# The Binomial Theorem

The Binomial Theorem is a fundamental theorem in algebra that is used to expand

expressions of the form

where n can be any number.

The Binomial Theorem is given as follows:

which when compressed becomes

or

The above equations are quite complicated but you’ll understand what each component

means if you look at the section on

combinations before you look at Binomial Theorem. The rest should become

clearer by the time you are through with this page.

The Binomial Theorem is important because as **n** gets larger, the expressions

tend to become a lot more complicated.

For example:

As you can see, the above is relatively complicated and it would take a while to

expand to that final form, so a need arises for some way of making expanding

much quicker and easier.

The coefficients of each term in the above expression are

{1, 6, 15, 20, 15, 6, 1}

and these are referred to as binomial coefficients. These are also numbers that

correspond to 6 in the Pascal’s Triangle!

## Pascal’s Triangle

Pascal’s Triangle refers to a triangle of numbers with each subsequent row corresponding

to the next whole number from zero onwards. These numbers also happen to be binomial

coefficients.

The Mathematics behind Pascal’s Triangle is a little more advanced but the triangle

itself is very simple. Below is Pascal’s Triangle for the first numbers zero to

eight.

n = 0: | 1 | |||||||||||||||||

n = 1: | 1 | 1 | ||||||||||||||||

n = 2: | 1 | 2 | 1 | |||||||||||||||

n = 3: | 1 | 3 | 3 | 1 | ||||||||||||||

n = 4: | 1 | 4 | 6 | 4 | 1 | |||||||||||||

n = 5: | 1 | 5 | 10 | 10 | 5 | 1 | ||||||||||||

n = 6: | 1 | 6 | 15 | 20 | 15 | 6 | 1 | |||||||||||

n = 7: | 1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 | ||||||||||

n = 8: | 1 | 8 | 28 | 56 | 70 | 56 | 28 | 8 | 1 |

The first thing you should observe is that for each number **n**, there corresponds

**(n + 1)** coefficients in Pascal’s Triangle. For example for **n** = 3,

there are 3 + 1 = 4 coefficients.

Next observe how the coefficients in the row are sums of the coefficients in the

row above with which they form an upside down triangle (except for zero and one).

For example, consider 3 and 4

for **n** = 3, the coefficients from Pascal’s triangle are {1, 3, 3, 1}

for **n** = 4: the coefficients from Pascal’s triangle are {1, 4, 6, 4, 1}

If you look closely at these two sets of coefficients you should observe the following;

The first coefficient in any set is 1,

the next coefficient is equal the sum of the first and second coefficients of the

previous set, i.e.

the next coefficient is equal to the sum of the second and third coefficients of

the previous set, i.e.

the next coefficient is then equal to the sum of the third and fourth coefficients

of the previous set, i.e.

the last coefficient in each set is always 1.

Thus from knowing the first 5 set of coefficients from Pascal’s Triangle, you can

always come up with subsequent sets.

For example, knowing that the coefficients corresponding for **n** = 4 are {1,

4, 6, 4, 1} It becomes a lot simpler to get the coefficients for **n** = 5 as

{1, 5, 10, 10, 5, 1} and for **n** = 6 as {1, 6, 15, 20, 15, 6, 1}

To prove that Pascal’s Triangle does indeed give the term by term coefficients in

any expansion, lets expand the following example:

First lets take **n** = 0

Any number or variable raised to the power zero is equal to one.

Next take **n** = 1

the coefficients of each term are 1 and 1, as in Pascal’s Triangle

Next take **n** = 2

the coefficients of each term are {1, 2, 1} as in Pascal’s Triangle

Next take **n** = 3

the coefficients of each term are {1, 3, 3, 1}

The same trend would be observed if you continued for **n** = 4,5,6,…

These same coefficients can be obtained using Mathematical Combinations of the form

which is also written as

such that for the equation

**n** is the exponent and **r** refers to the term of the expansion, keep

in mind that the first term is represented by **r** = 0. The reason for this

is because for any exponent **n** there are **n + 1** terms as we have seen

in Pascal’s Triangle, so in order to have **n + 1** terms, we need to start with

**r** = 0 as the first term.

Another reason is that **r** is also an exponent of one of the variables but

more on this in a moment.

We can also show this using the following example

From above, we have seen that the expansion is as follows

where we can see the coefficients of each term as {1, 3, 3, 1}

Next we need to test the Combinations formula to see if we get the same coefficients,

keep in mind that **n** = 3 and **r** represents each term of the expansion,

we start with r = 0 to represent the first term

for **r** = 0, first term, the coefficient is given by

for **r** = 1, second term, the coefficient is given by

for **r** = 2, third term, the coefficient is given by

for **r** = 3, fourth and final term, the coefficient is given by

So we get as coefficients from the above: {1, 3, 3, 1} which are the same coefficients

from Pascal’s triangle.

So now we have seen that the Binomial Theorem gives the coefficients of the expansion,

it doesn’t stop there, the theorem also provides a way of keeping track of the exponents.

Let’s take a look at the binomial theorem once again

Taking an example of **n** = 3,

Keeping in mind that any number or variable raised to the power zero is equal to

one, the above can also be written as

Observe the ascending order of the powers from 0 to 3 on **b** and the descending

order of the powers from 3 to 0 on **a**. That is how any expansion of the form

**(a + b)** raised to the power **n** will turn out for any positive whole

number **n**, as the exponent on one variable increases, the exponent on the

other variable decreases, such that for any term of the expansion, the sum of the

exponents is always equal to **n**.

Looking at the example above

Since **n** = 3, you should see the when you add the exponents of any term, they

equal to 3. This is another reason why we start with **r** = 0 as the first term.

This relationship can be represented mathematically as

such that at any point in the expansion

and r increases from 0 to n.

So now we have seen the different components of the binomial theorem and what they

mean, each term of the expansion (a + b) raised to the power n is given by the following

which we know to be the same as

Since we now know each term, we add them up and symbolically we use

which means that add up all the terms from **r** = 0 to **r** = **n**.

So the whole binomial theorem formula comes together as

and using this you can obtain the terms of any expansion for n is a positive whole

number.

## Binomial Theorem Examples

### Example 1

Expand the following using binomial Theorem

**Step 1**

We have seen that the coefficients of the expansion can be obtained using both the

Pascal’s triangle or from the binomial theorem. But since this question requires

us to use the theorem, lets see how it works.

The theorem states that the coefficients can be obtained from the following formula

**Step 2**

substituting in for the required variables:

**Step 3**

**Step 4**

**Step 5**

Now that we have obtained the exponents, lets proceed to solve for the coefficients

(keep in mind that 1 raised to any power is still 1, thats why it does not appear

below)

**Step 6**

**Step 7**

### Example 2

Expand the following using binomial Theorem

**Step 1**

We have already stated the formula given by the binomial theorem as:

**Step 2**

So now all that is left is to substitute for the variables and proceed to calculate

the exponents and coefficients. Observe the difference between this and the previous

example

**Step 3**

**Step 4**

In order to keep with the theorem’s requirements, we substitute a = 2x and b = -3y

such that (a + b) = (2x + -3y). A common mistake most people make is not to apply

the exponents to the entire term ie not 2x^{2} but (2x)^{2} as below

**Step 5**

**Step 6**

Now for the coefficients

**Step 7**

Now for coefficients

**Step 8**

**Step 8**

Observe how in the final answer the negative appears for every odd power of y, this

will always be the same format in any expression like the one in this example.