The calculations get longer and longer as we go, but there is some kind of pattern developing.
You are watching: What is the binomial expansion of (x + 2)4?
That pattern is summed up by the Binomial Theorem:
The Binomial Theorem
Don”t worry … it will all be explained!
And you will learn lots of cool math symbols along the way.
Exponents
First, a quick summary of Exponents.
An exponent says how many times to use something in a multiplication.
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, we get 1:
(a+b)0 = 1
Exponent of 1
When the exponent is 1, we 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 = (a2 + 2ab + b2)(a+b) = 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 = a3-0b0 = a3 | an-kbk = a3-1b1 = a2b | an-kbk = a3-2b2 = ab2 | an-kbk = a3-3b3 = 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 (compare to the top of the page).
We have success!
We can now use that pattern for exponents of 5, 6, 7, … 50, … 112, … you name it!
That pattern is the essence of the Binomial Theorem.
Now you can take a break.
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When you come back see if you can work out (a+b)5 yourself.
Answer (hover over): a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
As a Formula
Our next task is to write it all as a formula.
We already have the exponents figured out:
an-kbk
But 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.
The “!” means “factorial”, for example 4! = 4×3×2×1 = 24
You can read more at Combinations and Permutations.
And it matches to Pascal”s Triangle like this: (Note how the top row is row zero and also the leftmost column is zero!) |
Example: Row 4, term 2 in Pascal”s Triangle is “6”.
Let”s see if the formula works:
Yes, it works! Try another value for yourself.
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 (y+5)4
Start with exponents: | y450 | y351 | y252 | y153 | y054 |
Include Coefficients: | 1y450 | 4y351 | 6y252 | 4y153 | 1y054 |
Then write down the answer (including all calculations, such as 4×5, 6×52, etc):
(y+5)4 = y4 + 20y3 + 150y2 + 500y + 625
We may also want to calculate just one term:
Example: What is the coefficient for x3 in (2x+4)8
The exponents for x3 are 8-5 (=3) for the “2x” and 5 for the “4”:
(2x)345
(Why? Because:
2x: | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
4: | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
(2x)840 | (2x)741 | (2x)642 | (2x)543 | (2x)444 | (2x)345 | (2x)246 | (2x)147 | (2x)048 |
But we don”t need to calculate all the other values if we only want one term.)
And let”s not forget “8 choose 5” … we can use Pascal”s Triangle, or calculate directly:
n!k!(n-k)! = 8!5!(8-5)! = 8!5!3! = 8×7×63×2×1 = 56
And we get:
56(2x)345
Which simplifies to:
458752 x3
A large coefficient, isn”t it?
Geometry
The Binomial Theorem can be shown using Geometry:
In 2 dimensions, (a+b)2 = a2 + 2ab + b2
In 3 dimensions, (a+b)3 = a3 + 3a2b + 3ab2 + b3
In 4 dimensions, (a+b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
(Sorry, I am not good at drawing in 4 dimensions!)
Advanced Example
And one last, most amazing, example:
Example: A formula for e (Euler”s Number)
We can use the Binomial Theorem to calculate e (Euler”s number).
e = 2.718281828459045…
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(the digits go on forever without repeating)
It can be calculated using:
(1 + 1/n)n
(It gets more accurate the higher the value of n)
That formula is a binomial, right? So let”s use the Binomial Theorem:
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First, we can drop 1n-k as it is always equal to 1:
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And, quite magically, most of what is left goes to 1 as n goes to infinity:
Which just leaves:
With just those first few terms we get e ≈ 2.7083…
Try calculating more terms for a better approximation!(Try the Sigma Calculator)
Challenging 1 Challenging 2
Isaac Newton
As a footnote it is worth mentioning that around 1665 Sir Isaac Newton came up with a “general” version of the formula that is not limited to exponents of 0, 1, 2, …. I hope to write about that one day.
Polynomial Exponent Pascal”s Triangle Sigma Notation Algebra Index