/2

,

$\displaystyle \frac 1\sqrt 1+x=\textstyle 1-\frac 12x+\frac 38x^2-\frac 516x^3+\frac 35128x^4-\frac 63256x^5+\cdots$

1

1
+
x

=

1

1
2

x
+

3
8

x

2

5
16

x

3

+

35
128

x

4

63
256

x

5

+

\displaystyle \frac 1\sqrt 1+x=\textstyle 1-\frac 12x+\frac 38x^2-\frac 516x^3+\frac 35128x^4-\frac 63256x^5+\cdots

### Further generalizations[ edit ]

The generalized binomial theorem can be extended to the case where x and y are complex numbers. For this version, one should again assume |x| > |y| [Note 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.
The generalized binomial theorem is valid also for elements x and y of a Banach algebra as long as xy = yx, x is invertible, and ||y/x|| < 1.

A version of the binomial theorem is valid for the following Pochhammer symbol -like family of polynomials: for a given real constant c, define

$\displaystyle x^(0)=1$

x

(
0
)

=
1

\displaystyle x^(0)=1

and

$\displaystyle x^(n)=\prod _k=1^n[x+(k-1)c]$

x

(
n
)

=

k
=
1

n

[
x
+
(
k

1
)
c
]

\displaystyle x^(n)=\prod _k=1^n[x+(k-1)c]

for

$\displaystyle n>0$

n
>
0

\displaystyle n>0

. Then [13]

$\displaystyle (a+b)^(n)=\sum _k=0^n\binom nka^(n-k)b^(k).$

(
a
+
b

)

(
n
)

=

k
=
0

n

(

n
k

)

a

(
n

k
)

b

(
k
)

.

\displaystyle (a+b)^(n)=\sum _k=0^n\binom nka^(n-k)b^(k).

The case c = 0 recovers the usual binomial theorem.

More generally, a sequence

$\displaystyle \p_n\_n=0^\infty$

p

n

n
=
0

\displaystyle \p_n\_n=0^\infty

of polynomials is said to be binomial if

• $\displaystyle \deg p_n=n$

deg

p

n

=
n

\displaystyle \deg p_n=n

for all

$\displaystyle n$

n

\displaystyle n

,

• $\displaystyle p_0(0)=1$

p

0

(
0
)
=
1

\displaystyle p_0(0)=1

, and

• $\displaystyle p_n(x+y)=\sum _k=0^n\binom nkp_k(x)p_n-k(y)$

p

n

(
x
+
y
)
=

k
=
0

n

(

n
k

)

p

k

(
x
)

p

n

k

(
y
)

\displaystyle p_n(x+y)=\sum _k=0^n\binom nkp_k(x)p_n-k(y)

for all

$\displaystyle x$

x

\displaystyle x

,

$\displaystyle y$

y

\displaystyle y

, and

$\displaystyle n$

n

\displaystyle n

.

An operator

$\displaystyle Q$

Q

\displaystyle Q

on the space of polynomials is said to be the basis operator of the sequence

$\displaystyle \p_n\_n=0^\infty$

p

n

n
=
0

\displaystyle \p_n\_n=0^\infty

if

$\displaystyle Qp_0=0$

Q

p

0

=
0

\displaystyle Qp_0=0

and

$\displaystyle Qp_n=np_n-1$

Q

p

n

=
n

p

n

1

\displaystyle Qp_n=np_n-1

for all

$\displaystyle n\geqslant 1$

n

1

\displaystyle n\geqslant 1

. A sequence

$\displaystyle \p_n\_n=0^\infty$

p

n

n
=
0

\displaystyle \p_n\_n=0^\infty

is binomial if and only if its basis operator is a Delta operator . [14] Writing

$\displaystyle E^a$

E

a

\displaystyle E^a

for the shift by

$\displaystyle a$

a

\displaystyle a

operator, the Delta operators corresponding to the above “Pochhammer” families of polynomials are the backward difference

$\displaystyle I-E^-c$

I

E

c

\displaystyle I-E^-c

for

$\displaystyle c>0$

c
>
0

\displaystyle c>0

, the ordinary derivative for

$\displaystyle c=0$

c
=
0

\displaystyle c=0

, and the forward difference

$\displaystyle E^-c-I$

E

c

I

\displaystyle E^-c-I

for

$\displaystyle c<0$

c
<
0

\displaystyle c<0

.

### Multinomial theorem[ edit ]

Main article: Multinomial theorem

The binomial theorem can be generalized to include powers of sums with more than two terms. The general version is

$\displaystyle (x_1+x_2+\cdots +x_m)^n=\sum _k_1+k_2+\cdots +k_m=nn \choose k_1,k_2,\ldots ,k_mx_1^k_1x_2^k_2\cdots x_m^k_m.$

(

x

1

+

x

2

+

+

x

m

)

n

=

k

1

+

k

2

+

+

k

m

=
n

(

n

k

1

,

k

2

,

,

k

m

)

x

1

k

1

x

2

k

2

x

m

k

m

.

\displaystyle (x_1+x_2+\cdots +x_m)^n=\sum _k_1+k_2+\cdots +k_m=nn \choose k_1,k_2,\ldots ,k_mx_1^k_1x_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

$\displaystyle \tbinom nk_1,\cdots ,k_m$

(

n

k

1

,

,

k

m

)

\displaystyle \tbinom nk_1,\cdots ,k_m

are known as multinomial coefficients, and can be computed by the formula

$\displaystyle n \choose k_1,k_2,\ldots ,k_m=\frac n!k_1!\cdot k_2!\cdots k_m!.$

(

n

k

1

,

k

2

,

,

k

m

)

=

n
!

k

1

!

k

2

!

k

m

!

.

\displaystyle n \choose k_1,k_2,\ldots ,k_m=\frac n!k_1!\cdot k_2!\cdots k_m!.

Combinatorially, the multinomial coefficient

$\displaystyle \tbinom nk_1,\cdots ,k_m$

(

n

k

1

,

,

k

m

)

\displaystyle \tbinom nk_1,\cdots ,k_m

counts the number of different ways to partition an n-element set into disjoint subsets of sizes k1, …, km.

### Multi-binomial theorem[ edit ]

It is often useful when working in more dimensions, to deal with products of binomial expressions. By the binomial theorem this is equal to

$\displaystyle (x_1+y_1)^n_1\dotsm (x_d+y_d)^n_d=\sum _k_1=0^n_1\dotsm \sum _k_d=0^n_d\binom n_1k_1\,x_1^k_1y_1^n_1-k_1\;\dotsc \;\binom n_dk_d\,x_d^k_dy_d^n_d-k_d.$

(

x

1

+

y

1

)

n

1

(

x

d

+

y

d

)

n

d

=

k

1

=
0

n

1

k

d

=
0

n

d

(

n

1

k

1

)

x

1

k

1

y

1

n

1

k

1

(

n

d

k

d

)

x

d

k

d

y

d

n

d

k

d

.

\displaystyle (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_1k_1\,x_1^k_1y_1^n_1-k_1\;\dotsc \;\binom n_dk_d\,x_d^k_dy_d^n_d-k_d.

This may be written more concisely, by multi-index notation , as

$\displaystyle (x+y)^\alpha =\sum _\nu \leq \alpha \binom \alpha \nu x^\nu y^\alpha -\nu .$

(
x
+
y

)

α

=

ν

α

(

α
ν

)

x

ν

y

α

ν

.

\displaystyle (x+y)^\alpha =\sum _\nu \leq \alpha \binom \alpha \nu x^\nu y^\alpha -\nu .

### General Leibniz rule[ edit ]

Main article: General Leibniz rule

The general Leibniz rule gives the nth derivative of a product of two functions in a form similar to that of the binomial theorem: [15]

$\displaystyle (fg)^(n)(x)=\sum _k=0^n\binom nkf^(n-k)(x)g^(k)(x).$

(
f
g

)

(
n
)

(
x
)
=

k
=
0

n

(

n
k

)

f

(
n

k
)

(
x
)

g

(
k
)

(
x
)
.

\displaystyle (fg)^(n)(x)=\sum _k=0^n\binom nkf^(n-k)(x)g^(k)(x).

Here, the superscript (n) indicates the nth derivative of a function. If one sets f(x) = eax and g(x) = ebx, and then cancels the common factor of e(a + b)x from both sides of the result, the ordinary binomial theorem is recovered.

## Applications[ edit ]

### Multiple-angle identities[ edit ]

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,

$\displaystyle \cos \left(nx\right)+i\sin \left(nx\right)=\left(\cos x+i\sin x\right)^n.$

cos

(

n
x

)

+
i
sin

(

n
x

)

=

(

cos

x
+
i
sin

x

)

n

.

\displaystyle \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

$\displaystyle \left(\cos x+i\sin x\right)^2=\cos ^2x+2i\cos x\sin x-\sin ^2x,$

(

cos

x
+
i
sin

x

)

2

=

cos

2

x
+
2
i
cos

x
sin

x

sin

2

x
,

\displaystyle \left(\cos x+i\sin x\right)^2=\cos ^2x+2i\cos x\sin x-\sin ^2x,

De Moivre’s formula tells us that

$\displaystyle \cos(2x)=\cos ^2x-\sin ^2x\quad \textand\quad \sin(2x)=2\cos x\sin x,$

cos

(
2
x
)
=

cos

2

x

sin

2

x

and

sin

(
2
x
)
=
2
cos

x
sin

x
,

which are the usual double-angle identities. Similarly, since

$\displaystyle \left(\cos x+i\sin x\right)^3=\cos ^3x+3i\cos ^2x\sin x-3\cos x\sin ^2x-i\sin ^3x,$

(

cos

x
+
i
sin

x

)

3

=

cos

3

x
+
3
i

cos

2

x
sin

x

3
cos

x

sin

2

x

i

sin

3

x
,

\displaystyle \left(\cos x+i\sin x\right)^3=\cos ^3x+3i\cos ^2x\sin x-3\cos x\sin ^2x-i\sin ^3x,

De Moivre’s formula yields

$\displaystyle \cos(3x)=\cos ^3x-3\cos x\sin ^2x\quad \textand\quad \sin(3x)=3\cos ^2x\sin x-\sin ^3x.$

cos

(
3
x
)
=

cos

3

x

3
cos

x

sin

2

x

and

sin

(
3
x
)
=
3

cos

2

x
sin

x

sin

3

x
.

In general,

$\displaystyle \cos(nx)=\sum _k\text even(-1)^k/2n \choose k\cos ^n-kx\sin ^kx$

cos

(
n
x
)
=

k

even

(

1

)

k

/

2

(

n
k

)

cos

n

k

x

sin

k

x

\displaystyle \cos(nx)=\sum _k\text even(-1)^k/2n \choose k\cos ^n-kx\sin ^kx

and

$\displaystyle \sin(nx)=\sum _k\text odd(-1)^(k-1)/2n \choose k\cos ^n-kx\sin ^kx.$

sin

(
n
x
)
=

k

odd

(

1

)

(
k

1
)

/

2

(

n
k

)

cos

n

k

x

sin

k

x
.

\displaystyle \sin(nx)=\sum _k\text odd(-1)^(k-1)/2n \choose k\cos ^n-kx\sin ^kx.

### Series for e[ edit ]

The number e is often defined by the formula

$\displaystyle e=\lim _n\to \infty \left(1+\frac 1n\right)^n.$

e
=

lim

n

(

1
+

1
n

)

n

.

\displaystyle e=\lim _n\to \infty \left(1+\frac 1n\right)^n.

Applying the binomial theorem to this expression yields the usual infinite series for e. In particular:

$\displaystyle \left(1+\frac 1n\right)^n=1+n \choose 1\frac 1n+n \choose 2\frac 1n^2+n \choose 3\frac 1n^3+\cdots +n \choose n\frac 1n^n.$

(

1
+

1
n

)

n

=
1
+

(

n
1

)

1
n

+

(

n
2

)

1

n

2

+

(

n
3

)

1

n

3

+

+

(

n
n

)

1

n

n

.

\displaystyle \left(1+\frac 1n\right)^n=1+n \choose 1\frac 1n+n \choose 2\frac 1n^2+n \choose 3\frac 1n^3+\cdots +n \choose n\frac 1n^n.

The kth term of this sum is

$\displaystyle n \choose k\frac 1n^k=\frac 1k!\cdot \frac n(n-1)(n-2)\cdots (n-k+1)n^k$

(

n
k

)

1

n

k

=

1

k
!

n
(
n

1
)
(
n

2
)

(
n

k
+
1
)

n

k

\displaystyle n \choose k\frac 1n^k=\frac 1k!\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

$\displaystyle \lim _n\to \infty n \choose k\frac 1n^k=\frac 1k!.$

lim

n

(

n
k

)

1

n

k

=

1

k
!

.

\displaystyle \lim _n\to \infty n \choose k\frac 1n^k=\frac 1k!.

This indicates that e can be written as a series:

$\displaystyle e=\sum _k=0^\infty \frac 1k!=\frac 10!+\frac 11!+\frac 12!+\frac 13!+\cdots .$

e
=

k
=
0

1

k
!

=

1

0
!

+

1

1
!

+

1

2
!

+

1

3
!

+

.

\displaystyle e=\sum _k=0^\infty \frac 1k!=\frac 10!+\frac 11!+\frac 12!+\frac 13!+\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.

### Probability[ edit ]

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

$\displaystyle \X_t\_t\in S$

X

t

t

S

\displaystyle \X_t\_t\in S

with probability of success

$\displaystyle p\in [0,1]$

p

[
0
,
1
]

\displaystyle p\in [0,1]

all not happening is

$\displaystyle P\left(\bigcap _t\in SX_t^C\right)=(1-p)^=\sum _n=0^S(-p)^n$

P

(

t

S

X

t

C

)

=
(
1

p

)

|

S

|

=

n
=
0

|

S

|

(

|

S

|

n

)

(

p

)

n

\displaystyle P\left(\bigcap _t\in SX_t^C\right)=(1-p)^=\sum _n=0^ \choose n(-p)^n

A useful upper bound for this quantity is

$\displaystyle e^-pn$

e

p
n

\displaystyle e^-pn

. [16]

## The binomial theorem in abstract algebra[ edit ]

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[ edit ]

• 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.” [17]
• 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.

• Mathematics portal
• Binomial approximation
• Binomial distribution
• Binomial inverse theorem
• Stirling’s approximation

## Notes[ edit ]

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

## References[ edit ]

1. ^ a b Weisstein, Eric W. “Binomial Theorem” . Wolfram MathWorld.

2. ^ a b c d Coolidge, J. L. (1949). “The Story of the Binomial Theorem”. The American Mathematical Monthly. 56 (3): 147–157. doi : 10.2307/2305028 . JSTOR   2305028 .
3. ^ a b c Jean-Claude Martzloff; S.S. Wilson; J. Gernet; J. Dhombres (1987). A history of Chinese mathematics. Springer.
4. ^ a b Biggs, N. L. (1979). “The roots of combinatorics”. Historia Math. 6 (2): 109–136. doi : 10.1016/0315-0860(79)90074-0 .
5. ^ “Taming the unknown. A history of algebra from antiquity to the early ttwentieth century” (PDF). Bulletin of the American Mathematical Society: 727. However, algebra advanced in other respects. Around 1000, al-Karaji stated the binomial theorem
6. ^ Rashed, R. (1994-06-30). The Development of Arabic Mathematics: Between Arithmetic and Algebra . Springer Science & Business Media. p. 63. ISBN   9780792325659 .
7. ^ a b O’Connor, John J. ; Robertson, Edmund F. , “Abu Bekr ibn Muhammad ibn al-Husayn Al-Karaji” , MacTutor History of Mathematics archive , University of St Andrews .
8. ^ Landau, James A. (1999-05-08). “Historia Matematica Mailing List Archive: Re: [HM] Pascal’s Triangle” (mailing list email). Archives of Historia Matematica. Retrieved 2007-04-13.
9. ^ a b c Kline, Morris (1972). History of mathematical thought. Oxford University Press. p. 273.
10. ^ Bourbaki, N. (18 November 1998). Elements of the History of Mathematics Paperback. J. Meldrum (Translator). ISBN   978-3-540-64767-6 .
11. ^ a b Barth, Nils R. (2004). “Computing Cavalieri’s Quadrature Formula by a Symmetry of the n-Cube”. The American Mathematical Monthly. 111 (9): 811–813. doi : 10.2307/4145193 . ISSN   0002-9890 . JSTOR   4145193 , author’s copy , further remarks and resources
12. ^ Binomial theorem – inductive proofs Archived February 24, 2015, at the Wayback Machine .
13. ^ Sokolowsky, Dan; Rennie, Basil C. (February 1979). “Problem 352” (PDF). Crux Mathematicorum. 5 (2): 55–56.
14. ^ Aigner, Martin (1997) [Reprint of the 1979 Edition]. Combinatorial Theory. Springer. p. 105. ISBN   3-540-61787-6 .
15. ^ Seely, Robert T. (1973). Calculus of One and Several Variables. Glenview: Scott, Foresman. ISBN   978-0-673-07779-0 .
16. ^ Cover, Thomas M.; Thomas, Joy A. (2001-01-01). Data Compression. John Wiley & Sons, Inc. p. 320. doi : 10.1002/0471200611.ch5 . ISBN   9780471200611 .
17. ^ “Arquivo Pessoa: Obra Édita – O binómio de Newton é tão belo como a Vénus de Milo” . arquivopessoa.net.

• Bag, Amulya Kumar (1966). “Binomial theorem in ancient India”. Indian J. History Sci. 1 (1): 68–74.
• Graham, Ronald; Knuth, Donald; Patashnik, Oren (1994). “(5) Binomial Coefficients”. Concrete Mathematics (2nd ed.). Addison Wesley. pp. 153–256. ISBN   978-0-201-55802-9 . OCLC   17649857 .

 The Wikibook Combinatorics has a page on the topic of: The Binomial Theorem
• Solomentsev, E.D. (2001) [1994], “Newton binomial” , in Hazewinkel, Michiel , Encyclopedia of Mathematics , Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN   978-1-55608-010-4
• Binomial Theorem by Stephen Wolfram , and “Binomial Theorem (Step-by-Step)” by Bruce Colletti and Jeff Bryant, Wolfram Demonstrations Project , 2007.

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The
Binomial Theorem: Formulas
(page
1 of 2)

Sections: The formulas,
Worked
examples

The Binomial Theorem is
a quick way (okay, it’s a less slow way) of expanding (or multiplying
out) a binomial expression that has been raised to some (generally inconveniently
large) power. For instance, the expression (3x
� 2)10
would
be very painful to multiply out by hand. Thankfully, somebody figured
out a formula for this expansion, and we can plug the binomial
3x
� 2
and the power 10
into that formula to get that expanded (multiplied-out) form.

The formal expression of
the Binomial Theorem is as follows:

Yeah, I know; that formula
never helped me much, either. And it doesn’t help that different texts
use different notations to mean the same thing. The parenthetical bit
above has these equivalents:

Recall that the factorial
notation
"n!"
means " the product of all the whole numbers between
1
and
n",
so, for instance,
6!
= 1�2�3�4�5�6
. Then
the notation
"10C7"
(often pronounced as "ten, choose seven") means:

 Many calculators can evaluate this "n choose m" notation for you. Just look for a key that looks like "nCm" or "nCr", or for a similar item on the "Prob" or "Math" menu, or check your owner’s manual under "probability" or "combinations". The evaluation will probably look something like this:

 There is another way to find the value of "nCr", and it’s called "Pascal’s Triangle". To make the triangle, you start with a pyramid of three 1‘s, like this: Then you get the next row of numbers by adding the pairs of numbers from above. (Where there is only one number above, you just carry down the 1.)
 Keep going, always adding pairs of numbers from the previous row..  To find, say, 6C4, you go down to the row where there is a "6" after the initial "1", and then go over to the 5th (not the 4th) entry, to find that 6C4 = 15.

As you might imagine, drawing
Pascal’s Triangle every time you have to expand a binomial would be a
rather long process, especially if the binomial has a large exponent on
it. People have done
a
lot of studies
on
Pascal’s Triangle, but in practical terms, it’s probably best to just
nCr,
rather than using the Triangle. The Triangle is cute, I suppose, but it’s
not terribly helpful in this context, being more time-consuming than anything
else. For instance, on a test, do you want to evaluate
"10C7"
by calculating eleven rows of the Triangle, or by pushing four buttons

I could never remember
the formula for the Binomial Theorem, so instead, I just learned
how it worked. I noticed that the powers on each term in the expansion
always added up to whatever n
was, and that the terms counted up from zero to
n.
Returning to our intial example of
(3x
� 2)10
, the
powers on every term of the expansion will add up to
10,
and the powers on the terms will increment by counting up from zero to
10:

(3x
2)10 = 10C
0
(3x)10�
0(�2)0
+ 10C
1
(3x)10�
1(�2)1
+ 10C
2
(3x)10�
2(�2)2

+ 10C3
(3x)10�
3(�2)3
+ 10C
4
(3x)10�
4(�2)4
+ 10C
5
(3x)10�
5(�2)5

+ 10C6
(3x)10�
6(�2)6
+ 10C
7
(3x)10�
7(�2)7
+ 10C
8
(3x)10�
8(�2)8

+ 10C9
(3x)10�
9(�2)9
+ 10C
10
(3x)10�
10(�2)10

Note how the highlighted
counter number counts up from zero to 10,
with the factors on the ends of each term having the counter number, and
the factor in the middle having the counter number subtracted from
10.
This pattern is all you really need to know about the Binomial Theorem;
this pattern is how it works.

a binomial to expand, should be to plug it into the Theorem, just like
I did above. Don’t try to do too many steps at once. Only after you’ve
set up your binomial in the Theorem’s pattern should you start to simplify
the terms. The Binomial Theorem works best as a "plug-n-chug"
process, but you should plug in first; chug later. I’ve done my "plugging"
above; now "chugging" gives me:

(1)(59049)x10(1)
+ (10)(19683)x9(�2) + (45)(6561)x8(4)
+ (120)(2187)x7(�8)

+ (210)(729)x6(16)
+ (252)(243)x5(�32) + (210)(81)x4(64)

+ (120)(27)x3(�128)
+ (45)(9)x2(256) + (10)(3)x(�512) + (1)(1)(1)(1024)

= 59049x10
� 393660x9 + 1180980x8 � 2099520x7
+ 2449440x6 � 1959552x5

+ 1088640x4
� 414720x3 + 103680x2 � 15360x
+ 1024

As painful as the Binomial-Theorem
process is, it’s still easier than trying to multiply this stuff out by
hand. So don’t let the Formula put you off. It’s just another thing to
memorize, so memorize it, at least for the next test. The biggest source
of errors in the Binomial Theorem (other than forgetting the Theorem)
is the simplification process. Don’t try to do it in your head, or try
to do too many steps at once. Write things out nice and clearly, as I
did above, so you have a better chance of getting the right answer. (And
it would be good to do a bunch of practice problems, so the process is
fairly automatic by the time you hit the next test.)

Top
|  1 | 2  |   Return
to Index   Next >>

 Cite this article as: Stapel, Elizabeth. "The Binomial Theorem: Formulas." Purplemath. Available from     https://www.purplemath.com/modules/binomial.htm. Accessed [Date] [Month] 2016

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## Video: How to Use the Binomial Theorem to Expand a Binomial

In this video lesson, you will see what the binomial theorem has in common with Pascal’s triangle. Learn how you can use Pascal’s triangle to help you to easily expand a binomial.

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# How to Use the Binomial Theorem to Expand a Binomial

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Lesson Transcript

Instructor:
Yuanxin (Amy) Yang Alcocer

Amy has a master’s degree in secondary education and has taught math at a public charter high school.

In this video lesson, you will see what the binomial theorem has in common with Pascal’s triangle. Learn how you can use Pascal’s triangle to help you to easily expand a binomial.

## The Binomial Theorem

In math, we usually have formulas we can work from. Sometimes, these formulas can be somewhat complicated. But many times, these formulas have patterns to them. If we can spot the patterns, we can find an easier way to use the formula. This is what we are going to do in this video. We will learn the patterns in the binomial theorem and how we can use it to expand our binomials that are multiplied with themselves numerous times.

Our binomial theorem is a theorem that tells us what happens when you multiply a binomial by itself numerous times. Recall that a binomial is a polynomial made up of two terms. To keep things simple for us, though, we will label our binomial as (a + b) where a and b are your two terms in the binomial.

Think of the number of wheels a bicycle has and you will be able to remember the number of terms a binomial has. The formula that is connected with the Binomial Theorem is this one. We have a plus b to the nth power is equal to the summation of n choose k times a to the n minus k power times b to the kth power from k equals 0 to k equals n.

Yes, I know, it looks complicated. But it does produce patterns that you can easily remember.

## Understanding the Formula

Let’s dissect this formula so that we can understand it better. The first part, a plus b to the nth power, we understand to be the type of problem we can use our formula with. For example, we can use our formula with problems such as (x + y)^3 and (a + b)^5 because they follow the form of the first part of the formula.

The next part of the formula, the part after the equal sign, tells us our answer for these types of problems. The big symbol in front is called the summation symbol. It tells you to sum up the part of the formula that is to the right of it starting from k = 0 and going until k = n. We will usually see a k and/or an n in the formula. For each k = 0, 1, 2, etc., until n, we will plug in our values and then add up all the terms together to find our final answer. I will show you an example shortly so that you can better understand it.

The part of the formula that is inside the parentheses is notation for a factorial problem. We read it as n choose k. It is short for n!/(k!(nk)!). So, if n = 4 and k = 2, this part would equal 4 times 3 times 2 times 1 in the numerator and 2 factorial times 4 minus 2 factorial, which is 2 factorial times 2 factorial in the denominator. So, the denominator is 2 times 1 times 2 times 1. Multiplying all these out, we find that it equals 6.

Now, let’s see an example of how this binomial theorem formula works. We will follow the formula to expand (a + b)^3.

Looks long doesn’t it? If we do our subtractions and our factorial calculations, we end up with this:

Hey, our final answer doesn’t look so bad. Can we spot any patterns here? Yes, we can. Look at the exponents of our a and our b. Notice that as we go to the right, the exponent of our a starts at our n and decreases until we get to 0. The exponent of our b, on the other hand, starts at 0 and increases until it gets to n.

In our case, our n is 3, so our a exponent started with 3 and went down to 0, and our b exponent started at 0 and went up to 3. Remember that when the exponent is 0, it equals 1. So b^0 equals 1. And when it equals 1, we won’t write it since there are other things in the term that we write. So, what other patterns can we spot?

## Pascal’s Triangle

This is where Pascal’s triangle comes in. Pascal’s triangle is a triangle of numbers where the first and last terms of a row are 1 and all the other numbers are the sum of the two numbers directly above it. Pascal’s Triangle always begins with 1 at the tip and two 1’s underneath, forming a triangle. The next row has a 1 at both ends. The middle number is the sum of the two numbers above it, so 1 + 1 equals 2. The next row will also have 1’s at either end.

The numbers in between these 1’s are made up of the sum of the two numbers above it. So, the first number is 1 + 2, which is 3. The next number is 1 + 2, which is also 3. So, this row’s numbers are 1, 3, 3, and 1. We can keep going to make our Pascal’s Triangle even bigger by continuing our addition.

Notice the numbers in the row with the 1, 3, 3, and 1. Doesn’t our answer have those same numbers as coefficients? Yes, it does. As it turns out, the coefficients from the binomial theorem follow the numbers of Pascal’s triangle. Each row would give the coefficients a different n. The first row, the tip, is for n = 0.

The next one is n = 1. The one after that is n = 2. And the next is n = 3. And we can keep going that way. So, if we want to expand the binomial (a + b)^5, our coefficients according to Pascal’s Triangle are 1, 5, 10, 10, 5, and 1, in that order.

## Expanding a Binomial

Now that we know all the patterns involved, we can expand our binomials much easier. What are the patterns, again? They are that the exponents of the a term decrease to 0 starting from n, while the exponents of b increase starting from 0 and ending with n. Our coefficients follow a particular row of Pascal’s triangle.

Let’s see about using these patterns to expand the binomial (a + b)^5. Our n equals 5, so we will use the coefficients from the row of Pascal’s triangle that corresponds to n = 5. This particular row has the numbers 1, 5, 10, 10, 5, and 1. So this means that we will have a total of 6 terms in our answer. The other patterns that we will use are the ones for the exponents.

We know that for our a the exponent begins at n, 5 in our case, and goes down until it is 0, while our b, the exponent begins at 0 and goes up until it is 5 or n. So, using all these patterns, we find that our answer equals a to the fifth power plus 5 times a to the fourth power times b plus 10 times a to the third power times b to the second power plus 10 times a to the second power times b to the third power plus 5 times a times b to the fourth power plus b to the fifth power.

This is our answer. See how it follows all the patterns that we talked about? Not bad, eh?

## Lesson Summary

Let’s review what we’ve learned now. We’ve learned that the binomial theorem is a theorem that tells us what happens when you multiply a binomial by itself numerous times. The formula is a plus b to the nth power is equal to the summation of n choose k times a to the n minus k power times b to the kth power from k equals 0 to k equals n. While the formula looks kind of scary, we see that it gives us some very useful patterns that we can use to help us.

We see that the exponents of the first term of our binomial begins at our n and goes down until it is 0, while the exponents of our second term begins at 0 and goes up until it is n. We also see that our coefficients follow a particular row of Pascal’s triangle, which is a triangle of numbers where the first and last terms of a row are 1 and all the other numbers are the sum of the two numbers directly above it. Using these patterns, we can easily expand any binomial.

## Learning Outcomes

Learning the topics in this lesson could enable you to:

• Recall the binomial theorem
• Recognize and use Pascal’s triangle to expand a binomial
• Expand a binomial using the binomial theorem

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4:40

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4:44

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Algebra II: Sets

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Algebra II: Combinatorics

Ch 24. Algebra II: Calculations, Ratios,…

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Algebra II: Calculations, Ratios, Percent & Proportions Review

Ch 25. Algebra II: Statistics

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Algebra II: Statistics

Ch 26. Algebra II:…

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Algebra II: Trigonometry

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