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### Course: Precalculus > Unit 9

Lesson 3: The binomial theorem# Pascal's triangle and binomial expansion

Sal introduces Pascal's triangle, and shows how we can use it to figure out the coefficients in binomial expansions. Created by Sal Khan.

## Want to join the conversation?

- ¿What would happen when you have a negative term? Say (a-b)^3(115 votes)
- You can Make (a-b)^3 to (a+(-b))^3 and make the expantion(43 votes)

- Did you guys see this pattern? If i take the number 1.1 and raise it into those powers, i will get the same results of the Pascal's triangle. For example:

1.1^0 is equal to 1.

1.1^1 is equal to 1.1

1.1^2 is equal to 1.21

1.1^3 is equal to 1.331

1.1^4 is equal to 1.4641

And so on. Can anyone explain that?(83 votes)- Yeah, I observed it when I first saw the Pascal’s triangle. It also works with 11. That’s because 11^n =(10+1)^n. And 1 raised to any power is always 1. So for 11^4 it is

(10^4) + (4*10^3*1^1)+(6*10^2*1^2)+(4*10*1^3)+10^0.

As you can see, the powers of 1 make no difference and the answer is simply 14641. Powers of 1.1 are just powers of 11/10 and so they follow the same pattern.(8 votes)

- Can't you just add the numbers like for example 2+1=3 so thats the next number under it and again and again cause while you were explaining all this i noticed this way(33 votes)
- Yes, it works that way too. Can you prove that they generate the same triangle?(22 votes)

- how was pascals triangle discovered?(19 votes)
- http://en.wikipedia.org/wiki/Pascal%27s_triangle this is its story and explanation :)(31 votes)

- Where did this triangle come from?(9 votes)
- I've actually never heard Sal's explanation of the number of ways to get to each point in the triangle. An easier way to calculate the numbers is that each number is the sum of the two directly above it. The exceptions of course are all the 1s on the borders. The next row beyond what Sal wrote out is easy to calculate this way: it would be 1 5 10 10 5 1(24 votes)

- What does this Symbol mean ∑? THanks(5 votes)
- The symbol means "sum of."(15 votes)

- Is it just coincidence that the sum of the coefficients of (a+b)^n = 2^n? Does this have any connection to the open n-set of combinations?(7 votes)
- I will use aCb to mean a choose b. Notice that (a+b)^n expands as nC0 * a^n + nC1 * a^(n-1) * b + nC2 * a^(n-2) * b^2 + ... + nC0 * b^n. Now let a = b = 1. Notice that now
**all**powers of a and b disappear and become ones, which don't affect the coefficients. But now, (a+b)^n is equal to (1+1)^n = 2^n. So we have 2^n = nC0 + nC1 + nC2 + ... + nC(n-1) + nCn. This proves that the sum of the coefficients is equal to 2^n.(9 votes)

- what if you have terms with exponents inside the brackets?

And also what do you do if there are multiple variables with exponents multiplied within the brackets?

I encountered this one reacently and became uncertain:

(3x^3y^2z)^4

How would that be solved and could you use pascals triangle for this?

Thank you:)(3 votes)- First off, the example you provided wasn't a binomial. For your example, all you need are simple exponent rules.

If you meant to ask "what if there were multiple variables added/subtracted within the brackets" then you would use what is called Multinomial Theorem which is a generalized binomial theorem. When you are expanding a trinomial (3 variables) then you could actually use what is called "Pascal's pyramid" which is exactly what it sounds like.

You can read more on Pascal's pyramid and Multinomial Theorem here:

https://en.wikipedia.org/wiki/Multinomial_theorem

https://en.wikipedia.org/wiki/Pascal%27s_pyramid(5 votes)

- I have the equation (2x+1)^4, and it says to write in descending powers of x. What does this mean? I just did it normally using powers 1,4,6,4,1 and I got the right solution. Like this:

1((2x)^4)((1)^0) + 4((2x)^3)((1)^1) + 6((2x)^2)((1)^2) + 4((2x)^1)((1)^3) + 1((2x)1^0)((1)^4) = 16x^4 + 32x^2 + 8x +1

But the steps aren't the same. Instead, they took powers of 4,3,2,3,4 like this:

1((2x)^4)((1)^0) + 4((2x)^3))((1)^1) + 6((2x)^2)((1)^2) + 4((2x)^1)((1)^3) + (1)^4. They also get 16x^4 + 32x^3 + 24x^2 + 8x + 1.

Can someone please explain? Thank you!(2 votes)- You have two errors in your calculations:

First: 4((2x)^3)((1)^1)

(2x)^3 = 8x^3

4*8x^3 = 32x^3

I think you made it into 32x^2

Second: You lost your term created by 6((2x)^2)((1)^2) = 24x^2

Hope this helps.(7 votes)

- whats the difference between a^3 + b^3 and (a+b)^3 ?(1 vote)

## Video transcript

In the previous video we were able
to apply the binomial theorem in order to figure out what
a plus b to fourth power is in order to expand this out. And we did it. And it was
a little bit tedious but hopefully you appreciated it. It would have been useful
if we did even a higher power-- a plus b to the seventh power,
a plus b to the eighth power. But what I want to do
in this video is show you that there's another way
of thinking about it and this would be using
"Pascal's Triangle". And if we have time we'll also think about why these two ideas
are so closely related. So instead of doing a plus b to the fourth
using this traditional binomial theorem-- I guess you could say-- formula right over
here, I'm going to calculate it using Pascal's triangle
and some of the patterns that we know about the expansion. So once again let me write down
what we're trying to calculate. We're trying to calculate a plus b to the fourth power-- I'll just do this in a different color--
to the fourth power. So what I'm going to do is set up
Pascal's triangle. So Pascal's triangle-- so we'll start with a one at the top.
And one way to think about it is, it's a triangle where if you start it
up here, at each level you're really counting the different ways
that you can get to the different nodes. So one-- and so I'm going to set up
a triangle. So if I start here there's only one way I can get here and there's only one way
that I could get there. But now this third level-- if I were to say
how many ways can I get here-- well, one way to get here,
one way to get here. So there's two ways to get here. One way to get there,
one way to get there. The only way I get there is like that,
the only way I can get there is like that. But the way I could get here, I could
go like this, or I could go like this. And then we could add a fourth level
where-- let's see, if I have-- there's only one way to go there
but there's three ways to go here. One plus two. How are there three ways? You could go like this,
you could go like this, or you could go like that. Same exact logic:
there's three ways to get to this point. And then there's only one way
to get to that point right over there. And so let's add a fifth level because
this was actually what we care about when we think about
something to the fourth power. This is essentially zeroth power--
binomial to zeroth power, first power, second power, third power.
So let's go to the fourth power. So how many ways are there to get here? Well I just have to go all the way
straight down along this left side to get here, so there's only one way. There's four ways to get here. I could
go like that, I could go like that, I could go like that,
and I can go like that. There's six ways to go here. Three ways to get to this place,
three ways to get to this place. So six ways to get to that and, if you
have the time, you could figure that out. There's three plus one--
four ways to get here. And then there's one way to get there. And now I'm claiming that
these are the coefficients when I'm taking something to the-- if
I'm taking something to the zeroth power. This is if I'm taking a binomial
to the first power, to the second power. Obviously a binomial to the first power, the coefficients on a and b
are just one and one. But when you square it, it would be
a squared plus two ab plus b squared. If you take the third power, these
are the coefficients-- third power. And to the fourth power,
these are the coefficients. So let's write them down. The coefficients, I'm claiming,
are going to be one, four, six, four, and one. And how do I know what
the powers of a and b are going to be? Well I start a, I start this first term, at the highest power: a to the fourth. And then I go down from there. a to the fourth, a to the third, a squared, a to the first, and I guess I could write a to the zero which of course is just one. And then for the second term
I start at the lowest power, at zero. And then b to first, b squared, b to the third power, and then b to the fourth, and then I just add those terms together. And there you have it. I have just figured out the expansion of a plus b to the fourth power. It's exactly what I just wrote down. This term right over here,
a to the fourth, that's what this term is. One a to the fourth b to the zero:
that's just a to the fourth. This term right over here is equivalent to this term right over there. And so I guess you see that
this gave me an equivalent result. Now an interesting question is
'why did this work?' And I encourage you to pause this video
and think about it on your own. Well, to realize why it works let's just
go to these first levels right over here. If I just were to take
a plus b to the second power. a plus b to the second power. This is going to be,
we've already seen it, this is going to be
a plus b times a plus b so let me just write that down:
a plus b times a plus b. So we have an a, an a.
We have a b, and a b. We're going to add these together.
And then when you multiply it, you have-- so this is going to be equal to a times a.
You get a squared. And that's the only way. That's the
only way to get an a squared term. There's only one way of getting
an a squared term. Then you're going to have
plus a times b. So-- plus a times b. And then you're going to have
plus this b times that a so that's going to be another a times b. Plus b times b which is b squared. Now this is interesting right over here. How many ways can you get
an a squared term? Well there's only one way. You're
multiplying this a times that a. There's one way of getting there. Now how many ways are there
of getting the b squared term? How many ways are there
of getting the b squared term? Well there's only one way.
Multiply this b times this b. There's only one way of getting that. But how many ways are there
of getting the ab term? The a to the first b to the first term. Well there's two ways. You can multiply
this a times that b, or this b times that a. There are--
just hit the point home-- there are two ways,
two ways of getting an ab term. And so, when you take the sum of these two you are left with a squared plus
two times ab plus b squared. Notice the exact same coefficients: one two one, one two one. Why is that like that? Well there is only
one way to get an a squared, there's two ways to get an ab, and there's only one way to get a b squared. If you set it to the third power you'd say
okay, there's only one way to get to a to the third power. You just multiply
the first a's all together. And there is only one way
to get to b to the third power. But there's three ways to get to a squared b. And you could multiply it out,
and we did it. We did it all the way back over here. There's three ways to get a squared b. We saw that right over there. And there are three ways to get a b squared. Three ways to get a b squared. And if you sum this up you have the
expansion of a plus b to the third power. So hopefully you found that interesting.