 So, go ahead and drop your homework off there, too. There's a pile right there. Okay. All right. So we'll go ahead and move on. I'll kind of let you know what the plan is. So we're going to go on to section 1.2, the binomial theorem. We'll talk about that today. I'm going to give you another assignment. So again, because of Tuesday, they kind of shifted everything back by a day. We'll probably get back at some point soon, hopefully, where I'm going to have you start training assignments in on Tuesday. But the next assignment, then, we'll just be doing a week from today. And so I will be giving you that today. And then we'll kind of do what we did before. And on Tuesday, we'll talk about a couple problems and such and help you get kind of comfortable with this. If we have time today, I'll do a problem or two, but I don't know that we will. So like I said, let's go ahead and get started here. Let's see. Okay. It's just as bad as before. Okay. So as I said, at least we have lines here. This section is called the binomial theorem. You've certainly seen some special cases of this theorem before. If you haven't seen it in full generality, you've seen this before. So, like usual, I'm just going to try to sort of build everything up here from the basics. So we're going to start with a definition. And the book actually defines this in the first section, but I just waited until this section because this is when we're really actually going to start using these things. You've probably all seen this before. Something called a factorial. Okay. So what we're going to do is we're going to make a recursive definition. Okay. And remember, I talked about this word recursion. This is sort of the inductive version of a definition. And we're going to do this on the natural numbers. And it's just going to go like this. This may look a little different than what you're used to, but now that we've seen induction, this really is sort of, in some sense, the best way to make this definition. So one, again, I assume that you've all seen this before. One factorial equals one. And of course I'm going to write down the definition here in a second. How many of you have seen this? The exclamation point thing, okay. Yeah. And then assuming we've defined n factorial, we define n plus one factorial by, okay, n plus one times, and I'll put this in parentheses so there's just no confusion about this, times n factorial. Okay, so n plus one factorial is defined to be n plus one times n factorial. And then this, as I said, this symbol, n with the exclamation point is not a surprising n. This is red n factorial. So there's the definition. And we'll just go through a quick example here. So of course, and I'll explain this as we go through here, but for a natural number n in factorial, it's just the product of all of the natural numbers less than or equal to n. You just multiply them all together, right? Okay, so again, just to make sure that everybody's with me on this. Sorry, this is a little messy. So of course by definition, by one above one factorial is one. And but I want you to actually look at the definition here just because I want you to get used to this because there will be instances where more complicated expressions are going to be defined. Two factorial, of course you might say, oh well it's what you just said, it's two times one. Which it is, but let's just go through the definition and see how we can get that. Okay, so two factorial by definition is of course, just because one plus one is two, this is one plus one factorial, which is equal to, okay, I want you to look at the second definition. We're going to use this in proof. I'm also defining it this way because you really need to be able to understand that n plus one factorial really is n plus one times n factorial. So one plus one factorial then is just plug in, in the second definition, plug in one for n, right? So it's one plus one times one factorial when you plug in one for n, right? So this of course is just two times, I'll just do one more. Three factorial is two plus one factorial, which again, if you look at the second definition, right? This is just two plus one times two factorial, and we just figured out what two factorial was. So this is three times two times one. Of course we don't need the one, but that's what we get. And similarly we can just keep doing this, right? So four factorial is four times three times two times one, and you get the familiar definition that you're used to before you saw me write the inductive one, right? Okay, so we're also going to, I'm singling this out especially, but I could have just built this into the definition too. We're also going to, and you'll see, I think you'll actually appreciate why we're doing this here after this section has been lectured on. We're going to also define zero factorial to be equal to one. Okay, and this is really just done for convenience so that formulas work out without a bunch of caveats before the theorems and such. So it just makes everything uniform in this way. And again, you'll appreciate this. I think once you see the binomial theorem, why we would want to define it this way. Okay, so now we're going to look at sort of the main object of study here in this section. It's called binomial coefficients. And so we're going to suppose that zero is less than or equal to k, which is less than or equal to n. I may suppress this from time to time, but I'll be clear here. k and n are non-negative integers. They're not just arbitrary real numbers. We're still living in the realm of the integers now. So the inequality, of course, implies that they're both non-negative. So all I really need to say here is that k and n are integers, something called the binomial coefficient. And you may wonder why this terminology is coming from. And you'll see here in a minute, once we do the binomial theorem, where this terminology is coming from. So here's the notation. Big left parentheses and above a k, and then a big right parentheses. Okay, so there's a way to read this. Yes, thank you. Good. You've heard of this before. I don't think the author actually says this in the book, which is bizarre. But n choose k. This is what everyone else does, except for the author lives in his own world. But this is read n choose k. There's actually a reason for this, which I won't go into. But let me, in detail anyways, but let me get this down first. I can kind of give you an idea of where this comes from, where this terminology comes from. Okay, so n choose k is equal to n factorial in the numerator divided by k factorial times n minus k factorial. So everybody can see this. Okay, yeah. It's readable, I guess. So let me just say a couple of things about this. First of all, I want you to notice that k itself could be zero. There's nothing precluding k from being zero. But you all learned this a long time ago. You can't divide by zero. But the definition of zero factorial tells us that this denominator is never zero. Even if k is zero, this is not zero. And since k is bigger than or equal to n, n minus k is bigger than or equal to zero. So this factorial is defined for n minus k. And it's because k is less than or equal to n. If k was five and n was two, right? Sorry. So if, yeah, so that works. If k was five and n was two, then you'd have minus three factorial, which we haven't defined. So everything still is perfectly well defined given what we did above, right? So where does this n choose k business come from? Well, does anybody know this? Do you know where this comes from? Yeah. Joe. Right. So the best way to say it really is if you have a set that has n elements in it, then the number of k element subsets is n choose k. That's what this is coming from. So we're not going to go into that aspect of it. We're going to look at it from a different perspective, which is from the point of view of algebra, of multiplying out, taking powers of a sum of two numbers or two variables or two polynomials. It doesn't really matter. But that's sort of the focus. And that's where this binomial coefficient comes from. It's going to come from the theorem. The binomial theorem is dealing with how to find the coefficients in front of a and b when you're multiplying out, say, a plus b to the 50th. What are all the coefficients on the powers of a and b? So that's what we're going to get to here pretty soon. But before I do that, I'll go ahead and do another example just in case you haven't seen this stuff before, just to make sure everybody's with me. Yes, actually it is. You might notice that my writing is a little neater. I had two people come to my office today and both people said, your handwriting is just impeccable on the board. It's amazing. You wouldn't know that from this. If I was writing on the whiteboard, he would say, wow, he has the best handwriting of anyone I've ever seen. But it's slowly improving. Maybe it'll get there at some point. Okay, so let's just look again. This is not meant to be difficult. So this is just a computational example just to make sure that everybody is understanding the ideas here. 8 choose 3. Okay, well, I'm not going to go back to the other slide now, but the definition is that this is 8 factorial, right? divided by 3 factorial times 5 factorial, right? n choose k was n factorial over k factorial times n minus k factorial. The n is 8, the k is 3, so n minus k factorial is 5 factorial in this case, right? Okay, so let's just write this out. I'm going to write out more than what we have to, but okay, so this is 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1, right? Oops, sorry about that. Okay, 3 times 2 times 1 times 5 times 4 times 3 times 2 times 1. Easy enough, right? Just using the definition of factorial. Okay, well, what can we do? Certainly we can do that, right? So just canceling, I'm just canceling out 5 times 4 times 3 times 2 times 1 from the top and the bottom, which of course we can do more than that, right? So we can cancel the 6 and the 3 times 2, and so this just becomes 56. So hopefully that's easy enough. Everybody okay with this, right? Okay, so now we're going to actually start doing some more theoretical calculations here, and I'm just going to call this lemma 1. This is called, this is pretty uniform I think, this is called Pascal's Rule. Okay, and here's what it says. This is actually a lot more useful than you might think right now, and in fact you're going to use this in the exercises quite a bit. So we're going to assume that, I'm going to use a little shorthand here. I'm going to leave out a few things. Okay, so I should be very clear again on this. These factorials are only defined for non-negative integers. So everything, when I say 1 is less than or equal to k, less than or equal to n, I mean that k and n are integers. I'm not going to write that out every time, but it's understood that k and n are integers here, okay? Okay, and so here is the result. So 1 is less than or equal to k, which is less than or equal to n. Then the result is n choose k plus n choose k minus 1 equals n plus 1 choose k. Okay, and actually as it turns out, there's a nice little trick that makes this proof very short, very understandable. And this result is very useful too, and you'll see how this is used. And by the end of the homework, you'll hopefully have a really good appreciation for how nice this result actually is. Okay, so just basic algebra really. So the first thing to note is, okay, let me offset this with an asterisk. Let's take 1 over k and we're going to add 1 over n minus k plus 1. So 1 over k plus 1 over n minus k plus 1. Okay, so you know how to do this, I assume. So this is n minus k plus 1 plus k divided by k times n minus k plus 1. Okay, so what did I do here? Common denominator. That's it. That's all I'm doing. Multiply the top and bottom of the first fraction by n minus k plus 1, multiply the top and bottom of the second by k, and that's where this numerator comes from. Okay, so basic algebra is all this is. Okay, and what does this equal to? Well, you notice that we can cancel the k out on the top, right? So we just have n plus 1 divided by k times n minus k plus 1. Okay, so now what we're going to do is we're going to multiply both sides of this equation, n factorial divided by k minus 1 factorial times n minus k factorial. You may wonder where this is coming from, but you'll see actually by doing this that we're going to get the desired equation to fall out pretty quickly. I'm going to go through this slowly to make sure I don't lose anybody. If you just see where this is coming from, the rest of it is very, very simple. And actually this is really pretty easy too. Okay, almost done. This looks a little bit messy, but I want to just remind you that this is not anything that's complicated really. Okay, so the equation, the start equation, all I'm doing is multiplying both sides of that. I'm sort of ignoring the middle, okay? I'm multiplying both sides of it by n factorial over k minus 1 factorial times n minus k factorial. So if you just look at the first term, right, 1 over k, what do we get when we multiply that first term by this? Well, we get n factorial in the numerator, right? You see that? Just multiply again, 1 over k times this. So we have n factorial in the numerator in the bottom, then we have a k, k minus 1 factorial times n minus k factorial. That's where the first term comes from. Again, it's just multiplying it through by this guy. That's it. And then the same is true for the other two terms. I've rearranged them slightly just so that you can see where the factorials are going to pop out here in the next step. That's it. That's all it is. Okay. So let me try to at least get one line in here. Oh, yeah. Okay. So I can just scroll down instead of actually having to get a new page. Okay. Yeah. So I just used the sidebar here then. Okay. Great. Thank you. Okay. So this is equal to... Okay. And I want you guys to think about this as we go through this because it's really not, like I said, it's really not that bad. And I'm going to go through it slowly, maybe more slowly than I need to, but look at... I don't want to mess up the notes here, but look at the k times k minus 1 factorial. So what is... Okay. What's k minus 1 factorial? It's 1 times 2 times 3 all the way up to k minus 1. Right? So what happens when I multiply that by k? I get k factorial. And it just comes from the definition, really. But it is. It's just the definition. K factorial. Okay? So here I get k factorial times n minus k factorial. And remember... Now, I'm not going to write this down yet, but remember what we're trying to get to. You see the thing we're trying to prove at the top? You see that? That is n choose k. We do have the first term. We just need to get that for the other two. And then we're done. You see that? Okay. So plus... Okay. So n factorial over k minus 1 factorial. Okay. Now look at the last two terms in the denominator here, right here. What do those two terms multiplied together give us? What do we get? Yes. n minus k plus 1 factorial. It's the same reason, right? If some of you are getting confused by the terms, just think about what's happening. Look at this. What's the relationship here? Well, this guy is exactly one more than that guy. That's for sure. Right? And this is all the numbers up... All the product of all the positive integers up to it times the next guy is that guy factorial. That's it. That's all it is. Okay? Once you know you have all the numbers up to a certain point and then multiply together and then you're multiplying by the next positive integer, it's that next positive integer factorial by definition. Okay. And what do we get over here? So what does the numerator become on the right side? And plus 1 factorial, right? So now you can appreciate why I gave you the recursive definition because you sort of need to know that now, right? Okay. So in the bottom, what can I do if I want to combine these two terms together? What do I get? k factorial. Okay. And what is this? n minus k plus 1 factorial. You see that for these two terms here? We just talked about this, right? Yeah, I like this system much better. Okay. I'm just going to change this slightly now. Okay. So n factorial, I don't think I need to do anything to the first term. n factorial over k factorial times n minus k factorial plus, and like I said, I'm going to change this just a little bit. n factorial over k minus 1 factorial times n minus in parentheses k minus 1. You see what I did here? Grouping this, right? All I did is just change the second term of the denominator to n minus the quantity k minus 1. When you distribute the minus, what do you get? You get n minus k plus 1, which is exactly what we started with. I haven't changed anything, right? I'm doing this for a reason you'll see in the last step, and then we'll be done with this. Okay. And so this is, I'm going to do the same thing for the other term. n plus 1 factorial divided by k factorial times n minus the quantity k minus 1, all of that factorial. Oh, sorry, sorry. I'm going to, I'll say what? No, that's, it's correct, but that's not really what I wanted. Whoops, wrong one. Sorry. There we go. Sorry about that. k factorial, and this is just, and really haven't done anything, this is even easier, n plus 1 minus k factorial. You all believe that these two things are the same in the bottom? Right? I just changed the order. n minus k plus 1 is the same thing as n plus 1 minus k. I mean, there's nothing deep. Okay, so now, again, I think I'll just let you look at your notes for this part, and then we're done now. Remember what the notes said? Okay, do you all believe that? That's n choose k by definition. It's in your notes. n factorial over k factorial times n minus k factorial by definition is n choose k. And here's why I changed the form explicitly so that you can all follow where I'm getting this. Okay, think about this for a second. I want you to think about this. You're going to have to get these ideas down for the homework, so it's good to just think about it right now to see where these are coming from. The second is n choose k minus 1. What's the definition? n choose k minus 1. n factorial over k minus 1 factorial, right? The thing in the bottom, that factorial, times n minus the quantity k minus 1 factorial, right? You guys by that? That's why I wrote it out explicitly so you can see. This just comes right from the definition now. And what about this? Equals, and this is exactly also what we wanted. You may have forgotten what it is we're trying to prove, but this is what we were trying to prove. Think about it. Just take a second to think about it to make sure that you process this. What's n plus 1 choose k? Definition. It's the top guy factorial, n plus 1 factorial, divided by the bottom guy factorial, k factorial, times the top guy minus the bottom guy factorial, n plus 1 minus k factorial. That's exactly what we have. Okay? So there it is. So there's Pascal's rule. Okay, so this is going to be very useful. You're going to use this in the homework a decent amount. Or at least some consequence of this. So I'm going to show you what this is good for. So really what I want to try to do is get through two things. The first is not something that the book explicitly mentions, but I'm going to do this just to give you some added practice. So here's what we're going to work on now. Applications of Pascal's rule. So here's the first corollary, and it says this. For every non-negative integer n, that's an i, by the way. Sorry. Okay, let me just put an asterisk next to this. If 0 is less than or equal to k, which is less than or equal to n, then n choose k is a natural number. There are various ways that you can prove this. I'm doing it this way because I want to show you that how induction works, and this kind of has a slightly different flavor to it. So I want to do this inductively. And this is just going to use Pascal's rule. Now I should mention that it's not immediate that n choose k is an integer. Well, of course it's a whole number. Well, remember how n choose k is defined. n factorial over k factorial times n minus k factorial. Maybe you get five-ninths. You don't, but it's not immediately obvious that you're going to get a whole number that way. And actually, regardless of how you prove this, it's going to require a little bit of work. You can't do it in half a line. You can't. You need a little bit of machinery to do this. So here's how we're going to do it. So we're going to use induction, and this is also going to give me another avenue to explain that there's nothing magical about the number one in terms of starting the induction there. It depends on what it is you're trying to prove. Let me just say this really quickly. Hopefully I don't run out of time, but suppose you want to prove that something's true, not for the positive integers, but say for all the non-negative integers. So in other words, we're adding zero now to our set. The same principle still applies. If you can show that the statement's true for zero, and if you show that it's true for any non-negative number, and then it's true for n plus one, then it's true for all non-negative integers. There's nothing magical about the number one. If you want to prove something's true for all integers and bigger than or equal to negative 4,000, prove that it's true for negative 4,000, prove that if it's true for some integer and bigger than or equal to negative 4,000, then it's true for n plus one, then it's true for everything bigger than or equal to minus 4,000. There's nothing magical. The starting point, and then when you know it's true for something, it's true for the next guy, then all the dominoes fall. That's all you need to know. So just to be clear, if you ever in your life, I don't think that'll happen, but you need to prove something's true for every integer bigger than or equal to minus 4,000. Oh, no, it's not a natural number, so I can't use inductive. Yes, you can. It's just that your base step starts at minus 4,000 instead of one. That's it. So that's another reason, again, why I wanted to do this, so that you can see that the scope of induction might be a little bit broader than you may think at first. So what we're going to do is we're going to use induction, but instead of the natural numbers, we're going to use induction on the non-negative integers. Notice, remember, these two steps are not the same. Okay, so how do we do this? Well, it's the same basic two steps as before. The first step is, that should be a one, or an I really, okay. And what we need to establish is that this claims, and what I mean by the claim, I mean the starred claim, is true for n equals zero. So I want you to think about this for a second. Here's this starred claim. When n is equal to zero, what is it that we are claiming? We're claiming that if zero is less than or equal to k, which is less than or equal to zero, then zero choose k as a natural number. That's what we're trying to prove. That's what we get when we replace n with zero. You guys with me on this? It doesn't change. The n changes to zero. So here's what we have to check. And again, I'm writing it out in gory detail here just to make sure that I don't lose anyone. If zero less than or equal to k, which is less than or equal to zero, then zero choose k as a natural number. Okay, remember, just like I said before, the k in the end, these are integers now. K is not 0.2, okay? Everything's an integer. So of course, it doesn't matter, really. But what do you know about k? If zero is less than or equal to k, which is less than or equal to zero, what do you know about k? Well, you don't need to squeeze theorem. K has to be zero. There's nowhere for it to go, right? So, yeah, k is certainly equal to zero. If it wasn't, then zero would be less than zero. If k was positive, then you'd have zero less than k less than or equal to zero, so zero would be less than zero, and it's not possible. Okay, negative zero is the same as zero. Negative zero equals zero, so it doesn't matter. So any such k is necessarily equal to zero. There's only one case to check here. Zero choose k has to be the same thing as zero choose zero. And let's just use the definition again, okay? This is zero factorial over zero factorial times zero minus zero factorial, right? Again, I'm writing it all out just to make sure that everyone sees all the details. You should be able to do this in your head, but, you know, certainly by definition this is zero factorial over zero factorial times zero factorial. And what's zero factorial equal to? What was the definition? One, so this is equal to one, which is certainly a natural number, right? Okay, so I'm going to have to go down now. So let me make sure everybody has this first. Okay, so everybody have this now? Okay, so that's the first part. That was easy enough. The second part is we're going to suppose this claim holds some non-negative integer in. Okay, so what is it that we have to prove? Okay, so we must, I'm just going to write it down, so we must show. Okay, so you have this in your notes, so you put a star next to it. If k is an integer with zero less than or equal to k less than or equal to n plus one, then n plus one choose k is a natural number, right? So we're going to look at some cases, and I'll tell you why we're doing that here in just a second. For now I just want you to follow along here. What if k is zero? Can we prove it if k is zero? Well, we proved that zero to zero is a natural number. We haven't done this yet. n plus one choose k is n plus one choose zero, right? And what's the definition? This should be n plus one factorial over what? What do I want down here? Zero factorial times, right? Okay, do you guys see where I'm getting this? Yeah, no. I'm not saying, I never said that zero is a natural number. No, just that k is between. n choose k is a natural number. The k itself could be zero. I'm not saying that k has to be a natural number. I'm saying n choose k has to be a natural number. All I'm saying about k is that zero is less than or equal to k, which is less than or equal to n. k itself could not be a natural number, but I'm claiming that n choose k is a natural number. But n choose k is one. There's nothing contradicted here. I assure you, nothing's contradicted. n and k are not assumed to be natural numbers. They're assumed to be non-negative integers. And choose k is claimed to be a natural number. Okay, that's a good question. The choose, the binomial coefficient is a natural number. And n and k don't necessarily have to be. Is everybody clear on this point? Okay, so what is this equal to? This is pretty easy, right? What's it equal to? One. So that case was easy. What about this case? Case two, these are sort of the extreme cases. k is equal to n plus one. That's certainly possible, right? The only restriction on k is that it's between zero and n plus one inclusive. So k could be zero and k could be n plus one. Definitely those are possibilities here. Then in this case, n plus one choose k equals n plus one choose n plus one, right? Which is the top guy factorial times the bottom guy factorial times the top guy minus the bottom guy factorial, which is zero because they're both the same, right? You guys with me here? And what is this equal to? Well, it's the same thing as before, right? It's equal to one. Okay, and in the last case, and now you'll see if you're wondering why I chose these, you know, I isolated these. I think you'll see why here. Now, case three. What do we know about k? It's between zero and n plus one. The first case was k equals zero. The second case was k equals n plus one. Suppose that k doesn't satisfy either of those two conditions. Then what do we know about k? Well, it's not zero, it's not the first guy, it's not the last guy. So it's squeezed in the middle somewhere, okay? So then the third condition, and think about it for a second, or the third case, right, if neither of these hold, is that one is less than or equal to k, which is less than or equal to n. Think about it for a second. You believe that? And then it's squeezed between zero and n plus one. It's got to be squeezed between one and n. Okay, so what's the big deal? Why did I do this? Then look back to Pascal's rule. Okay, I'm not, again, I just, I'll let you look in your notes here. I would encourage you to do that. Look at Pascal's rule. Look at the hypothesis of Pascal's rule, and look at what our conclusion was. What was the hypothesis? The hypothesis that was made in Pascal's rule, right? k and n are integers with one less than or equal to k, less than or equal to n, right? That was a hypothesis, okay? That's why I isolated those other two as separate cases, because if those two don't hold, then we do know that one is, that k is sandwiched between one and n. So that does, so now we're in the, so now we've satisfied the antecedent in Pascal's rule. Satisfied, because we know that k is between one and n now. And what's the conclusion? The conclusion is that n plus one choose k is equal to n choose k plus n choose k minus one, right? Okay, now, now you might say, and this is a point I really want to be clear on, and this is another reason why I did this problem. We're going to use the inductive hypothesis now. So we're assuming, okay, now I'm going to, I'll wait until you copy this down, but I'm going to go back up here for a second. You guys have this? We're assuming for n that this asterisked sentence holds for n, okay? What does it say? What does the sentence say? It says that if k is any, I didn't write all this out, but if k is any integer with zero less than or equal to k less than or equal to n, then n choose k as a natural number. That's what this is saying. And we are assuming that for our inductive hypothesis. That is true for n. It's not, and then that doesn't just say one k works. It says any k between zero and n works, okay? That's the thing that you have to remember. We're not using strong induction here. That's what a lot of you are going to think, oh, strong induction because there's a k and a k minus one. We have to use strong induction. No, we don't. We do not need strong induction. So what do we have? What do we have over here? Okay, so this is assumed. That's the inductive hypothesis. So we know for sure, because certainly since one is less than or equal to k, which is less than or equal to n, and certainly zero is less than or equal to k, which is less than or equal to n. That's certainly true. So by the inductive hypothesis, n choose k as a natural number, okay? So this guy right here is a natural number. What can we say about k minus one? Here's my question. I'll write it all this down, but I want you to think about it for a second. One is less than or equal to k, which is less than or equal to n. So what can you say about k minus one? Zero is less than or equal to k minus one, which is less than or equal to n minus one, which is less than or equal to n. So my point is that zero is less than or equal to k minus one, which is less than or equal to n. That's certainly true. So just subtract one off everywhere in this inequality. Zero is less than or equal to k minus one, which is less than or equal to n minus one, which is less than or equal to n. So zero is less than or equal to k minus one, which is less than or equal to n. By the inductive hypothesis up here, therefore n choose k minus one is also a natural number. The sum of two natural numbers is natural, therefore n plus one choose k is a natural number. There's the proof. There's still one or two of you, when you think about this, are going to be convinced that this is wrong and you have to use strong induction. No, that is not correct. You should trust my brain, not yours. I'm sorry. You should trust me. When it comes to this, you should trust me on this. So these are natural numbers by the inductive hypothesis. So their sum is also a natural number. So therefore, since it's a sum of two natural numbers, let me not be repetitive here. Oops. OK. That's all right. These wands instead of hands. OK. Done. That's it. We consider all the cases, and we've proved it in every case. Now we're done by the first principle of finite induction. OK. The last thing we're going to talk about, hopefully I'll have time to get this done. I hope I do, is the binomial theorem. OK. So I'm just going to go, I'm going to sort of just change gears because I want to try to get this done. Here's the setup. I'm going to squeeze it in over here. OK. OK. A plus B. Now you're going to notice that I'm not telling you what A and B are. OK. And I'll go into that later. A plus B to the first is A plus B. You all know that. A plus B squared. You learned this a while ago. A squared plus 2AB plus B squared. And you can all do this, I'm sure. You can multiply this out, foil it, what not. In combined like terms, you get A cubed plus 3A squared B plus 3AB squared plus B cubed. And you can actually continue this and you can figure out there is a certain pattern that emerges. Notice that as we go through, I've only done three of them, but A plus B to the first has two terms added together. A plus B squared has three terms added together. A plus B cubed has four terms added together. And so if you write these down, you've probably heard this before, the Pascal's Triangle. Maybe you've heard this. So they just keep increasing by one and you end up with this triangular shape if you sort of center everything. And the coefficients follow a certain pattern. They kind of go up and then they come back down again. I'm not going to go into that because that's not really the point of what we're going to do now. But this Pascal's Triangle business comes from this. It's the fact when you combine all the like terms, you just keep going up by one in terms of number of terms of every successive one. OK, now we're going to look at something called the binomial theorem. And this is a nice formula for computing A plus B to any power. It involves factorials. And I'm going to go through. The proof of this is actually not very long. I'm going to try to get through that. If I don't, we'll finish it up on Tuesday. OK, so I guess I'm going to need another page. OK, so let me go ahead and state this. And I'm going to go ahead and call this. I think I called the previous result corollary one, I think. Didn't I? I think I did. Maybe I didn't. I did? Yes, OK. So this is kind of too big of a theorem to be called a corollary. But really, in some sense, it really is just a corollary of this principle, this Pascal's principle that we saw before. OK, and it just says this. So again, just to be clear, this is called the binomial theorem. Let N be in capital N. In other words, N is a positive integer. Then, first, if you haven't seen this before, this may look a little nasty. But it's not as bad as it looks. A plus B to the N is equal to the sum from 0, K equals 0 to N. N choose K. A to the N minus K times B to the K. Now, just I think you've all seen this at some point in calculus, for example, when you're dealing with series, you certainly saw this sigma sign before. I'm hoping you've all seen this sign before and know what this is. If you don't, it's not very difficult to explain. What this means, this sigma K equals 0, and then N on the top just means that expression you see to the right of the sigma, you plug in 0 for K and you see what you get. You plug in 1 for K, you see what you get. You plug in 2 for K, you see what you get. You plug in everything up to N, see what you get, and you add everything together. That's what that means. So N is fixed. N is not moving in some sense. K is the thing that's moving here. K is ranging from 0 to N. OK, so let's take a look at this. I'll see how far I get on this. If I get caught in the middle of this, well, I may finish it next time, or I may just go and do a couple of problems next time. I'll just have to see how much I have left to do. So what we're going to do is we're just going to use induction. And in case this isn't clear, we're going to use the first principle. We don't actually need strong induction here. So what is the first thing that we have to prove? So there are two things that we need to show. We must, the base case, we must prove this claim, star, when N equals 1. OK, that's the base part now. And remember, now we're proving something for all natural numbers. So we're trying to prove this is true for all natural numbers. 0 is not a natural number. So we're not using including 0 in this case. So we're going to start with N equals 1. OK, so what is the claim when N equals 1? Well, just to save time, I'm not going to write this out. Let's just do it. It's not very hard. When N equals 1, I'm going to start on the right side for now. I think it's easier to just do it this way. On the right side, when N equals 1, it's just the sum from k equals 0 to 1. We're just replacing N with 1 now. And I'm just going to show you that we get the left-hand side in this case. 1 choose k, a to the 1 minus k times b to the k. You all buy that? That's what we get on the right side when N equals 1. And what we need to do is check that it's equal to what? a plus b. That's what we want, because when N is 1, then a plus b to the first is just a plus b. OK, so let's see what we get. All we have to do is plug in two values of k here. Remember what I was saying before? Whatever k equals, that's where we start. And then we keep plugging in values of k until we get to this top number. And the top number is 1. So it's just k equals 0 and k equals 1. We're just plugging those two numbers in for k. Adding them up, we should get a plus b. OK, so what happens when we plug in 0? So this is 1 choose 0, a to the 1 minus 0 times b to the 0. OK, now might be a good time to say something about a and b. a and b can be, they don't have to be natural numbers. a and b can be real numbers. They can actually be even more than that. They could be polynomials. In general, if any of you have taken abstract algebra, a and b can just be any elements of a commutative ring. That's it. If you don't know what I'm talking about, it doesn't matter. This theorem holds in general in a very more general setting than what you're used to if you haven't taken abstract algebra. That's why I didn't say what a and b were. They don't have to be numbers. They don't have to be polynomials. They could be all sorts of weird things. OK, so what do we get? OK, I'll probably go through this a little more quickly than I did before. What's 1 choose 0? 1. What's a to the 1 minus 0? What's b to the 0? So you buy that? OK. Sorry. I didn't let me put this parenthetically now. Equals a. This isn't actually what this is equal to. It's this plus the next guy, right? OK, which is just we're plugging in now 1 for k, right? 1 choose 1. a to the 1 minus 1 times b to the first. And what's this equal to? Right. And then so we get a plus b, which is a plus b to the first, and which is what we wanted, right? So the theorem is definitely true when n equals 1. We just verified that. Second part. And this is a little trickier. It's just going to use Pascal's rule again, though. We're going to assume for some n and n, some little n and big n. And again, I'm just going to use the same notation as I used before. We'll just put an asterisk next to this. a plus b to the n is equal to the sum from k equals 0 to n. And choose k a to the n minus k times b to the k. So because I'm going to have to go down another page anyways, I'm not going to write out exactly what we have to prove. If you want to put this in your notes, you can. I mean, I'm going to prove it, but I'm just not going to say it beforehand. I think I'm going to run out of time here. But what's the statement when we replace n with n plus 1? It's a plus b to the n plus 1 equals the sum from k equals 0 to n plus 1, n plus 1, choose k, a to the n plus 1 minus k times b to the k, right? OK, so that's what it is that we have to prove. And we can assume that this is the inductive hypothesis. We assume that we have it for n. And we're going to use that fact to get it for n plus 1. OK, here is what I am going to do, because this part is just a little bit technical, and I don't want to get stuck in the middle of this, actually. I'd rather just do this in one class period, because otherwise I'm going to have to do it again anyways to refresh your memory on Tuesday. So I'm going to just stop, I think, at this point. What I'm going to do is I'm going to give you a homework assignment for this section. Again, Tuesday I'll finish this up. I will definitely do a couple of problems, again, like I did last Tuesday, to get you up to speed. I'd like another week or two to flip the assignments due date back to Tuesday, so that you have a weekend to kind of work on things. And you've got a little bit more time than just two days after I do a couple of examples. But we're going to try to get there soon. So let me give you, like I said, give you the homework assignment. Sorry I didn't quite get through the rest of this, but we'll finish this up on Tuesday. OK, so this is to be continued. Although I will say this, OK, the homework that I'm giving you, I would say the majority of the problem, the book actually gives you hints to these. And the book also works out a couple of examples. So really, you, I think, have the tools you need to be able to work on the homework already. Some of this is going to involve induction again. But you're just going to assume the binomial theorem is true, and then just kind of work through the homework, just pretending that you know what's going on. OK, so here's the homework. 3A, 3C, 4C, 5A, 5B, and 5C. Could have given you a lot more than this. But that is all that you have to turn in for next Thursday. And if anyone, I don't think anyone came in that late, but if anyone didn't turn their homework in, definitely get that in before you leave. So I think that's all I wanted to say. So all right, we'll have a good one.