Lesson 9 - Binomial Theorem
What happens when you multiply a binomial by itself ... many times?
Here is the answer:
Don't worry ... I will explain it all!
And you will learn lots of cool math symbols along the way.
Binomial
A binomial is a polynomial with two terms
example of a binomial
Multiplying
The Binomial Theorem shows what happens when you multiply a binomial by itself (as
many times as you want).
It works because there is a pattern ... let us see if we can discover it.
Exponents
But first you need to know what an Exponent is.
Here is a quick summary:
An exponent says how many times to use something in a
multiplication.
In this example: 82 = 8 × 8 = 64
An exponent of 1 means just to have it appear once, so you get the original value:
Example: 81 = 8
An exponent of 0 means not to use it at all, and we have only 1:
Example: 80 = 1
Exponents of (a+b)
Now on to the binomial.
We will use the simple binomial a+b, but it could be any binomial.
Let us start with an exponent of 0 and build upwards.
Exponent of 0
When an exponent is 0, you get 1:
(a+b)0 = 1
Exponent of 1
When the exponent is 1, you get the original value, unchanged:
(a+b)1 = a+b
Exponent of 2
An exponent of 2 means to multiply by itself (see how to multiply polynomials):
(a+b)2 = (a+b)(a+b) = a2 + 2ab + b2
Exponent of 3
For an exponent of 3 just multiply again:
(a+b)3 = (a+b)(a2 + 2ab + b2) = a3 + 3a2b + 3ab2 + b3
We have enough now to start talking about the pattern.
The Pattern
In the last result we got:
a3 + 3a2b + 3ab2 + b3
Now, notice the exponents of a. They start at 3 and go down: 3, 2, 1, 0:
Likewise the exponents of b go upwards: 0, 1, 2, 3:
If we number the terms 0 to n, we get this:
k=0 k=1 k=2 k=3
a3 a2 a 1
1 b b2 b3
Which can be brought together into this:
an-kbk
How about an example to see how it works:
Example: When the exponent, n, is 3.
The terms are:
k=0: k=1: k=2: k=3:
an-kbk an-kbk an-kbk an-kbk
= a3-0b0 = a3-1b1 = a3-2b2 = a3-3b3
= a3 = a2b = ab2 = b3
It works like magic!
Coefficients
So far we have: a3 + a2b + ab2 + b3
But we really need: a3 + 3a2b + 3ab2 + b3
We are missing the numbers (which are called coefficients).
Let's look at all the results we got before, from (a+b)0 up to (a+b)3:
And now look at just the coefficients (with a "1" where a coefficient wasn't shown):
They actually make Pascal's Triangle!
Each number is just the two numbers above it
added together (except for the edges, which are
all "1")
(Here I have highlighted that 1+3 = 4)
Armed with this information let us try something new ... an exponent of 4:
a exponents go 4,3,2,1,0: a4 + a3 + a2 + a + 1
b exponents go 0,1,2,3,4: a4 + a3b + a2b2 + ab3 + b4
coefficients go 1,4,6,4,1: a4 + 4a3b + 6a2b2 + 4ab3 + b4
And that is the correct answer.
We have success!
We can now use that pattern for exponents of 5, 6, 7, ... 50, ... 112, ... you name it!
As a Formula
Our last step is to write it all as a formula.
But hang on, how do we write a formula for "find the coefficient from Pascal's
Triangle" ... ?
Well, there is such a formula:
It is commonly called "n choose k" because it is how many
ways to choose k elements from a set of n.
You can read more at Combinations and Permutations
The "!" means "factorial", for example 4! = 1×2×3×4 = 24
Example: Row 4, term 2 in Pascal's Triangle is "6". Let's see if the formula works:
Yes. Correct.
Putting It All Together
The last step is to put all the terms together into one formula.
But we are adding lots of terms together ... can that be done using one formula?
Yes! The handy Sigma Notation allows us to sum up as many terms as we want:
Sigma Notation
Now it can all go into one formula:
The Binomial Theorem
Use It
OK ... it won't make much sense without an example.
So let's try using it for n = 3 :
BUT ... it is usually much easier just to remember the patterns:
The first term's exponents start at n and go down
The second term's exponents start at 0 and go up
Coefficients are from Pascal's Triangle, or by calculation using n!/(k!(n-k)!)
Like this:
Example: What is (x+5)4
Start with exponents: x450 x351 x252 x153 x054
Include Coefficients: 1x450 4x351 6x252 4x153 1x054
Then write down the answer (including all calculations, such as 4×5, 6×52, etc):
(x+5)4 = x4 + 20x3 + 150x2 + 500x + 625
You may also want to calculate just one term:
Example: What is the coefficient for x3 in (2x+4)8
The exponents for x3 are:
(2x)345
The coefficient is "8 choose 3". We can use Pascal's Triangle, or calculate directly:
n! 8! 8! 8×7×6
= = = = 56
k!(n-k)! 3!(8-3)! 3!5! 3×2×1
And we get:
56(2x)345
Which simplifies to:
458752 x3
A large coefficient, isn't it?
QUESTION/ANSWER SESSION
Q1. Expand (x + 1)6
Ans. x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1
Q2. Expand (2 + x)7
Ans. 128 + 448x + 672x2 + 560x3 + 280x4 + 84x5 + 14x6 + x7
Q3.
Expand (2x - 3)5
Ans.
32x5 - 240x4 + 720x3 - 1,080x2 + 810x - 243
