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Lecture Comments (7)

1 answer

Last reply by: Professor Selhorst-Jones
Tue Mar 3, 2015 5:34 PM

Post by Jamal Tischler on March 1, 2015

It is great you focus on proofs, it realy helps me.

0 answers

Post by Saadman Elman on December 23, 2014

It was very helpful. Thanks.

3 answers

Last reply by: Professor Selhorst-Jones
Wed Nov 20, 2013 1:02 PM

Post by Linda Appling on July 31, 2013

could you slow down in your lectures?

The Binomial Theorem

  • A binomial is an expression that has two terms:
  • The binomial theorem tells us what happens when we raise a+b to some arbitrary power n: (a+b)n.
  • Notice that when we expand (a+b)n, it will always come out in the form
    an + an−1b + an−2b2 + …+ a2 bn−2 + a bn−1 + bn .
    We call each of the blanks a binomial coefficient since they are the coefficients of the binomial expansion.
  • Binomial theorem: When expanding (a+b)n, the coefficient of the term an−kbk is





    [Remember, (n || k) is spoken as `n choose k'. We can also write it as nCk or C(n,k). We first learned about it in the lesson Permutations and Combinations. Also remember, `!' means `factorial': the value we get when we multiply a number by all the positive integers below it: 5! = 5·4 ·3 ·2 ·1. We define 0! = 1 for ease and other fiddly reasons.]
  • We can also find the binomial coefficients with Pascal's triangle. We start with a 1 at the very top of the triangle for n=0, then each row below is created by diagonal addition. [This idea is hard to explain with words but makes a lot more sense in pictures. Check out the video for a diagram of what's going on and how to use it.]
  • If you're interested, this lesson has a proof of the binomial theorem at the end. Don't worry about watching it if you don't want to, but if you do, great! Working through proofs like this is good experience if you're interested in eventually studying higher mathematics.

The Binomial Theorem

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:06
    • We've Learned That a Binomial Is An Expression That Has Two Terms
  • Understanding Binomial Coefficients 1:20
    • Things We Notice
    • What Goes In the Blanks?
    • Each Blank is Called a Binomial Coefficient
  • The Binomial Theorem 6:38
    • Example
    • The Binomial Theorem, cont.
    • We Can Also Write This Expression Compactly Using Sigma Notation
  • Proof of the Binomial Theorem 13:22
    • Proving the Binomial Theorem Is Within Our Reach
  • Pascal's Triangle 15:12
    • Pascal's Triangle, cont.
    • Diagonal Addition of Terms
    • Zeroth Row
    • First Row
    • Why Do We Care About Pascal's Triangle?
    • Pascal's Triangle, Example
  • Example 1 21:26
  • Example 2 24:34
  • Example 3 28:34
  • Example 4 32:28
  • Example 5 37:12
  • Time for the Fireworks! 43:38
  • Proof of the Binomial Theorem 43:44
    • We'll Prove This By Induction
    • Proof (By Induction)
  • Proof, Base Case 47:00
  • Proof, Inductive Step - Notation Discussion 49:22
    • Induction Step
  • Proof, Inductive Step - Setting Up 52:26
    • Induction Hypothesis
    • What We What To Show
  • Proof, Inductive Step - Start 54:18
  • Proof, Inductive Step - Middle 55:38
    • Expand Sigma Notations
    • Proof, Inductive Step - Middle, cont.
  • Proof, Inductive Step - Checking In 1:01:08
    • Let's Check In With Our Original Goal
    • Want to Show
    • Lemma - A Mini Theorem
  • Proof, Inductive Step - Lemma 1:02:52
    • Proof of Lemma: Let's Investigate the Left Side
  • Proof, Inductive Step - Nearly There 1:07:54
  • Proof, Inductive Step - End! 1:09:18
    • Proof, Inductive Step - End!, cont.

Transcription: The Binomial Theorem

Hi--welcome back to Educator.com.0000

Today, we are going to talk about the binomial theorem.0002

Long ago, when we studied polynomials, we learned that a binomial is an expression that has two terms.0005

For example, each of the below is a binomial, because it has two separate terms:0011

x + 7; 2l3 - 5; 3y2 + √2z.0015

A binomial is anything that we can put in the form a + b, some a and some b.0023

All we are looking for to be a binomial is just having two terms.0028

What if we wanted to see what happens if we raise a binomial to some power?0032

If it was small, like (a + b)2, we could FOIL it; we could put one factor next to another factor,0036

and then we would just multiply them out by distribution and just work it out.0041

We are used to doing that; we have been doing that for years--it wouldn't be that difficult.0044

But what if it was really large, like (a + b)7, or even larger?0047

We could do it by just putting all of the factors out, but it is going to get really big and really messy really fast.0052

It is not going to be easy for us to multiply out (a + b)7, and it would be even worse if it was a larger value than 7.0057

So, in this lesson, we are going to learn how to expand a binomial for any arbitrary power n.0064

If we have any integer n, we will be able to use this to figure out how to expand (a + b) raised to the n.0068

To help us understand what we are working with, let's look at (a + b)n expanded for various values of n.0076

(a + b)0 will just come out to be 1, because if you raise anything to the 0, you get 1.0082

(a + b)1 comes out to be a + b; (a + b)2 would come out as a2 + 2ab + b2.0087

(a + b)3 would be a3 + 3a2b + 3ab2 + b3.0094

(a + b)4 equals a4 + 4a3b + 6a2b2 + 4ab3 + b4.0100

And (a + b)5 would equal a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5.0106

We could keep going in this way; but we are starting to see a pattern at this point.0115

We can realize that there is a pattern going on, and we can use this pattern to let us figure out general things about a + b, to just any n whatsoever.0118

Let's look specifically at (a + b)4 and (a + b)5 above0127

to help us see some properties about how binomials work.0131

We will be able to talk about (a + b)n, the general expansion of a binomial, by looking at these in specific.0134

So, looking at these two as stand-ins for the general way that (a + b)n expands, we notice various things.0142

The first thing is that the expansion of (a + b)n has n + 1 terms.0149

What I mean here is that (a + b)4--well, that would be n = 4;0153

we count how many terms come out in the expansion: 1, 2, 3, 4, 5: there are 5 terms total.0157

We started with n = 4; 4 + 1, n + 1, gets us 5; so we have 5 terms coming out of it.0165

The same thing happens with (a + b)5; we have 1, 2, 3, 4, 5, 6 terms total: 5 + 1 is 6 terms.0174

So, we end up getting n + 1 as the number of terms that come out.0183

The expression also has symmetry; powers of a decrease by 1 each step, every time we move forward a term, while powers of b will increase by 1.0188

If we have (a + b)4, for example, we start at a4, because (a + b)4...0198

well, the largest kind of a that it will be able to produce (largest in the sense that it will have the largest exponent) will be a4.0203

We manage to get a4 out of it; and then the next one--we will end up having a30210

(it has just stepped down 1), and then the next term would be a2 (step down again), a (step down again),0214

and finally it turns blank; and a blank is another way of saying a0, because a0 is just 1, so it disappears.0219

The same thing happens with the b's, except that it does the reverse process.0227

It increases; since it is blank, it doesn't appear here; that is b0, and then we have b1;0230

and then b2, and then b3, and then finally b4.0235

Our a's are decreasing by 1 each step, at the same time that our b's are increasing.0240

a starts at n; b starts at 0; and then click, click, click, click, click, click...until finally a is at 0 and b is at n,0245

and we have finished their expansion in terms of the a and b exponents.0253

The sum of the powers for each term comes out to be n.0258

Here, our n is 4; if we grab any term inside of this...we will end up seeing a3b1:0261

if we add up the powers on these terms, 3 + 1 gets us 4.0268

If we grab another one, like a2b2, 2 + 2 gets us 4.0272

Even at the extremes...a0 and b4: well, 4 + 0 gets us 4.0276

This ends up happening with (a + b)5, as well; if we grab any one of the terms here...0283

this is a 4; this is a 1; 4 + 1 gets us 5; so if we add the powers, we always end up getting n.0288

Whatever our (a + b)n is, adding the powers of the a and the b will always come out to be equal to the value of n.0295

And finally, the coefficients vary symmetrically; there is symmetry in how the coefficients are.0305

If we start in the middle, there is a 6 here by itself, because it is in the very middle, so it can't be symmetric with anything else.0310

And then, we have 4's matching on either side, and then we have 1's (effectively, the coefficient here is a 1,0316

because it doesn't appear at all, so its coefficient is 1); and the 1's match on either side, as well.0323

The same thing here: the 10's match, and the 5's match, and the invisible 1's match.0328

All right, so we end up seeing that the coefficients vary symmetrically; great.0339

With all of these ideas in mind, these properties, we see that any expansion of a binomial, (a + b)n,0344

is going to have this form of an, an - 1, an - 2,0351

until we finally work down to a2, a, and then just a0 in here.0356

And we reverse that with the b's; we have b0 at the beginning, and then it works up:0362

b1, b2, working up...bn - 2, bn - 1, and bn, finally.0366

So, we are always going to have this form; the only question is what goes in the blanks in front of it.0372

We call these blanks binomial coefficients, because they are the coefficients of the binomial expansion.0377

They are the number multiplying against that binomial having been expanded--0383

each one of the coefficients for the expansion of some a, some b, and powers on those a's and b's.0388

The only question we have at this point is what those coefficients will be.0393

We get this with the binomial theorem; if we want to know the binomial coefficients of some binomial expansion, we can use the binomial theorem.0398

The binomial theorem says that expanding some (a + b)n, the coefficient of the term an - kbk0405

will be n choose k (and n choose k is just the same thing as n! over (n - k)! times k!)...remember, we talked about this n choose k,0414

this parentheses with the values n and k above each other; this is n choose k--we have talked about this before...0426

You can also write it as nCk, or c of n,k; but we still pronounce all of these spoken aloud the same way: "n choose k."0434

We first learned about this idea in the lesson Permutations and Combinations.0443

So, if you really want a lot more experience with how that works, you can go and look that up.0447

But really, we have enough, just for what we are looking for here, from what we will be able to get in just this lesson.0450

Also, remember: the exclamation mark means factorial, and factorial is the value that we get0455

when we multiply a number by all of the positive integers below it.0460

For example, 5! is equal to 5 times 4 times 3 times 2 times 1; we multiply those all together, and we get the value of 5!.0464

Finally, we define 0! as equal to 1; we just say 0! = 1; that is the end of the story, because it will help us out.0473

It makes things easier, and there are also some other reasons that we don't really need to go into.0480

You have to memorize that 0! = 1; that is just that.0484

All right, let's see the binomial theorem get used in an example, so we can see how it works.0488

For any (a + b)n, the coefficient of the an - kbk term is n choose k0493

(which is the same thing as saying n!/(n - k)!(k!)).0501

For example, if we had (a + b)7, and we wanted to know the term containing a2b5,0506

when expanded; that is going to show up, because 2 + 5 is 7;0512

so we know that a2b5 will show up in the expansion.0516

We could figure out what the term would be with the binomial theorem.0520

We know that it is going to be n choose k; so the only thing that we have to do is figure out what our n is;0523

well, it is (a + b)7, so it means that n = 7.0527

What is our k? Well, it is b5, so our k equals 5.0530

an - k, bk...we can also check, because it is n - k here, and this is 2;0535

well, if we take 7 - 5, sure enough, we get 2, which is this value right here.0542

So, it checks out; we are following the method of the binomial theorem.0551

We see how the binomial theorem works: we have a7 - 5 times b5; that is a2b5.0555

We know that the coefficient is going to be n choose k; our n is 7; our k is 5; so we have 7 choose 5.0562

That expands to 7!/2!(5!) times a2b5.0570

And that will simplify to 21 times a2b5; so we have been able to figure out what the coefficient is,0577

what the term would look like when it is fully expanded: we get 21a2b5.0583

If you are not quite sure how we get the value of 21 out of that, we can figure it out,0587

because 7 choose 5...well, we wrote here already that that would be 7! divided by 7 - 5 (that is 2), factorial, times 5!.0590

7! is the same thing as 7 times 6 times 5 times 4 times 3 times 2 times 1.0603

We can also write 5 times 4 times 3 times 2 times 1 as just 5!.0610

And now we have things we can cancel out; 2! is just 2 times 1, so we can just leave that as 2; so it is 2 times 5!.0614

We have 5! and 5!; they cancel out; 7 times 6 over 2...6 cancels into 3 times 2, so our 2's cancel out; we have 7 times 3, or 21;0620

and that is how we end up getting the 21 for our coefficient.0631

And we know it will be a2b5, because we figured out that it was the coefficient of the a2b5 term.0634

Great; at this point, we can now know the coefficient of any term, which means that we can expand (a + b)n with relative ease.0640

(a + b)n will end up being n choose 0 an, because it is ak of 0 there,0649

because b hasn't even shown up, so it is b0.0654

And then, next we would have n choose 1: an - 1b1; our k there is 1.0657

Next, we would have n choose 2 times an - 2b2.0663

And we would continue on in this fashion, until we have n choose n - 2; n minus (n - 2) comes out to be +2;0668

the minus on the -2 becomes positive when we get just +2 on our a; and k is n - 2 bn - 2.0677

n choose n - 1: abn - 1; and finally, n choose n: it is now an - n, or a0, so it just disappears; and bn.0684

We can now expand out the whole thing with relative ease.0695

It will definitely still take some effort; we will have to figure out what each of these "choose" values comes out to be.0698

But we can write the whole thing out, expanded, and we will be able to get it out.0702

And it will still be easier than doing each term, and then multiplying (a + b) and (a + b) and (a + b) and (a + b), and trying to FOIL the whole thing.0705

Expanding factors like that would just be a real pain, if it was a large thing, like (a + b)10.0713

So, this is really useful if we are trying to expand a very large exponent, n.0718

Also, we can condense this in a really nice way.0723

This is a monster of an expansion; it would be a pain if we had to write this repeatedly.0726

We can use sigma (or summation) notation to condense this into a much smaller thing.0729

(a + b)n is equal to the sum of k = 0 to n of n choose k times an - kbk.0735

If you want to check and make sure that that works--well, if we plug in k = 0, sure enough, we end up getting this term.0744

n choose 0: an - 0 is an; b0 just disappears, so we end up getting that.0750

And we work all the way through; here would be k = n; n choose n; an - nbn; we end up getting n choose n bn.0757

And we end up seeing all of the terms here show up; there would be k = 1, k = 2, k = 3 next, k = 4...until we finally get to k = n - 2, k = n - 1, k = n.0766

We can write the whole thing with sigma/summation notation.0778

If this is really confusing, if you really haven't done much work with sigma/summation notation,0780

remember to check out the lesson Introduction to Series; we will talk about this.0785

We will have a much better understanding of sigma/summation notation (that is a mouthful) and how it works through.0788

So, make sure that you check out that lesson if this is really confusing and you have to use it.0796

Proof of the binomial theorem: Proving the binomial theorem is within our reach.0802

We could prove this; it is quite challenging, but it is an interesting proof.0806

And it will give us the chance to try out our shiny new proof-by-induction.0810

We will use proof by induction to prove this thing.0814

Working through it will really deepen our understanding of how mathematics works.0816

We will get a much better understanding of how some really complicated stuff in math works, by just working through it.0819

However, proving it won't directly help us see how to use the theorem.0825

If we work through it, it won't directly help us in any way for the kind of problems that you are likely to have to work on right now.0830

It will help us understand why the theorem works, but it is not likely to actually make anything easier that we actually have to do in class.0836

As such, the proof for the binomial theorem is in this lesson; it is here; but it is at the end, after the examples.0843

If you are interested, I encourage you to go ahead and check it out; it is some cool stuff.0852

You will get the chance to really flex your math muscles.0855

We are basically going to be looking at a college-level proof.0858

But I believe that, if you are interested in this sort of thing, and you put a little bit of effort0860

into watching it and working through it, you will totally be able to understand it.0863

So, if you are interested in this kind of math, go ahead and check it out; it is a really cool thing.0867

We will spend a while working through it and explaining it carefully.0870

But at the same time, if you are busy, or you don't really care--you are not that interested in math--0873

you are just working through this so that you can get a grade in your class, well, I would like for you to be interested in math;0877

but I am not going to be able to always make you interested in math.0882

So, I just want you to do well; but to be honest, you don't actually have to understand how this proof works0884

to do well in a math class at this level, or even in the next level.0888

This is more something for if you think you might be really interested in serious math or serious-level science courses.0891

You probably want to check this out; but that is still even going to be a couple of years away, if you are currently in a precalculus class like this.0896

All right, so check it out if you are interested; but if you are not, don't worry about it.0903

Pascal's triangle: we can arrange the binomial coefficients in a triangular pattern to make a thing called Pascal's triangle.0907

It is named after one of the people to have found it and discovered it.0914

We can have n = 0 as our first row (well, the 0th row, we will call it), n = 1 as the next row (the first row),0917

n = 2 as the next row (what we will call the second row), n = 3, n = 4, and so on, and so on.0925

Notice: we have our n value at the top each time; so it is n = 0 and then n = 1, n = 2, n = 3, and so on and so on...0930

And then, we always have 0, and then 0, 1, 0, 1, 2, 0, 1, 2, 3...until it finally ends up having the number on top match the number on bottom.0940

And that is how we end up making the triangle.0952

Why is this interesting? We are going to see a really interesting pattern emerge when we replace each n choose k,0954

each of these like this, with what the value is that ends up coming out of it.0961

What does 2 choose 0 come out to be? We replace it, and we start to see a pattern emerge.0965

We replace it, and we end up getting this: n = 0 gets 1 for its row; n = 1 gets 1 and 1; n = 2 gets 1 2 1.0970

n = 3 is 1 3 3 1; n = 4 is 1 4 6 4 1; you might have seen this before.0978

Notice how each row can be created from the row above through diagonal addition of the terms.0983

For example, if we go out from n = 0, and we go out diagonally left and diagonally right,0989

1 added down would end up getting us 1 and 1.0995

Then, if we add diagonally again, down left, down right, and then the same thing for the other one--0999

down left and down right, well, 1 +...only by itself would come out as 1.1005

1 + 1 would get us 2; it is this one and this one, combined together, because it has both of the diagonals coming in through here.1009

And then, 1 only on itself would get us 1 here.1018

And we can keep going along with this pattern: 1 diagonally out, 2 diagonally out, 1 diagonally out...1 just by itself would get 1;1020

1 + 2 would get us 3; 2 + 1 would get us 3; 1 by itself would get us 1.1028

And we can keep going in this pattern: 1 on itself would get 1; 1 on 3 is 4; 3 on 3 is 6; 3 on 1 is 4; 1 by itself is 1.1033

So, by doing this, we can extend the triangle down to whatever value of n we are interested in considering.1045

Whatever value of n we are interested in considering, we can just get to by diagonal addition of terms.1052

For example, if we are interested in knowing what n = 5 is, then we just expand out diagonally like this, through diagonal addition.1058

And we can work it out: 1 added to itself just gets 1; 1 added to 4 gets us 5; 4 added to 6 gets us 10;1066

6 added to 4 gets us 10; 4 added to 1 gets us 5; 1 added to itself gets us 1.1073

And we have been able to create the row n = 5, the fifth row.1079

Since the first row is based off of n = 0, it is called the 0th row.1084

So, here is the 0th row; this will help keep us from having confusion later on.1089

The next row is called the first row, because it is based off of n = 1; so here is the first row.1094

We might think that the first row would be the one on top, but not with Pascal's triangle,1101

because it is really based off of n = 0, and then n = 1.1104

So, the second row would be the one connected to n = 2.1109

It is based on the number of the n, not the actual location of the row, so much as if you were just counting location of rows.1114

It is based on the n that it is connected to: n = 4 would produce the fourth row, and so on and so forth.1121

Why do we care about Pascal's triangle--why is this thing useful?1128

Because the nth row gives the coefficients to expanding (a + b)n.1131

The way that we set this up in the first place was by using those binomial coefficients.1138

Remember: all of those binomial coefficients, the n choose k--we had that set up based off of which n we were on.1142

So, what we did was put down all of the binomial coefficients for a given n value,1148

which means what (a + b)n is...we have just talked about all of the coefficients for expanding some (a + b)n.1152

It is really handy, if we have to expand the entire thing.1159

For example, since the nth row gives the coefficients1162

to expanding (a + b)n, we can expand (x + y)4 using the triangle.1164

If it is (x + y)4, then that means we are dealing with n = 4; so we care about this row of the triangle.1169

We work this out, and that means that it is going to be (x + y)4.1176

Well, we know that it is going to be some blank, and it will start with x4y0...1179

plus _x3y1 (y goes up by 1; x goes down by 1) + _x2y21184

+ _x1y3 + _x0y4.1194

At this point, we can now plug in each of our binomial coefficients: 1 will go here; 4 here; 6 here; 4 here; 1 here.1202

And we can also write the thing a little bit more simply.1213

1x4y0: well, 1...we normally don't write a coefficient of 1; so we can just write this as x4,1215

because y0 also just turns into a 1, and we don't normally write out the 1's.1221

x4 + 4x3y1...we normally just write that as y, so we can write that like that;1225

plus 6x2y2 + 4x1 (we normally just write it as x) y3,1232

plus 1x0...well, we normally don't write 1's; x0 just becomes 1, so it will disappear as well;1239

and we are left with just y4; so we end up being able to expand all of (x + y)4,1247

using Pascal's triangle, which tells us the binomial coefficients for the associated n row.1252

We can use that triangle, and we can expand things pretty easily.1259

If we had tried to expand (x + y)4 by hand, we would probably still be doing it.1262

It takes a while; and this was even through explaining it.1266

If we were just punching it out, we would be able to do this really quickly.1270

So, Pascal's triangle is really useful for when we have to expand some binomial quickly and easily, without having to take a whole bunch of time--pretty useful.1273

All right, we are ready for some examples.1282

Use the binomial theorem to give the coefficients for the term containing s4t7 from (s + t)11.1283

The first thing that we want to identify is what the n we are working with is.1291

The n that we are working with is 11.1294

All we care about is s4t7; we are asked to find the binomial coefficient for that term when we expand (s + t)11.1297

We don't really care about all of the other terms; all we really care about is s4t7.1308

What is that going to be? That is going to end up being some n choose k.1312

And if we wanted to have that, it would also be s4t7.1318

So really, at this point, all that we need to find is what k is.1321

k is what? Well, it is normally...from the binomial theorem, we would have n choose k; an - kbk.1325

So, in this case, we have n choose k, but we are not using a and b; we are using s and t.1338

So, it will be sn - ktk; what is our n?1346

We have 11 choose k, times s11 - ktk.1351

Well, if we have t7 and s4, then it must be that k has to be equal to 7.1358

11 - 7 gets us 4, and k = 7 gets us t7.1368

So, we now know what k is; we now know what n is; so we can plug this in.1373

We have 11 choose 7, times s4t7; that will be the s4t7 term.1376

What comes out to be 11 choose 7? Let's go do that in a sidebar.1383

11 choose 7: well, that is going to be 11! over (11 - 7), 4, factorial, times 7!.1386

11!...well, we can write that as 11 times 10 times 9 times 8 times 7....1398

But we can also just write it as times 7! now; that is convenient, because we have 7! on the bottom.1405

So, 4! times 7!...we cancel out the 7!'s, and we have 11 times 10 times 9 times 8.1410

On the bottom, we can expand 4! as 4 times 3 times 2 times 1.1419

We will just leave the 1 off; 2 and 4 become 8, which cancels the 8 on top; 3 cancels with part of the 3 of the 9,1424

so we have times 3, because we cancel out one of the 3's from the 9.1431

11 times 10 times 3 comes out to be 330; 330 is our coefficient, so we get 330 times s4t7.1434

That is what we get as the coefficient of the term containing s4t7.1446

The coefficient is just the 330 part, and the whole term would be 330 times s4t7, when we expand (s + t)11.1450

Now, could you imagine how difficult that would be to do by hand?1460

That would be practically impossible for us to do by hand, because it would take so much time and effort.1462

But by using this n choose k business, we are able to figure out what the coefficient is relatively easily.1467

Use the binomial theorem to give the coefficient for the term containing x10y12 from (x2 + y3)9.1473

The first thing that I want to point out: x10y12: well, what we talked about before was:1480

if you take the exponents, and you add them together, well, what is our n?1486

Our n here is n = 9; but if we add 10 and 12 together, 10 + 12 is 22; what?1491

Wait, that doesn't make sense; what is going on here?1500

The issue isn't that it is x10y12, because what we really have is a and b.1502

But what is our a and b? Our a and b are x2 and y3; a = x2; b = y3.1508

So, it is (a + b)9; so when we expand this out, what we are looking at is n choose k on an - kbk.1518

But now we have called out a and b as x2 and y3, because that is what the actual two terms of it are.1535

It is not just x, and it is not just y; it is x2 and y3.1541

So, we have some n choose k; we will plug those in later.1546

And so, it is x2 to the n - k, and b--what is b?--b is y3, to the k.1549

So now, we need to figure out what value k has to be for us to be able to get x10y12.1561

Well, if we have y3 to the k, and we know that that ends up being the same thing as y12,1569

well, then k has to be equal to 4, because (y3)4 would be 3 times 4, so y12.1577

So, we end up getting y12 from setting k equal to 4.1584

And similarly, 9 minus 4, since it is an - k here...9, our n value here, minus 4, our k value here, would get us 5 for n - k.1587

And x2 to the fifth does come out to be x10; 2 times 5 is 10.1601

At this point, we now see what our k is; our k is 4; our n is 9; so we can plug those in.1607

And what we are looking at is 9 choose 4, times x2 to the n - k, so in this case 5, y3, to the fourth.1615

So, that is x2 to the 5; that does become n; y3 to the fourth does become 12.1630

And 5 and 4 put together do become 9; so everything checks out--this makes sense.1635

At this point, we only need to figure out what 9 choose 4 is.1639

9! over (9 - 4), the top minus the bottom, that is 5, factorial, times just the bottom, factorial:1643

9!--we can write that as 9 times 8 times 7 times 6 times 5!, over 5!; look, that is convenient; they cancel;1653

times 4; and now let's expand 4!, as well: 4 times 3 times 2...we will omit the 1, because it doesn't do anything.1663

5! and 5! cancel; 3 and 2 cancel out the 6; 8 cancels with a 4 to make a 2 on top.1670

So, we have 9 times 2 times 7; 9 times 2 times 7 comes out to be 126.1677

There is our coefficient right there; or if we wanted to write out the whole thing, we would end up getting 126,1684

times x...not to the fifth; not squared; but x10, what it was, because it said the term containing1690

x10y12, not the one where we have raised that part of the binomial to the fifth, but x10y12.1696

And what it asked for was the coefficient; the coefficient is 126; times x10y12.1704

Great; the third example: Find the value of the coefficient c on the term cx8 in the expansion of (2x2 - 3)7.1711

Once again, we need to realize that this is 8, but what we have is an n = 7.1720

So, we realize that it is 2x2; that is the thing we are dealing with.1728

So, we have a = 2x2, and b equals...not just 3, but it has to be including the negative as well,1732

because normally it is a + b; so if it is a + b normally, but now it is a - b, then it must be that the b also has a negative inside of it.1743

So, our b is equal to -3; OK.1751

At this point, we want to figure out what number we have to raise 2x2 to, to get x8 to pop out of that.1754

What number do we raise that to? If we have x2 (we will consider just the x2 for a moment,1764

because the 2 won't affect what exponent the x has) to the ?, is equal to x8;1771

well, then our question mark has to be 4, because 2 times 4 comes out to be 8.1781

So, x2, raised to the 4, equals x8.1786

We know that whenever we do this expansion, n choose k, times 2x2, our whole a, to the n - k,1791

times our whole b, -3, to the k; now we need to figure out what our k is.1801

Well, our n was 7, so our k must be 3, if we are going to produce a 4 there.1807

n - k has to be 4, because we know that this value here has to match up with n - k here.1815

n - k there has to match up, as well; so our 7 is what we have for n; 7 - k = 4; 7 - k = 4 means that k has to be 3.1823

Now, we have figured out that k has to be 3; we are ready to plug everything in.1834

n choose k: 7 choose 3, times 2x2 to the n - k (comes out to be 4); -3 to the 3.1839

n - k at 4...2x2 to the fourth; sure enough, that would make some constant times x8.1855

-3 to the 3: that is just going to also be part of that constant term, c.1862

At this point, let's figure out what it is; let's solve 7 choose 3; what does that value come out to be?1866

7! over (7 - 3), 4, factorial, times 3!...we have 7 times 5 times 6...sorry, 7 times 6 times 5; I'm not sure why I confused that order;1872

it doesn't matter, but I don't want to make you think that something weird happened;1885

7 times 6 times 5, over 4!, cancelled out with all of the rest of those...we have 3! on the bottom; that is just 3 times 2 times 1.1888

We will omit that; 3 times 2 cancels out with the 6; 7 times 5 gets us 35.1895

We can now swap that in for our 7 choose 3, and we have 35 times...1900

2x2 becomes 2 to the fourth, x2 to the fourth; 24 is 16; x2 is x8.1905

-3 cubed is -27 (-3 times -3 times -3); we use a calculator to multiply 35, 16, and -27, and we get -15,120x8.1913

That means that our c must be equal to -15,120, because what we are looking for1930

was the coefficient c on the expansion of cx8, once the thing is fully expanded.1937

All right, great; we are ready to move on to Pascal's triangle, the fourth example.1943

Use Pascal's triangle to expand (p - q)6.1947

If we are going to the sixth, then that means n = 6; we are going to have to get down to the sixth row,1951

which is really the row that is 7 down (if we say that the row at the very top is one down); but we will just write n = 0,1956

so we don't get confused by this; so our 0th is at a 1, and then n = 1 is the next one: 1 1;1965

n = 2: 1 2 1; n = 3: 1 3 3 1; n = 4: 1 4 6 4 1; n = 5: 1...1 + 4 is 5...4 + 6 is 10; 6 + 4 is 10; 4 + 1 is 5; 1 on itself is 1.1975

And finally, the one we care about, n = 6: 1 on itself is 1; 1 + 5 is 6; 5 + 10 is 15; 10 + 10 is 20; 10 + 5 is 15; 5 + 1 is 6; 1 on itself is 1.1999

Those are all of the coefficients that we care about at this expansion.2011

All right, if we are going to expand (p - q)6, we know that it is going to end up having some expansion2016

that looks like blank, this one first...well, this one is p; and this one, though, is actually -q; so it is going to be _p62023

times (-q)0; and then we work the q part up and the p part down, one step at a time:2037

p6 becomes p5 on our next step; (-q)0 becomes (-q)1;2045

plus _p4(-q)2...and just continue on the next line...+ p3 times (-q)3...2050

oh, there should be a blank after that plus, right here; + _p2(-q)4 +2064

(I will just bring this down to the next line, once again) _p1(-q)5 +2077

_p0(-q)6; great--at this point, we can now swap in our values for our binomial coefficients.2086

So, we know 1 will go here; a 6 will go here; a 15 will go here; a 20 will go here; a 15 will go here; a 6 will go here; a 1 will go here.2094

See how they come in symmetrically?2108

At this point, we will now write it all on this line around here.2110

And we will also simplify it: 1p6(-q)0: well, 1 just disappears;2115

we have p6; and (-q)0--well, if you raise anything to the 0,2120

it just becomes 1; so we will have that disappear, as well.2124

6p5(-q)1: well, that -q is going to show up now; it is going to come in.2126

So, it is not plus; it is now minus, because it is -q to the 1; so - 6p5q.2132

That -q here doesn't get cancelled out, because there is just one negative.2141

And next, we will have + 15p4(-q)2; negative on negative cancels out, so we do have positive q2.2145

Plus 20p3(-q)3; negative, negative, negative means we still have a negative.2156

So, it is not a plus; it is a minus, because of the (-q)3.2167

So, we have q3; next, plus 15...that was the last one...this is what we are working on now:2171

15p2(-q)4: well, the negatives will end up canceling there, because it is an even number, q4.2179

Plus 6p1(-q)5; well, (-q)5 is an odd, so it will be minus 6pq5.2186

And then finally, (-q)6, with a coefficient of 1, is our very last term.2196

We just did this one here as our last term: 1p0...those things both disappear, and we have +q6.2201

And the negative disappears, because it has an even exponent; great.2207

And there is our full expansion; it is still not super fast, but it is much, much better than if we had tried to expand that whole thing out by hand.2212

If we did that whole thing by hand, it would be taking forever; but this goes relatively quickly, by using Pascal's triangle.2221

The final example: Use Pascal's triangle to expand (√3u2 + 1/2√p)4.2227

At this point, what is our n? Our n equals 4, so we need to get down to the n = 4 row.2235

We start at n = 0: 1; n = 1: 1 1; n = 2: 1 2 1; n = 3: 1 3 3 1; n = 4 (finally, the row that we actually care about): 1 4 6 (3 + 3 is 6); 3 + 1 is 4; and 1.2241

Great; so that is the row that we will actually end up using, because our n equals 4.2264

So that we don't end up getting cramped, I am not going to end up writing here.2268

I am going to end up doing the expansion down here.2271

Our first thing will be some blank, times...well, what is our a?2275

That is the good thing to identify: √3u2 is what our a is equal to; a + b to the some n...2279

so in this case, the a is all of √3u2, and the b is all of 1/2√p; OK.2287

We are now ready to do this; the first one will be √3u2 to the n = 4.2296

And then, 1/2√p to the 0...I won't even write that one there;2304

plus _(√3u2)3(1/2√p)1 + _(√3u2)2(1/2√p)22308

+ _...let me break that down to the next line, just so that we can have plenty of room...2326

+ _(√3u2)1(1/2√p)3 + _(√3u2)0,2331

so we can just make that entire thing disappear; times (1/2√p), all to the 4.2348

Great; I want you to notice here: Don't get confused by the fact that it is u2 or √p.2357

The fact that there are exponents on stuff inside of the binomial doesn't matter.2362

We still do this part where the one exponent steps down with each step, and the other exponent,2366

the one that starts at 0, steps up with each step of the terms that we work through.2374

Great; at this point, we can now plug in our binomial coefficients:2379

a 1 for our first blank, a 4 for the next blank, a 6 for the next blank, a 4 for the next blank, and a 1 for our final blank.2382

At this point, we can now finally actually simplify this one; well, let's do it in two steps.2390

1...we will just have that disappear; (√3u2)4: well, √3 squared is 3; so √3 so the fourth is 32.2395

So, √3 to the fourth is the same thing as 9.2403

u2 to the fourth is u8; plus 4 times...√3 cubed is going to be 3√3.2406

(u2)3 is u6; (1/2√p)1 is just 1/2√p.2414

Plus...6(√3u2)2: √3 squared is 3; u2 squared is u4.2421

Times...1/2√p squared will be 1/4 (1/2 squared is 1/2 times 1/2, so 1/4); √p squared is p.2430

Break the line once again; plus...next is 4(√3u2)1; well, that just stays as √3u2.2441

Times...(1/2√p)3: 1/2 to the 3 is 1/8; times √p cubed is p√p.2452

And then finally, plus 1 times 1/2√p to the fourth; that is going to be 1/16; (√p)4 is going to come out as p2; great.2462

Now, we can simplify this whole thing and do the last of the simplification.2473

9u8 +...4 times 3√3 times 1/2...1/2 cancels the 4 down to a 2;2477

we are left with 6√3u6√p, plus...we have 1/4 here, so this will become 1/2;2486

this will become 3; 6 times 1/4 becomes 3 times 1/2; so 3 times 3 is 9, divided by 2...we have 9/2 as the coefficient there, u4p.2499

Plus...4√3u21/8p√p...that is a lot of things; so 1/8...that will cancel to 2,2514

when it knocks out this 4; and so we have (√3/2)u2p√p.2523

And finally, our last term is 1/16p2.2531

And there is the full expansion; it is about as difficult as any expansion with Pascal's triangle will end up being.2537

But still, notice how much faster that ended up making it than if we had tried to do this whole thing by hand--2544

doing √3u2 + 1/2√p, times √3u2 + 1/2√p,2549

times √3u2 + 1/2√p, times √3u2 + 1/2√p...2553

We would still be working through it, and still have a lot more to go.2557

So, it makes things faster.2559

We have to be careful, and make sure that...there are a couple of things that we have to pay attention to.2560

We really have to make sure that our n, the row that we are using of Pascal's triangle, matches to the exponent we are raising to.2564

We have to pay attention to what our whole a is and what our whole b is.2570

It has to be the entire term, not just the u, not just the p, but the entire term of something + something.2575

And if it is a minus, it has to be something plus a negative something.2582

And then, we set it up with this step down/step up pattern, the blanks; and then, we can just slot everything in.2586

If it is a simpler problem, you might be able to do it without having to take such careful steps.2591

But if it is a big, complicated problem, I really recommend doing the careful steps.2594

It will make it easy to not make a mistake and make the problem just slow and easy.2598

All right, that finishes the examples; if you are leaving now, we will see you at Educator.com later; goodbye!2602

But if you are staying around for the proof to the theorem, it is time for the "bonus round."2608

All right, we are ready to tackle a proof of the binomial theorem.2620

First, let's phrase the theorem as, "For any n = 0, 1, 2...and so on and so on" so that is all natural numbers,2623

"(a + b) to the n is equal to the sum, going from k = 0 to n, of n choose k times an - kbk."2632

So, notice that that is just putting all of the binomial coefficients added together, for a fully expanded term.2641

We are going to prove this by induction; so, just in case you haven't watched the previous lessons,2646

but you are really interested in this proof, we will be using mathematical induction.2651

And we will also be using a fair bit if sigma (that is, summation) notation.2655

So, if you are not familiar with mathematical induction or sigma/summation notation, watch the lesson on mathematical induction,2659

and watch the lesson on Introduction to Series, where we will talk about sigma/summation notation,2664

because those things will definitely be necessary prerequisites.2669

And before we get into this, I really want to stress: what we are about to go through is a pretty challenging proof.2671

This is definitely a college-level proof for a reasonably good math class.2676

So, don't be shocked if you find this a little bit challenging to work through.2682

This isn't going to be super easy to understand the first time; and that is perfectly OK.2688

Math is a puzzle to be worked through and understood over time.2692

So, watch through this; if you have difficulty understanding what is going on, moment-to-moment,2696

I highly recommend that you take a piece of paper and actually write what is happening on-screen2699

as we are working through it in the lesson.2705

By working through it yourself, you will be able to understand the steps better,2707

because you will have to internalize them, because you yourself will have to be understanding and working through it.2711

If at some point a step happens where you have no idea how that step happens,2715

pause the video and try to understand what just happened on your paper.2718

Figure out how you can get from one step to the next step.2722

If you still can't figure it out, back it up; hopefully I will explain through it.2724

I will explain a lot of this; we will be working through it slowly and carefully.2727

But a really great trick is to work through it on paper.2731

It really makes things make a lot more sense, just getting it down on paper, so that is not just "rattling around" in your head.2733

You can actually see it, think about it, and have it in front of you.2739

It is a really great way to work.2742

I really recommend this any time that you are trying to work through a math textbook and things get confusing in the math textbook.2744

Just take a piece of scratch paper and write what is going on in the math textbook.2748

Writing it out for yourself, understanding it step-by-step, when you write it out in your own hand,2751

will end up making most issues that you end up having clear up so much faster,2756

because you are having to work through it yourself; you are not just trying to read someone else's language.2760

You are putting it in your own language before you are moving on.2764

All right, let's get started; this is really good--this is what math is really about, in my opinion.2768

Math lets us do all sorts of things; but for me, the cool thing about math is the fact that is puzzles,2774

that it is logic; it is something that we can believe in, and it is just there--real logic, real proof.2779

This stuff actually comes together and can be shown very clearly.2786

All right, let's get to it: proof by induction: begin by noticing how nicely the theorem lends itself to being written2789

as some pn statement, some statement for some value of n.2796

We have n = 0, 1, 2...so it is just stepping up, one at a time, at (a + b)n equals this sum here.2801

That is what the binomial theorem says; so we can just say that the nth statement is (a + b)n equals that sum; great.2808

It is really easy to write out in this statement idea.2815

Our base case: now, before we prove the base case, notice that there is this slight issue.2819

The theorem doesn't start at n = 1; it starts at n = 0.2823

When we first talked about mathematical induction, we always set our base case at n = 1.2829

But with this one, we can't set our base case at n = 1; we actually need to start at p0.2833

So, it is not p1, but p0, because the theorem starts at n = 0.2838

If we think about this, though, we realize that this isn't really an issue.2843

It is just that we need a starting point to work out of.2846

It needs to be the first stepping-stone that we then walk forward from, using that inductive step.2849

That is what the inductive step is: that is taking one step after another,2854

and guaranteeing that future stepping-stones will be there, as long as we have this first stepping-stone.2857

So, it doesn't really matter if this stepping stone starts at n = 0 or starts at n = 15 or starts at n = 1.2862

It just needs to be some integer value that we can then step forward from.2867

n = 0 is a perfectly fine thing to set our base case at.2871

We work through this: a base case of p0 would end up giving us (a + b)0.2876

Is that equal to (we are saying this as a question; we don't know if this is true yet;2881

we are checking to make sure that the two sides are true)...it would be equal to the sum2885

(if this is true) of k = 0 summed to 0 of 0 choose k, a0 - kbk.2889

So, our left side...well, (a + b)0...anything raised to the 0 just comes out to be 1.2896

Our sum, k = 0, going up to 0...well, that means that where we start is where we stop.2902

So, there is only one term in this series when we expand this summation, this sigma notation.2908

It just turns into 0 choose 0; we plug in our k; we plug in the k here, 0 - 0, and plug in the k here, b to the 0.2912

So, what does that come out to be?2920

Well, a0 - 0 just comes out to be 1; b0 just comes out to be 1.2922

0 choose 0 is going to be 0!/(0 - 0)!, so 0! times 0!.2927

So, we have this here, 0!/0!(0!): well, remember: when we set up 0!, we just stated 0! = 1.2932

We simply define it that way; that means 0!/0!(0!) turned out to be 1, so our entire right side is 1.2941

Our entire left side is 1, as we already figured out early on.2950

So that means, indeed, that this does check out; the base case is true; we are good with the base case.2953

On to the inductive step: now, here is where things get a little bit tricky, just for understanding notation.2959

There is a minor issue with notation here: when we first saw induction, we always had some general statement, pn, this (a + b)n = stuff.2966

Now, what we did for the inductive step was assumed...we said, "If pk is true,2978

then we have to show that pk + 1 is true."2984

If this thing is true, then the next one must be true.2987

But we have this problem; the way that we usually write the binomial theorem, we use k.2991

k shows up as the index in the sum; we are using this all over.2998

So, we cannot use pk; this means we cannot use pk...3003

if we use pk, we will end up having to re-index and change around variables.3010

So, we cannot use pk without re-indexing and changing around variables.3016

If we had to use pk and then pk + 1, we would have to come out with some new variable3019

to swap out for all of these k's up here, and then it would get confusing,3023

because that is not how we are already used to ending up working with the theorem.3026

And so, it is going to change its ways, and it just makes things confusing.3029

What we are going to do, to avoid confusing ourselves with new variables:3033

to avoid confusion, we are going to prove the following for this step:3037

if pn is true for some n, then we will show that pn + 1 is true, as well.3041

So, up here, this is the general statement of pn, and we are trying to show that it is true for all n.3048

Any natural number n is what we are trying to show for our general statement.3054

But what we are going to do is say, "Sure, let's assume that it is true for some specific n."3058

We don't need to name what that specific n is; but it is just one value of n.3063

And then, we are going to show that, if it is true for that one value, it has to be true for the next value, n + 1, as well.3066

This is OK; before, we talked about pk and then pk + 1.3072

But it doesn't really matter what symbol we use; all that matters is that, if it is true at one step, it must be true at the next step.3076

If it is true here, it has to be true at the next step.3085

If it is true at pk, then it has to be true at pk + 1.3088

If it is true at pz, it has to be true at pz + 1.3091

If it is true at pn, it has to be true at pn + 1.3095

That is different than our general statement of "it is true for all n, forever."3098

We are saying, "Let's say that it is true for some specific n; and then let's show that it has to be true for the next one in line."3102

This might be a little bit confusing; but really, it is OK to end up using this pn, then pn + 1,3109

because what we are not doing is assuming that it is always true for all n.3113

We are just saying, "Let's say it is true for some n," and then we will show what would happen for the next n, n + 1.3117

And that is what we will do to avoid having to get all of this notation confusion3122

of having to use letters other than n, of having to use letters other than k,3125

because then it would be not what we are used to seeing.3129

And while we could work through it, it makes it a lot easier if it is something that we are used to seeing, used to working with.3131

We will end up avoiding notation confusion by keeping n and k in this,3136

and just sort of knowing what is going on, on a personal level.3140

All right, we are ready to actually work through this.3144

Let's set this up: for our inductive step, we start by assuming our inductive hypothesis.3146

We assume that, for some n, pn is true; that is, (a + b)n is equal to the sum of k = 0 to n of n choose k times an - kbk.3151

Now, what we want to show is that pn + 1 must also be true of pn is true.3163

Assuming pn, pn + 1 must be true; that is what we are working towards.3170

We don't know it yet; we are working towards it.3174

So, how would pn + 1 be stated? Well, that would be stated as (a + b)n + 13176

equals...and now we swap out the n here for n + 1, and the n here for n + 1, and the n here for n + 1;3182

and that would be the pn + 1 version; up here is the pn version.3190

So, what we want to show is that this here is true.3194

Thus, if we can show that the left side of pn + 1 is just the same thing as the right side--3199

that they are just two different ways...writing the left side and the right side, that they are clearly equivalent,3207

by manipulating it through "Well, we can do this, and we can do this, and we can do this..."3211

and they are all clearly equal, and we can make it from one side to the other side; we will have shown that this statement is true.3215

If we can show that the left side is the same thing as the right, we will have shown that, indeed, the statement is true.3221

So, what we want to do is see how we can manipulate (a + b)n + 1.3227

Well, in a proof by induction, we always want to toss our inductive hypothesis into this.3231

So, our first question is, "Is there a way to turn (a + b)n + 1 into something where we can use our hypothesis?"3234

We want to see if there is a way to turn (a + b)n + 1 into the right side of pn + 1.3244

Our first question is going to be, "How can we apply our hypothesis--how can we apply pn, (a + b)n = that sum?"3249

We look at it: if we look at (a + b)n + 1, it is pretty easy to get (a + b) to the n to show up; we just split it:3257

(a + b)n + 1 is just the same thing as (a + b)(a + b)n.3264

At this point, we can now swap out (a + b)n for the other side of our inductive hypothesis.3269

We swap it out; we now have (a + b) times the sum k = 0 to n of n choose k (an - kbk).3276

We have just swapped out for our inductive hypothesis.3282

Now, at this point, we have a + b over here; we can distribute that: a + b distributes onto that sum.3285

And we will have a times the sum, plus b times the sum.3293

Well, a and b don't vary once we have expanded the thing; they just end up being the same thing.3300

So, what we can do is actually bring them inside of the sum.3304

They have no direct connection to the index, so we can bring them inside of the sum.3307

So, a will apply onto the an - k; an - k times a will be an - k + 1, which we can write as an + 1 - k.3312

The same thing happens over here: b applied to bk becomes just bk + 1; great.3327

At this point, let's expand these sigma notations.3335

The sigma notations are a great way of keeping things compact, but it is a little hard to tell what is going on inside of the notation.3338

So, let's expand them, so we can see what is going on better.3344

If we expand this, we will have k = 0 here; we will plug that in here first, so we will have n choose 03346

times an + 1 - 0, so an + 1, and b0 (that just disappears).3352

And then, next, we will go up one step, and choose 1; anb...great.3358

And we will work our way up, until finally we get up to n; and so, we will have n choose n.3362

And it will be an + 1 - n when we are plugging in n.3367

an + 1 - n gets us just 1, so we have a to the 1 here; and bn now, because we are swapping in n for k; and we have bn here.3372

Great; so that is what we end up getting from this sum on this side.3381

And we can do basically the same thing and expand this sum over here.3386

We start at k = 0; we will get n choose 0, anb1 (if we plug in k at 0, we will have 0 + 1, so b1).3388

We work our way up; we get to n - 1, one before our last one.3397

n choose n - 1; we will have an - n - 1; if you are not sure why that comes out as n - n - 1,3401

well, this becomes minus n plus 1, so we get a1 here.3408

k + 1...well, if we are plugging in n - 1 for k, n - 1 + 1 gets us abn.3412

And then finally, our last one, when we plug in n: swap that in for our k; n - k (n - n for our a here) will end up just disappearing entirely.3422

We will have a to the 0, so we just omit that.3430

We will have n choose n, and b, plugging in k for n...n + 1 here, so we get n + 1 here.3432

So now, we have expanded these two sums; now we can start working with them.3439

What we want to do first is notice that there are some similarities between these things.3443

In fact, we have this really great similarity going on here: anb and anb match up.3447

abn and abn match up; in fact, we would end up having this match-up3454

between all of the things inside of those ellipses; the pattern of matching will continue throughout.3459

The only things who don't fit this matching pattern are the ones at the far ends, the an + 1 and the bn + 1.3464

Notice: the only terms that don't match up of these a to the something, b to the something terms are the first and last terms, respectively.3274

If we separate them out, we can relate the sums; we can have this connection.3481

We pull them out: n choose 0, an + 1; n choose n, bn + 1 on the two ends.3484

And then, we end up having the matching going on here: n choose 1, anb;3492

that will match up to n choose 0, anb.3496

And then, n choose n, abn, will match up to abn with an n choose n - 1 times abn.3499

We have this nice matching going on here.3507

The matching will also continue in here with the same thing of the top3510

being above the bottom by 1 on its bottom and above the bottom by 1 on its bottom, every time; great.3514

So, if that is the case, we can combine the parts in brackets.3520

If n choose 1, n choose 0, are both multiplied by anb, well, then we can combine them: anb...3523

and we end up having n choose 1 plus n choose 0, all of that times anb.3532

We pull out the anb when we put them together.3539

The same thing happens over here: n choose n, abn, n choose n - 1, abn:3541

we pull out the abn, and we end up getting n choose n plus n choose n - 1, that quantity, times abn here.3547

And inside of it, we will end up having the same pattern go on with the n choose something plus n choose (1 less than that something),3553

times the ab stuff; so that pattern will end up continuing throughout.3561

And we still have our things on the end; if you are not sure how they transformed from n choose 0 and n choose n,3565

to just the a and b, well, n choose 0 is just 1 (if you are not sure about that n choose 0, well,3571

that is equal to n! over (n - 0)!, so n!, times 0!; 0! is just 1; n! and n! cancel out;3578

so we end up just getting a 1 here, so it disappears and we are just left with an + 1).3587

The exact same reasoning works for n choose n becoming 1, as well, on the bn + 1; and we get bn + 1 here, as well.3591

At this point, we can put this giant sum in sigma notation.3600

an + 1 just stays around; bn + 1 just stays around.3604

The thing we care about is this giant sum here, which we broke on lines, because it is too much to write on a single line.3607

We can do that as the summation from k = 1 to n of n choose k plus n choose k - 1, as a quantity, times an + 1 - k times bk.3614

Let's check and make sure that this works.3627

If we did, indeed, plug in k = 1, yes, that checks out; 1 for k here, k - 1, 1 - 1; that does get us 0.3628

an + 1 - 1 would get us an; b1 would get us b; so that checks out there.3637

If we were to do n, that would end up checking out here, as well.3644

This is probably one of the hardest steps to see.3646

If you work through it slowly, you can see that, yes, what we are doing is that this whole sum3649

is just the same thing as what we have this in these giant brackets, this giant bracketed-off portion.3654

That is the same thing here; so we can put the whole thing in sigma notation.3663

At this point, let's check in with our original goal.3667

We have gotten pretty far with this thing; what we have is an + 1 plus the big sum plus bn + 1.3669

Now, compare that to what we are trying to get the whole thing to look like.3677

We started with the left side, and we are trying to get it to turn into what its right side was.3681

And its right side was a summation: k = 0 to n + 1 of n + 1 choose k, times an + 1 - kbk.3684

This stuff here matches really closely to this.3693

And n + 1, k is not too far off from what is here.3697

And n and k = 1...that is not too far off, either.3700

There are some similarities; we are starting to get there.3704

What would be great is if we could somehow show that what we have here is the same thing as what we have here,3706

that n choose k + n choose k - 1 is just the same thing as n + 1 choose k.3714

If we could somehow show that that is the same thing, we would be really close3721

to showing what we are trying to show-- what we want to show--getting to our goal.3726

So, since that would be so useful if it were true--this n choose k plus n choose k - 1 equals n + 1 choose k-- let's go ahead and try to prove it.3730

What we will do is create a lemma; a lemma is basically just a mini-theorem.3739

A mini-theorem is something that we use to prove another larger theorem that we are interested in.3744

It is basically like a thing that we can use to make a short hop to some more useful conclusion.3749

It deserves being proven on its own, because it is not just something we can do in a single line.3754

But it makes sense, and it is something that we will use towards another theorem.3759

It is something that we make on the way towards another theorem.3762

That is a lemma; so what we want to do is show that that is true, as a lemma.3765

Our lemma is that, for any natural numbers r and s, r choose s plus r choose s - 1 is equal to r + 1 choose s.3769

Now, if we want to use our lemma, we have to prove anything that we want to use.3777

This is a proof, after all; so let's get to proving it.3781

Let's investigate the left side, r choose s plus r choose s - 1.3784

We break that down: our r choose s becomes r! over (r - s)!(s!), plus...r choose s - 1 becomes r! over (r + 1 - s)!(s - 1)!.3789

And the only thing that you might be confused about is that r + 1 - s.3802

Well, remember: r - (s - 1), because it is this quantity here, would end up coming out to be...3804

and that part will be factorial; that is what matches up here...r - (s - 1)...that becomes plus; that becomes minus; that becomes positive.3812

So, that is why we end up getting the r + 1 and the - s there; that is how we get that.3818

What we want to do now is get these things over a common denominator.3824

Any time we are trying to simplify and work with expressions, if we have two fractions,3828

and we are trying to show that they are something else, well, we put them together, and we see what happens.3831

If we are going to get these over a common denominator, let's look at them for a second.3835

Well, we have (r - s)!(s!) and (r + 1 - s)!(s - 1)!; well, notice: (r + 1 - s)! is just one more than (r - s)!.3838

r - s + 1 is the same thing as r + 1 - s.3849

Similarly, (s - 1)! is the same thing as one more going up to s.3852

So, we are trying to get s - 1 up by 1, and r - s up by 1; we want to get this to go up, and we want to get this to go up.3858

If we can get them to each increase by 1 inside of the factorial, we will be able to have a common denominator.3867

And notice: based on how a factorial works, we have the following way to increase a factorial by multiplication: x! times (x + 1) equals (x + 1)!.3871

In case you don't see that, consider if we had 6!.3883

Well, that would be the same thing as 6 times 5 times 4 times 3 times 2 times 1.3885

Well, if we came along and multiplied 6 by 6 + 1, that is, 7, times 6!; well, that would be the same thing as 7 times...3891

the expansion of 6!, 6 times 5 times 4 times 3 times 2 times 1.3898

Well, now we have 7 all the way down to 1; so we can write that as 7!.3903

So, if we have x!, and we multiply it by one more than that x, we will end up increasing.3908

we will hop up one more step for our factorial, and we will go (x + 1)!.3912

That idea is what we will use to increase (s - 1)! up to s!, and to increase (r - s)! up to (r - s + 1)!, or (r + 1 - s)!.3916

All right, to do this, we will multiply the left side by (r + 1 - s)/(r + 1 - s).3929

We can't just multiply the bottom; we have to multiply with a fraction on top and bottom.3934

And we will multiply over here by s and s, because they match up, too; increasing s - 1 + 1 gets us s; r - s + 1 gets us r + 1 - s.3938

OK, we work that out; on the top, r! times r + 1 - s will just be r! times (r + 1 - s).3946

On the bottom, the (r - s) times one more will end up increasing to (r + 1 - s)!; the s! remains the same.3954

On the other side, the r! times s is just still r! times s; they can't really combine.3962

On the bottom, though, we have s - 1 times s; that is s - 1 + 1, so that becomes s!.3969

And the left side doesn't change: r + 1 - s!.3976

At this point, we now have a common denominator; the denominator here is the exact same as the denominator here.3979

So, we can combine them; we combine them, and we have r! times (r + 1 - s), r! times s, up here;3984

our denominators are just the same thing, because they just combined by addition.3992

But on the top, what we can do is pull out the r!'s; they pull out to r! times what the factors they were being multiplied against were.3997

That is (r + 1 - s), and then + s; the - s cancels with the + s, and we are left with r!(r + 1).4007

And r + 1 times r!...well, that just means it bumps up by 1; so now we have (r + 1)!.4017

We can also toss some fancy parentheses around this to help us see it.4023

We have (r + 1)! on the top, and (r + 1 - s)!(s!), which is just the exact same thing as r + 1 choose s.4026

(r + 1)! on the top, times (r + 1 - s)!(s!)--that is how r + 1 choose s works.4039

So, what we have done is: we have now shown that the left side and the right side are exactly the same thing.4048

We have set out to show that the left side and right side of our original thing are equivalent.4054

Our original lemma is that they are the same thing: so r choose s plus r choose s - 1 is equal to r + 1 choose s.4059

We have now completed our lemma; great; let's go and apply that lemma.4066

So, here is our lemma, right here; and where were we before?4071

Where we left off with our proof was: we had an + 1 and bn + 1 on the two extremes,4074

and then the big one that we really care about the most, the sum in the middle:4081

the sum of k = 1 to n of (n choose k plus n choose k - 1) times an + 1 - kbk; great.4085

But at this point, we now have that nice, handy lemma; we have r choose s plus r choose s - 1.4096

Well, that matches up to n choose k and n choose k - 1; so we will get n + 1 choose k out of it.4105

We just used our lemma; we are on our way to ending this thing--just a little bit more effort, and we will be done.4111

At this point, we are really, really close; this one here looks practically just like what we want to show in the end.4117

The only difference is that we need to have our upper and lower summation limits be 0 and n + 1.4123

We want to get this to become a 0, and we want to get this to become n + 1.4129

How are we going to get that to happen?4134

Well, on the way to doing that, we also have to deal somehow with this an + 1 and this bn + 1.4136

They are also in the way; they weren't in our finished thing.4142

Maybe these things can be done together; maybe we can somehow shove these into that sigma notation--4146

shove them into the sum, and in doing so, get the 0 and get the n + 1 that we want to appear in that.4151

Indeed, we can figure out a way to do that.4157

Notice: 1 is equal to n + 1 choose 0; also, 1 is equal to n + 1 choose n + 1.4159

Now, you can go along and multiply anything you want by n.4165

So, since you can multiply anything by 1, then we can multiply a 1 here and a 1 here.4170

Well, we know that 1 is just the same thing as n + 1 choose 0; so we swap that out for n + 1 choose 0.4174

And 1 is also just the same thing as n + 1 choose n + 1, so we swap that out here, as well.4179

So now, we have something that we can mash into our summation.4185

They fit the format of the summation; so we can fit them into the sigma notation, to obtain k = 0 to n + 1.4188

If we plug in k = 0 on this n + 1 choose k, well, that would be n + 1 choose 0 for k = 0.4198

And an + 1 - k, an + 1 - 0, would be an + 1bk.4205

b0 would be something that doesn't exist.4211

And then, the same exact thing happens over here: if we had k equal to n + 1, then we would have n + 1 choose n + 1.4215

And we would have a to the n + 1 minus...k is n + 1, so n + 1 - (n + 1).4223

Well, that would cancel; a0...and so we have the thing that doesn't show up here.4229

It doesn't show up, either; it just disappears.4232

And bn + 1 for our k...4235

And so, we see that they mash in; we can pull them in.4237

Everything else ends up staying the same; n and k = 1 aren't affected, because they do that.4241

It is just that these fit the format of the summation, so we can pull them in on the bottom and pull them in on the top.4245

We can fit them into the sigma notation, and we obtain this.4251

We have managed to show what we were trying to show from the beginning.4253

After much effort, we have finally shown that, assuming our inductive hypothesis, pn, we now know that pn + 1 is true.4256

That is, (a + b)n + 1 is equal to the sum of k = 0 to n + 1 of n + 1 choose k times an + 1 - kbk.4265

That checks out; we have finished showing our inductive step.4275

At this point, we know that the base case is true--we showed that from the beginning;4278

and we have now finished showing that the inductive step is true.4282

So, with some pn, we know that pn + 1 must be true.4284

Since that works as our inductive step, we now know, combining that with our base case, that pn is true for all n.4288

In other words, for any n equal to 0, 1, 2...we have the binomial theorem completely finished proving.4295

We have finished proving the binomial theorem; we are done--pretty cool.4302

So, once again, I want to say: what you just saw is not easy.4307

From the precalculus-level work that you are currently doing (or whatever the class that you are currently working in is),4312

this is 2 or 3 years ahead--doing these kinds of proofs.4317

But I honestly believe that, if you are really interested in this sort of stuff, you should be exposed to it now.4321

You totally have the chance to be able to understand this stuff and work through it.4326

It won't be easy like the sort of work that you are used to doing.4329

If you are in the kind of position where you want to understand this sort of thing,4332

the work that you are normally doing in your precalculus class probably isn't terribly difficult for you.4336

But this right here is a really interesting thing that lets us see how far math gets out--4340

what kind of stuff math is really like at the college level, when we are really studying serious-level math.4345

I think that this stuff is really cool, absolutely beautiful, and a great chance to just see logic in its purest form--4351

being able to make an argument that is really ironclad; I think that is really cool.4358

It is not for everybody; if you think this is just absolutely awful, and you watched the whole thing,4362

just because...I don't know why you watched the whole thing...but if you watched through it,4365

and you really didn't like it, that is OK; there are lots of things to do out there in the world.4368

But if you thought this was really cool, that is awesome; I think it is really cool, too.4371

And you can go ahead and study math, if you are interested.4375

You can study something like computer science, where you talk about algorithms like that, as well.4377

There are lots of things that will let you have the same idea of well-argued proof, of things building on top of each other-- these really interesting conceptual ideas.4383

All right, we will see you at Educator.com later--goodbye!4390