# 4. The Binomial Theorem

by M. Bourne

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## Definition: binomial

A **binomial ** is an algebraic expression containing 2 terms. For example, (*x* + *y*) is a binomial.

We sometimes need to expand binomials as follows:

(*a* + *b*)^{0} = 1

(*a* + *b*)^{1} = *a* + *b*

(*a* + *b*)^{2} = *a*^{2}
+ 2*ab* + *b*^{2}

(*a* + *b*)^{3} = *a*^{3}
+ 3*a*^{2}*b* + 3*ab*^{2} +
*b*^{3}

(*a* + *b*)^{4} = *a*^{4}
+ 4*a*^{3}*b* + 6*a*^{2}*b*^{2}
+ 4*ab*^{3} + *b*^{4}

(*a* + *b*)^{5} = *a*^{5}
+ 5*a*^{4}*b* + 10*a*^{3}*b*^{2}
+ 10*a*^{2}*b*^{3} + 5*ab*^{4}
+ *b*^{5}

Clearly, doing this by direct multiplication gets quite tedious and can be rather difficult for larger powers or more complicated expressions.

## Pascal's Triangle

We note that the coefficients (the numbers in front of each term) follow a pattern. [This was noticed long before Pascal, by the Chinese.]

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

You can use this pattern to form the coefficients, rather than multiply everything out as we did above.

## The Binomial Theorem

We use the **binomial theorem** to help us expand binomials
to any given power without direct multiplication. As we have seen, multiplication can be
time-consuming or even not possible in some cases.

Let's consider the properties of a binomial expansion first.

### a. Properties of the Binomial Expansion (*a* + *b*)^{n}

- There are `n + 1` terms.
- The first term is
*a*and the final term is^{n}*b*.^{n}

- Progressing from the first term to the last, the exponent of
*a*decreases by `1` from term to term while the exponent of*b*increases by `1`. In addition, the sum of the exponents of*a*and*b*in each term is*n*.

- If the coefficient of each term is multiplied by the exponent of
*a*in that term, and the product is divided by the number of that term, we obtain the coefficient of the next term.

Continues below ad ⇩

### b. General formula for (*a* + *b*)^{n}

First, we need the following definition:

**Definition: ***n*! represents the product of the first *n* positive integers i.e.

n! =n(n− 1)(n− 2) ... (3)(2)(1)

We say *n*! as "*n* factorial".

### Example 1 - factorial values

Here are some factorial values:

(a) `3! = (3)(2)(1) = 6`

(b) `5! = (5)(4)(3)(2)(1) = 120`

(c) `(4!)/(2!)=((4)(3)(2)(1))/((2)(1))=12`

**Note:** `(4!)/(2!)` cannot be cancelled down to `2!`.

### c. Factorial Interactive

**Instructions:** You can use the following interactive to find the factorial of any positive integer **up to 30**.

! =

For numbers greater than `22!` (and less than `31!`), you'll see output something like this: `2.652528e+32`. The "`e`" stands for exponential (base `10` in this case), and the number has value `2.652528 xx 10^32`.

### d. Binomial Theorem Formula

Based on the binomial properties, the binomial theorem states
that the following **binomial formula** is valid for
all positive integer values of *n*:

`(a+b)^n=` `a^n+na^(n-1)b` `+(n(n-1))/(2!)a^(n-2)b^2` `+(n(n-1)(n-2))/(3!)a^(n-3)b^3` `+...+b^n`

This can be written more simply as:

(*a* + *b*)* ^{n}* =

*C*

^{n}_{0}

*a*+

^{n}*C*

^{n}_{1}

*a*

^{n−1}

*b*+

*C*

^{n}_{2}

*a*

^{n−2}

*b*

^{2}+

*C*

^{n}_{3}

*a*

^{n−3}

*b*

^{3}+ ... +

*C*

^{n}

_{n}b^{n}

We can use the `{::}^nC_r` button on our calculator to find these values.

This can also be written *nCr*.

### e. Binomial Expansion Interactive

The following interactive lets you expand your own binomial expressions. It shows all the expansions from `n=0` up to the power you have chosen.

In the **first line** of each expansion, you'll see the numbers from Pascal's Triangle written within square brackets, [ ].

The **second line** of each expansion is the result after tidying up.

**Instructions:** You can use letters or numbers within the brackets. The **maximum power** you can use is 6.

### Example 2

Using the binomial theorem, expand (*x* + 2)^{6}.

Answer

In using the binomial formula, we let

`a=x,` `b=2,` and `n=6`.

Substituting in the binomial formula, we get :

`(x+2)^6`

`=(x)^6 +6(x)^5(2)` `+(6(5))/(2!) (x)^4(2)^2` `+(6(5)(4))/(3!)(x)^3(2)^3` `+(6(5)(4)(3))/(4!) (x)^2(2)^4 ` `+(6(5)(4)(3)(2))/(5!) (x)(2)^5+(2)^6`

`=x^6+12x^5` `+60x^4+160x^3+240x^2` `+192x+64`

Easy to understand math videos:

MathTutorDVD.com

### Example 3

Using the binomial theorem, expand (2*x* + 3)^{4}

Answer

In using the binomial formula, we let

`a = 2x,` `b = 3` and `n = 4.`

Substituting in the binomial formula, we get:

`(2x+3)^4`

`=(2x)^4+4(2x)^3(3)` `+(4(3))/(2!)(2x)^2(3)^2` `+(4(3)(2))/(3!)(2x)(3)^3+(3)^4`

`=16x^4+96x^3+216x^2+216x+81`

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### Example 4

Using the binomial theorem, find the first four terms of the expansion `(2a-1/x)^11`

Answer

In using the binomial formula, we let

`text(")atext(")=2a,` `b=-1/x` and `n=11`.

Substituting in the binomial formula, we get:

`(2a-1/x)^11`

`=(2a)^11+11(2a)^10(-1/x)` `+(11(10))/(2!)(2a)^9(-1/x)^2` `+(11(10)(9))/(3!)(2a)^8(-1/x)^3+...`

`=2048a^11-11264(a^10)/x` `+28160(a^9)/(x^2)-42240(a^8)/(x^3)+...`

Easy to understand math videos:

MathTutorDVD.com

## Binomial Series

From the binomial formula, if we
let *a* = 1 and * b =
x,* we can also obtain the **binomial
series** which is valid for any real number *n* if
|*x*| < 1.

(1 + *x*)^{n} = 1 + *nx *+ `(n(n-1))/(2!)x^2` `+(n(n-1)(n-2))/(3!)x^3``+...`

**NOTE (1):** This is an **infinite series**, where the
**binomial theorem** deals with a finite expansion.

**NOTE (2):** We cannot use the
`{::}^nC_r` button for the **binomial series.** The `{::}^nC_r` button can only be used with positive
integers.

### Don't miss...

The Binomial series graph interactive on the next page illustrates this concept:

### Example 5

(a) Using the binomial series, find the first four terms of the expansion `root3(1+x)`.

(b) Use your result from part (a) to approximate the value of `root3(1.2)`.

Answer

(a) We know `root3(1+x) = (1+x)^(1//3)`.

Substituting `n=1/3` into the binomial series, we get:

`(1+x)^(1/3)`

`~~ 1+1/3 x` `+((1/3)(1/3-1))/2 x^2` `+ ((1/3)(1/3-1)(1/3-2))/6 x^3+...`

`= 1+x/3-(x^2)/9+(5x^3)/81-...`

(b) We now use the above expression to approximate `root3(1.2)`.

Substituting `x=0.2` into our expansion for `root3(1+x)`, we have:

`root3(1.2)` `~~ 1+(0.2)/3` `-((0.2)^2)/9` `+(5((0.2)^3))/81-...` `=1.06271604938`

Is it correct?

The calculator value for `root3(1.2)` is `1.06265856918`, so our approximation using binomial series is accurate to the 4th decimal place. Taking more terms of the series would give us a more accurate result.

Here is the graph of what we just did in the answer for Example 5. The darker colored curve is

`y_1=root3(1+x)`

The lighter colored one is the first 4 terms of the series we found, that is:

`y_2=1+x/3-(x^2)/9+(5x^3)/81`.

Binomial series approximation for `y=root3(1+x)`

The approximation is quite good between −0.5 < *x* < 1, but we would need to take many more terms for a good approximation beyond these bounds.

The pink dot on the curve is the point `(0.2, 1.0627)`, representing the value we obtained for `root3(1.2)`.

Explore this example (and Example 6) in Binomial Series Interactive Applet.

### Example 6

(a) Using the binomial series, find the first four terms of the expansion `sqrt(4+x^2)`.

(b) Use your result from part (a) to approximate the value of `sqrt(4.25)`.

Answer

(a) To use the binomial **series**, we need to change the expansion to the form of (1 + *u*)^{n}.

So we perform the following steps to get it in the required form.

`sqrt(4+x^2)`

`=(4+x^2)^(1/2)`

`=[4(1+x^2/4)]^(1/2)`

`=4^(1/2)(1+x^2/4)^(1/2)`

`=2(1+x^2/4)^(1/2)`

Hence, if we let the "*u*" term be `x^2/4` and `n=1/2`, and then substituting in the binomial series,

(1 + *u*)^{n} = 1 + *nu *+ `(n(n-1))/(2!)u^2` `+(n(n-1)(n-2))/(3!)u^3` `+...`

we get:

`2(1+(x^2)/4)^(1/2)`

`~~2[1+(1/2)((x^2)/4)` `+\ ((1/2)(1/2-1))/(2!)((x^2)/4)^2` `+\ ((1/2)(1/2-1)(1/2-2))/(3!)((x^2)/4)^3` `{:+...]`

`=2+x^2/4-x^4/64+x^6/512+...`

(b) We now use the above expression to approximate `sqrt(4.25)`.

If we use `x=0.5`, then `x^2=0.25` and we'll have `4+x^2 = 4.25`.

Substituting `x=0.5` into our expansion for `sqrt(4+x^2)`, we have:

`sqrt(4.25)` `~~2+(0.5)^2/4-(0.5)^4/64+(0.5)^6/512` `~~2.06155395508`

Is it correct?

The calculator value for `sqrt(4.25)` is `2.06155281281`, so we've achieved 5 decimal place accuracy using our binomial series. We'd get a better answer if we were to take more than 4 terms.

Here is the graph of what we just did in the answer for Example 6. The darker colored curve is

`y_1=sqrt(4+x^2)`

The lighter colored curve is the first 4 terms of the series we found, that is:

`y_2=2+x^2/4-x^4/64+x^6/512`.

Binomial series approximation for `y=sqrt(4+x^2)`

The approximation is quite good between −2 < *x* < 2, but we would need to take many more terms for a good approximation beyond these bounds.

The pink dot on the curve is the point `(0.5, 2.06155)`, representing the value we obtained for `sqrt(4.25)`.

Explore this example (and Example 5) on the next page, Binomial Series Interactive Applet.

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