WEBVTT mathematics/pre-calculus/selhorst-jones 00:00:00.000 --> 00:00:02.100 Hi--welcome back to Educator.com. 00:00:02.100 --> 00:00:06.100 Today, we are going to talk about arithmetic sequences and series. 00:00:06.100 --> 00:00:11.000 Now that we have an understanding of sequences and series, we are ready to look at specific kinds of sequences. 00:00:11.000 --> 00:00:17.700 The first that we will consider is an arithmetic sequence, a sequence where we add a constant number each step. 00:00:17.700 --> 00:00:22.200 We will add some number, and we keep adding the same number every time we go forward a term. 00:00:22.200 --> 00:00:27.300 Sequences of this form pop up all the time in real life, and we often need to add up their terms. 00:00:27.300 --> 00:00:33.000 We will explore the creation of a formula for arithmetic series that will allow us to quickly and easily add up those terms. 00:00:33.000 --> 00:00:40.700 Plus, by the end of this lesson, we will be able to add all of the numbers between 1 and 1000 in less time than it takes to put on a pair of shoes. 00:00:40.700 --> 00:00:48.500 I think that is pretty cool; we will be able to do something really, really fast that seems like it would take a long time, just like that. 00:00:48.500 --> 00:00:55.600 All right, a sequence is arithmetic if the difference between any two consecutive terms is constant. 00:00:55.600 --> 00:01:01.300 We can show that with the recursive relationship a<font size="-6">n</font> - a<font size="-6">n - 1</font> = d. 00:01:01.300 --> 00:01:09.200 Notice: the nth term minus the n - 1 term (that is, the term one before the nth term) equals d. 00:01:09.200 --> 00:01:13.600 Some term minus the term before it equals d. 00:01:13.600 --> 00:01:19.500 And d is just some constant number; we call d the common difference. 00:01:19.500 --> 00:01:27.000 Here are two examples of arithmetic sequences: they are arithmetic because every step in the sequence has the same change. 00:01:27.000 --> 00:01:38.700 For example, here, 1 to 4, 4 to 7, 7 to 10, 10 to 13...it is + 3, + 3, + 3, + 3. 00:01:38.700 --> 00:01:43.100 It is always the same amount that we change; it is always adding a constant number. 00:01:43.100 --> 00:01:56.400 Over here, it is 5 to 3, 3 to 1, 1 to -1, -1 to -3...it is - 2, - 2, - 2, - 2; and that pattern would continue, as well. 00:01:56.400 --> 00:02:03.800 In this case, our common difference is -2; we are adding -2 each time; or we can think of it as subtracting 2 each time. 00:02:03.800 --> 00:02:08.500 The important thing is that it is always the same; the difference can be positive; the difference can be negative; 00:02:08.500 --> 00:02:15.500 but it always has to be the same for each one--that is what makes it an arithmetic sequence. 00:02:15.500 --> 00:02:20.800 How can we find the nth term? The definition for an arithmetic sequence is based on a recursive relation. 00:02:20.800 --> 00:02:27.800 It is based on a<font size="-6">n</font> = a<font size="-6">n - 1</font> + d, that some term is equal to its previous term, with d added to it. 00:02:27.800 --> 00:02:33.800 So, how can we turn this formula into something for the general term--how can we get a general term formula out of this? 00:02:33.800 --> 00:02:38.600 Remember: a recursive relation needs an initial term--we have to have some starting place. 00:02:38.600 --> 00:02:44.000 There is nothing before our starting place to refer back to, so we actually have to be given the initial term directly. 00:02:44.000 --> 00:02:50.700 Now, we don't know its value yet; so we will just call it a₁; we will call it the first term--we will leave it as that. 00:02:50.700 --> 00:02:57.300 Now, from a<font size="-6">n</font> = a<font size="-6">n - 1</font> + d, we see that a₁ relates to later terms. 00:02:57.300 --> 00:03:02.900 At the most basic level, we have that a₂ is equal to a₁ + d. 00:03:02.900 --> 00:03:08.200 The second term is equal to the first term, adding d onto it; that is what it means for it to be an arithmetic sequence. 00:03:08.200 --> 00:03:11.500 We can take this out and continue looking at later terms. 00:03:11.500 --> 00:03:15.900 a₃ would be equal to a₂ + d, based on this recurrence relation. 00:03:15.900 --> 00:03:22.900 But we just figured out that a₂ is equal to a₁ + d, so we can swap out for a₂: 00:03:22.900 --> 00:03:30.600 a₁ + d, which now gets us a₁ + 2d when we add this d to that one there. 00:03:30.600 --> 00:03:35.600 So, we have a₁ + 2d for a₃; when we work on a₄, 00:03:35.600 --> 00:03:40.400 well, a₄ is going to be a₃ + d, the previous term, adding d. 00:03:40.400 --> 00:03:44.900 But once again, we have just figured out that a₁ + 2d is what a₃ is. 00:03:44.900 --> 00:03:51.000 So, we can plug in for a₃; we have a₁ + 2d, and now we can add that d onto the 2d. 00:03:51.000 --> 00:03:57.800 So, we end up getting that a₁ + 3d is what a₄ is equal to. 00:03:57.800 --> 00:04:04.700 So, we notice that this pattern is going to keep going; we are just going to keep adding on more and more d's to our number. 00:04:04.700 --> 00:04:13.300 So, a₁ = a₁; a₂ = a₁ + d; a₃ = a₁ + 2d; 00:04:13.300 --> 00:04:17.000 a₄ = a₁ + 3d, and this pattern will continue down. 00:04:17.000 --> 00:04:21.800 We see that the nth term is n - 1 steps from a₁. 00:04:21.800 --> 00:04:30.000 The first term is at 1, and the nth term is at n; so to get from the first term to the n term, we have to go forward n - 1 steps. 00:04:30.000 --> 00:04:35.800 Since it is n - 1 steps away from a₁, we will have added d for each of those steps; 00:04:35.800 --> 00:04:39.900 so we will have added d that many times, or n - 1 times d. 00:04:39.900 --> 00:04:47.800 That means that the a<font size="-6">n</font> term is equal to a₁ plus all of those steps times d. 00:04:47.800 --> 00:04:53.900 So, the nth term, a<font size="-6">n</font>, equals a₁, our first term, plus n - 1 times d. 00:04:53.900 --> 00:05:05.800 Thus, to find the formula for the general term of an arithmetic sequence, we only need to figure out the first term, a₁, and the common difference, d. 00:05:05.800 --> 00:05:14.200 With those two pieces of information, we automatically have the general term; we automatically have that nth term formula--pretty great. 00:05:14.200 --> 00:05:16.600 How about if we want an arithmetic series formula? 00:05:16.600 --> 00:05:27.400 Consider if we were told to add up all of the integers from 1 to 100; how could we find 1 + 2 + 3 + 4 + 5 + 6...+ 98 + 99 + 100? 00:05:27.400 --> 00:05:34.400 How could we add that whole thing up? Well, we could do it by brute force, where we would just sit down 00:05:34.400 --> 00:05:37.400 with a piece of paper or a calculator and just punch the whole thing out. 00:05:37.400 --> 00:05:40.900 We could do it by hand; but that is going to take a long time. 00:05:40.900 --> 00:05:45.300 And any time that we end up seeing something that is going to take us forever to do, we want to ask ourselves, 00:05:45.300 --> 00:05:53.500 "Is there a way to be clever--is there an easier way that I can do this that will be able to take away some, or a lot, of the time and effort?" 00:05:53.500 --> 00:05:57.400 How are we going to do that? We want to look for some sort of pattern that we can exploit. 00:05:57.400 --> 00:06:01.500 We want to find a pattern that we can exploit--something that will keep happening-- 00:06:01.500 --> 00:06:04.800 something that we can rely on, that will keep us from having to add up all of these numbers, 00:06:04.800 --> 00:06:08.600 because we can instead use this pattern to give us a deeper insight to what is going on. 00:06:08.600 --> 00:06:14.200 So, if we look at this for a while, we might start to realize that there is a pattern in the numbers. 00:06:14.200 --> 00:06:16.000 But that doesn't help us, because that is just adding the numbers. 00:06:16.000 --> 00:06:19.100 But is there a way that addition itself has a pattern? 00:06:19.100 --> 00:06:23.200 There is something that we could match up--something that we could create--and this is where we are getting really clever. 00:06:23.200 --> 00:06:26.500 This is the hard part, where you really have to sit down and think about it for a long time. 00:06:26.500 --> 00:06:29.900 And hopefully you just end up getting some "lightning bolt" of insight. 00:06:29.900 --> 00:06:35.800 And hopefully, at some point, we will notice that here is 100; here is 1; if we add them together, we get 101. 00:06:35.800 --> 00:06:44.600 But not only that--if we had 99 (let's use a new color)...if we use 99 and 2, we get 101. 00:06:44.600 --> 00:06:54.900 If we add 98 and 3, we get 101; if we keep doing this, working our way in, we are going to keep adding things up to 101, 101, 101... 00:06:54.900 --> 00:07:05.500 So, if we notice that we can add 1 and 100 to get 101; 2 and 99 to get 101; 3 and 98...we get 101; and so on and so on and so on... 00:07:05.500 --> 00:07:10.600 what we can do is pair up each number from 1 to 50 with a number from 51 to 100. 00:07:10.600 --> 00:07:13.500 And we will always be able to make 101 out of it. 00:07:13.500 --> 00:07:26.200 We start out at the extremes, 1 and 100, and we work in: 2 and 99; 3 and 98; 4 and 97; until we finally make our way to 50 and 51. 00:07:26.200 --> 00:07:28.900 So, we were able to figure out this pattern; there is something going on. 00:07:28.900 --> 00:07:34.700 Now, we are finding something; now we have something that we can pull into a formula that will make this really easy. 00:07:34.700 --> 00:07:38.000 With this realization in mind, let's look for an easy way to pair up the numbers. 00:07:38.000 --> 00:07:42.300 The first thing: it is nice to give names to things in algebra--it lets us work with them more easily. 00:07:42.300 --> 00:07:51.900 So, let us have s denote the sum of 1 to 100; so s is equal to 1 + 2 + 3...+ 99 + 100; it is all of those numbers added up together. 00:07:51.900 --> 00:07:56.200 Now, notice: we can rewrite the order of those numbers, since order of addition doesn't matter. 00:07:56.200 --> 00:08:04.600 1 + 2 + ... + 99 + 100, here, is the same thing as 100 + 99 + ... + 2 + 1. 00:08:04.600 --> 00:08:08.900 We can swap the order, and we still have the same value in the end. 00:08:08.900 --> 00:08:12.300 That is one of the nice things about the real numbers: order of addition doesn't matter. 00:08:12.300 --> 00:08:17.000 Furthermore, we can add two equations together--that is elimination. 00:08:17.000 --> 00:08:21.000 Remember: when we worked on systems of linear equations, if you have an equation, you can just add it 00:08:21.000 --> 00:08:23.400 to another equation, because they are both working equations. 00:08:23.400 --> 00:08:27.700 You can add the left sides and the right sides, and you know that everything works out; there is nothing wrong with doing that. 00:08:27.700 --> 00:08:36.000 What we have is: we can add the top equation there and the bottom equation, the normal order and the reversed order; we can add them together. 00:08:36.000 --> 00:08:43.400 What do we end up getting? Well, here we have a hundred and one, so we get 101; here we have 99 and 2, so we get 101; 00:08:43.400 --> 00:08:47.400 here we have 2 and 99, so we get 101; here we have 1 and 100, so we get 101. 00:08:47.400 --> 00:08:52.200 And we know that we are going to end up having 101 show up for every one of the values inside of here, as well. 00:08:52.200 --> 00:09:04.600 How many terms are there total? Well, we had 1, 2, 3...99, 100; so we had 100 terms, left to right. 00:09:04.600 --> 00:09:12.800 So, if we had that many terms total, well, even after we add them up, and each one of them becomes 101, then we have 100 terms total. 00:09:12.800 --> 00:09:20.800 We have 100 terms on the right side; so if we have the same number appearing 100 times, we can just condense that with multiplication. 00:09:20.800 --> 00:09:29.100 We can condense all of that addition with multiplication: 3 + 3 + 3 + 3 is the same thing as 3 times 4, 4 times 3. 00:09:29.100 --> 00:09:35.800 So, if we have 101 appearing 100 times, then we can turn that into 100 times 101. 00:09:35.800 --> 00:09:40.500 Our left side is still just 2s; so we have 2s = 100(101). 00:09:40.500 --> 00:09:48.400 What we are looking for is the sum, s = ...up until 100; so we just divide both sides by 2 to get rid of this 2 here. 00:09:48.400 --> 00:09:56.000 Divide both sides by 2; 100 divided by 2 gets us 50, so 50 times 101 means we have an answer of 5050. 00:09:56.000 --> 00:10:00.800 So, that probably took about as much time as if we had added up 1 + 2, all the way up to 100. 00:10:00.800 --> 00:10:03.300 If we had done that whole thing by hand, it would have taken a while. 00:10:03.300 --> 00:10:08.800 And now, we have the beginning kernel to think, "We can just do this for anything at all, and it will end up working out!" 00:10:08.800 --> 00:10:10.600 Indeed, that is what will work out. 00:10:10.600 --> 00:10:16.000 We have this method in mind of being able to string all of the things in our arithmetic sequence together, 00:10:16.000 --> 00:10:19.300 and then flip it and add them together and see what happens. 00:10:19.300 --> 00:10:23.300 We can now figure out a general formula for any finite arithmetic series. 00:10:23.300 --> 00:10:31.700 Let s<font size="-6">n</font> denote the nth partial sum--that is, the first n terms of the sequence, added together, of some arithmetic sequence. 00:10:31.700 --> 00:10:38.800 So, s<font size="-6">n</font> = a₁ + a₂ + a₃ + ... up until we get to + a<font size="-6">n - 1</font>, 00:10:38.800 --> 00:10:44.700 up until, finally, a<font size="-6">n</font> is our end, because we have the nth partial sum; great. 00:10:44.700 --> 00:10:52.300 Earlier, we figured out the general term for any arithmetic sequence is a<font size="-6">n</font> = a₁ + (n - 1)d. 00:10:52.300 --> 00:10:57.800 So, we can swap out a₁ for what it is in the general form, a₂ for what is in the general term, 00:10:57.800 --> 00:11:01.700 a<font size="-6">n - 1</font> for what it is in the general term, a<font size="-6">n</font> for what it is in the general term. 00:11:01.700 --> 00:11:07.100 This will get everything in terms of a₁ and d and that n; great. 00:11:07.100 --> 00:11:13.000 Thus, we can write out s<font size="-6">n</font> if we want to; we can write it out as s<font size="-6">n</font> = a₁, 00:11:13.000 --> 00:11:18.800 and then a₂ would be a₁ + d (2 minus 1, so 1 times d...a₁ + d). 00:11:18.800 --> 00:11:23.400 We work our way out: a<font size="-6">n - 1</font> would be a₁ + (n - 2)d; 00:11:23.400 --> 00:11:29.500 we plug in n - 1 for the general n term, so n - 1, minus 1...n minus 2 times d. 00:11:29.500 --> 00:11:33.600 And finally, the a<font size="-6">n</font> would be n plus n minus 1 times d. 00:11:33.600 --> 00:11:38.000 Great; so we have this thing where the only thing showing up there is a₁, n, and d. 00:11:38.000 --> 00:11:42.000 We have far fewer things that we have to worry about getting in our way. 00:11:42.000 --> 00:11:46.400 Furthermore, we can write s<font size="-6">n</font> in the opposite order; we are allowed to flip addition order. 00:11:46.400 --> 00:11:53.200 So, we write it in the opposite order as s<font size="-6">n</font> =...the last thing now goes first...a₁ + (n - 1)d. 00:11:53.200 --> 00:11:59.000 a₁ + (n - 2)d goes next; and then finally, we work our way down: a₁ + d...a₁... 00:11:59.000 --> 00:12:03.100 so now, we have the equation in its normal order and the equation in its opposite order. 00:12:03.100 --> 00:12:07.900 We can add these two equations for s<font size="-6">n</font> together; they are both equal; they are both fine equations. 00:12:07.900 --> 00:12:12.100 There is nothing wrong with them, so we are allowed to use elimination to be able to add equations together. 00:12:12.100 --> 00:12:17.700 We add them together, and we have our normal way of writing it, s<font size="-6">n</font> = a₁ + ... 00:12:17.700 --> 00:12:21.500 + up until our a<font size="-6">n</font> term, a₁ + (n - 1)d. 00:12:21.500 --> 00:12:27.900 And then, the opposite order is s<font size="-6">n</font> = a₁ + (n - 1)d + ... up until a₁. 00:12:27.900 --> 00:12:36.600 We add these together; a₁ + a₁ + (n - 1)d ends up getting us 2a₁ + (n - 1)d. 00:12:36.600 --> 00:12:46.000 Over on the far end, we will end up having the exact same thing: a₁ + (n - 1)d + a₁ will get us 2a₁ + (n - 1)d. 00:12:46.000 --> 00:12:49.800 And we are going to end up getting the same thing for every term in the middle, as well. 00:12:49.800 --> 00:12:56.500 All of those dots will end up matching up, as well, for the same reason that we added 1 and 100, then 2 and 99, then 3 and 98. 00:12:56.500 --> 00:12:58.900 They all ended up matching up together; the same thing happens. 00:12:58.900 --> 00:13:05.100 We will always end up having that be the value for each of the additions through our elimination. 00:13:05.100 --> 00:13:13.600 So, notice, at this point, that we can do the following: we can write this 2a₁ + (n - 1)d here: 2a₁ + (n - 1)d. 00:13:13.600 --> 00:13:18.300 Well, that is the same thing: we can split the 2a₁ into two different parts. 00:13:18.300 --> 00:13:23.300 So, we have a₁ plus...and then we can just put parentheses: a₁ + (n - 1)d. 00:13:23.300 --> 00:13:28.900 Well, we already have a way of writing this out: a₁ + (n - 1)d is the a<font size="-6">n</font> term. 00:13:28.900 --> 00:13:33.800 So, what we have is a₁ + a<font size="-6">n</font>; so we can write this as a₁ + a<font size="-6">n</font>. 00:13:33.800 --> 00:13:41.800 We swap each one of them out; we now have that 2s<font size="-6">n</font> is equal to a₁ + a<font size="-6">n</font> + ... + a₁ + a<font size="-6">n</font>. 00:13:41.800 --> 00:13:48.500 How many terms are there total? There are n terms here total, because we started at a₁ here, 00:13:48.500 --> 00:13:55.300 and we worked our way up until we finally got to a<font size="-6">n</font> here: first term, second term, third term...up until the nth term. 00:13:55.300 --> 00:13:58.800 The first term to the nth term--that means that we have a total of n terms. 00:13:58.800 --> 00:14:05.500 So, a₁ + a<font size="-6">n</font> gets added to itself n times (n terms, so n times, since they are all identical). 00:14:05.500 --> 00:14:10.500 At that point, we have 2s<font size="-6">n</font> = n(a₁ + a<font size="-6">n</font>). 00:14:10.500 --> 00:14:17.400 And since what we wanted on its own was just s<font size="-6">n</font>, we divide 2s<font size="-6">n</font> by 2 on both sides of our equation. 00:14:17.400 --> 00:14:23.500 And we get n/2(a₁ + a<font size="-6">n</font>); great. 00:14:23.500 --> 00:14:28.100 Thus, we now have a formula for the value of any finite arithmetic series. 00:14:28.100 --> 00:14:44.500 Given any arithmetic sequence, a₁, a₂, a₃...the sum of the first n terms is n/2(a₁ + a<font size="-6">n</font>). 00:14:44.500 --> 00:14:50.000 This works for any finite arithmetic sequence, starting at the first term and working up to the nth term. 00:14:50.000 --> 00:15:03.800 So, we can find the sum by only knowing the first term, a₁, the last term, a<font size="-6">n</font>, and the total number of terms, n. 00:15:03.800 --> 00:15:10.900 That is all we need, and we can just easily, just like that, find out what the value of a finite arithmetic series is--that is pretty great. 00:15:10.900 --> 00:15:18.300 Before we go on, though, one little thing to be careful about: be careful to pay attention to how many terms are in the series. 00:15:18.300 --> 00:15:25.000 It can be easy to get the value of n confused and accidentally think it is one higher or one lower than it really is. 00:15:25.000 --> 00:15:28.400 We will see why that is the case in the examples; so just pay really close attention. 00:15:28.400 --> 00:15:33.800 If you are working from a₁ up until a<font size="-6">n</font>, then that is easy, because it is 1, 2, 3, 4...up until the n. 00:15:33.800 --> 00:15:35.500 So, it must be that there are n things there. 00:15:35.500 --> 00:15:38.800 But it can start getting a little bit more confusing if you start at a number that isn't 1-- 00:15:38.800 --> 00:15:43.500 if you start at 5 and count your way up to 27, how many things did you just say out loud? 00:15:43.500 --> 00:15:46.800 We will see what we are talking about there as we work through the examples. 00:15:46.800 --> 00:15:52.700 All right, let's see some examples: Show that the sequence below is arithmetic; then give a formula for the general term, a<font size="-6">n</font>. 00:15:52.700 --> 00:15:56.500 First, to show that it is arithmetic, we need to show that it has a constant difference. 00:15:56.500 --> 00:16:08.400 To get from 2.6 to 3.3, we add 0.7; to get from 3.3 to 4, we add 0.7; to get from 4 to 4.7, we add 0.7; 00:16:08.400 --> 00:16:12.100 and we can see that this is going to keep going like this, so it checks out. 00:16:12.100 --> 00:16:19.000 It is an arithmetic sequence, because there is a common difference; its common difference is 0.7. 00:16:19.000 --> 00:16:23.300 To figure out the general term, a<font size="-6">n</font>, we want to figure out what our a₁ is. 00:16:23.300 --> 00:16:30.400 a₁ is just the first term, which is 2.6; so our general term, a<font size="-6">n</font>, always ends up working like this. 00:16:30.400 --> 00:16:35.000 It is the first term, plus (n - 1) times the common difference. 00:16:35.000 --> 00:16:41.600 So now, we can just plug in our values: a<font size="-6">n</font> =...we figured out that a₁ is 2.6, plus (n - 1)... 00:16:41.600 --> 00:16:46.900 that is just going in because it is the general term...times our difference of 0.7. 00:16:46.900 --> 00:16:51.300 And there we are; there is our general term; there is the formula for the nth term. 00:16:51.300 --> 00:16:57.200 Alternatively, if we wanted to, we could also simplify this a little bit more, so it isn't n - 1 (that part doesn't show up). 00:16:57.200 --> 00:17:00.900 Sometimes it is useful to have it in this format; but other times we might want to simplify it. 00:17:00.900 --> 00:17:13.400 So, if we decided to simplify it, we would have a<font size="-6">n</font> = 2.6 + n(0.7), so 0.7n, minus 1(0.7), so minus 0.7; 00:17:13.400 --> 00:17:20.600 so the 2.6 and the -0.7 interact, and we have 1.9 + 0.7n. 00:17:20.600 --> 00:17:25.100 Alternatively, we could write it like this: either of these two ways is perfectly valid. 00:17:25.100 --> 00:17:28.300 Either one of these two things is a formula for the general term. 00:17:28.300 --> 00:17:31.500 Sometimes it will be more useful to write it one way, and sometimes it will be more useful to write it the other way. 00:17:31.500 --> 00:17:37.700 So, don't be scared if you see one written in a different way than the other one; they are both totally acceptable. 00:17:37.700 --> 00:17:42.000 The second example: Find the value of the arithmetic series below. 00:17:42.000 --> 00:17:45.700 What is our difference? That will help us understand what is working on here. 00:17:45.700 --> 00:17:50.000 The difference will not actually be necessary to use our formula for an arithmetic series, 00:17:50.000 --> 00:17:53.600 but it will help us see what is going on just a little bit on our way to using it. 00:17:53.600 --> 00:18:00.100 We have a difference of 5 each time; so it is + 5, + 5, + 5...difference = 5. 00:18:00.100 --> 00:18:06.100 We need three things to know what the series' value is. 00:18:06.100 --> 00:18:11.000 We need to know the first term; that is easy--we can see it right there: a₁ = 7. 00:18:11.000 --> 00:18:16.100 We need to know the last term; that is easy, as well: a<font size="-6">n</font> = 107. 00:18:16.100 --> 00:18:20.900 And we need to know what the number of terms is, n = ?. 00:18:20.900 --> 00:18:24.200 So, how can we figure out how many terms there are? 00:18:24.200 --> 00:18:34.000 We might be tempted to do the following: 107 - 7 comes out to be 100; and then we say, 00:18:34.000 --> 00:18:43.600 "Oh, our difference is 5, so let's divide by 5," and so we get 20; so n must be 20...NO, that is not the case. 00:18:43.600 --> 00:18:47.800 Now, to understand why this is not the case, we need to look at something. 00:18:47.800 --> 00:18:52.900 Let's create a little sidebar here to understand what is going on a little better. 00:18:52.900 --> 00:19:00.000 Look at...if we wanted to talk about the number from 1 to 25, if we wanted to count how many numbers there are between 1 and 25, 00:19:00.000 --> 00:19:08.000 we count: 1, 2, 3, 4...25--pretty obvious: that means that the number of numbers is 25. 00:19:08.000 --> 00:19:10.600 There are 25 things there; great--that makes sense. 00:19:10.600 --> 00:19:15.500 What if we were talking about going from 25 to 50? 00:19:15.500 --> 00:19:25.300 Well, we might say that we can count by hand...50 - 25...so then, there are a total of 25 terms, because 50 - 25 is 25...No. 00:19:25.300 --> 00:19:30.200 Wait, what? Well, let's count it by hand: how does this work. 00:19:30.200 --> 00:19:48.900 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50; that is 26 things that we just counted out there. 00:19:48.900 --> 00:19:55.400 So, what is going on? 25 counts as something we have to count. 00:19:55.400 --> 00:20:02.500 When we count from 1 to 25, if we just subtracted 25 minus 1, if we subtracted 1 from 25, 25 - 1, that is 24. 00:20:02.500 --> 00:20:07.400 But we don't say, "Oh, from 1 to 25, there must be 24 numbers, because 25 - 1 is 24"--no, we don't think like that. 00:20:07.400 --> 00:20:11.100 We know that counting from 1 to 25, it is 25 things there. 00:20:11.100 --> 00:20:20.200 So, counting from 25 to 50 means a difference of 25; that means that there are 25 steps to get to 50 from 25. 00:20:20.200 --> 00:20:23.400 But we also have to count the first location that we started on. 00:20:23.400 --> 00:20:27.300 We have to actually count to 25, as well; so that is a total of 26. 00:20:27.300 --> 00:20:40.300 What we have is: we have 50 - 25 = 25; but then we have to have 25 + 1 = 26. 00:20:40.300 --> 00:20:45.400 So, our number is 26 for how many things we ended up doing. 00:20:45.400 --> 00:20:49.800 It is the same thing with 7 to 107; how many steps? 00:20:49.800 --> 00:20:55.900 If we have a distance of 5 for each step, how many steps do we have to take from the 7? 00:20:55.900 --> 00:21:00.300 Well, we have to take 20 steps, because 20 times 5 is 100. 00:21:00.300 --> 00:21:07.900 So, if we take 20 5-distance steps from 7, we will make it to 107; so we have 20 steps that we take. 00:21:07.900 --> 00:21:18.400 But we also have to count the 7, so we end up actually have n = 21. 00:21:18.400 --> 00:21:20.400 And so, that is our value for n. 00:21:20.400 --> 00:21:23.100 This is why you really have to think about this stuff carefully. 00:21:23.100 --> 00:21:27.300 It is really easy to just say, "OK, I took that many steps, so that must be my value." 00:21:27.300 --> 00:21:31.600 No, you have to really pay attention to make sure that you are counting also where you started. 00:21:31.600 --> 00:21:34.700 But sometimes, you have to pay attention and think about it: "Did I already count where I started?" 00:21:34.700 --> 00:21:36.600 So, you really have to be careful with this sort of thing. 00:21:36.600 --> 00:21:40.500 It is easy, as long as you can get a₁, a<font size="-6">n</font>, and the number of steps, n. 00:21:40.500 --> 00:21:44.100 But sometimes, it is hard to really realize just how many steps you have precisely. 00:21:44.100 --> 00:21:45.700 So, be careful with that sort of thing. 00:21:45.700 --> 00:21:53.300 All right, at this point, we are ready to use our formula: n/2(a₁ + a<font size="-6">n</font>) is the value 00:21:53.300 --> 00:21:57.600 of the sum of all of those numbers, the sum of that finite arithmetic series. 00:21:57.600 --> 00:22:11.500 Our n was 22, over 2; our a₁ was 7; our a<font size="-6">n</font> was 107, so + 107; so we get 21/2(114). 00:22:11.500 --> 00:22:21.200 We punch that into a calculator, and we end up getting 1197; 1197 is our answer for adding that all up. 00:22:21.200 --> 00:22:25.800 All right, the third example: here is my thing that I said at the beginning, when we talked through the introduction. 00:22:25.800 --> 00:22:33.800 At this point, we are now able to add the numbers from 1 to 1000 in less time than it takes to put on a pair of shoes and tie them up. 00:22:33.800 --> 00:22:40.500 If you are going to take off your shoes and test if that is really the case, now is the time to do it, before we start looking at this problem. 00:22:40.500 --> 00:22:48.800 Are you ready? OK, let me read the problem, and then we will have things be a fair challenge between shoe-putting-on and massive addition. 00:22:48.800 --> 00:22:56.300 Add all of the integers from 1 to 1000 (so 1 + 2 + 3...+ 999 + 1000). 00:22:56.300 --> 00:23:00.500 All right, are you ready? Ready, set, shoes on now! 00:23:00.500 --> 00:23:08.500 The first term is equal to 1; our last term is equal to 1000; the total number of terms we have from 1 to 1000 is simply 1000. 00:23:08.500 --> 00:23:21.400 So, it is 1000, the number of terms, divided by 2 times the first term, plus the last term...1000, so 500 times 1001...is equal to 500,500; I am done! 00:23:21.400 --> 00:23:28.700 It's pretty amazing how fast we can end up adding everything from 1 to 1000 in that little time. 00:23:28.700 --> 00:23:33.600 This is the power of the series formula; this is the power of studying series-- 00:23:33.600 --> 00:23:38.500 the fact that we can add things that would take so long to work out by hand, like that. 00:23:38.500 --> 00:23:41.800 We can do this stuff really, really quickly, once we work through this. 00:23:41.800 --> 00:23:47.900 Our first term was 1; our last term was 1000; so a₁ = 1; a<font size="-6">n</font> = 1000. 00:23:47.900 --> 00:23:52.100 How many things are there from 1 up to 1000? Well, that one is pretty easy; that one is 1000. 00:23:52.100 --> 00:24:04.700 So, we have n/2, 1000/2, times 1 + 1000, so 500 times (I accidentally made a little bit of a typo as we were writing that out) 1001. 00:24:04.700 --> 00:24:09.600 Multiply that out, and you get 500 thousand, 500; it is as simple as that. 00:24:09.600 --> 00:24:13.400 The fourth example: Find the value of the sum below. 00:24:13.400 --> 00:24:18.500 To do this, let's write out what this sigma notation ends up giving us. 00:24:18.500 --> 00:24:27.800 i = 4 is our first place, so that is going to be 53 minus...oops, if it is going to be an i here and a k here, they have to agree on that. 00:24:27.800 --> 00:24:34.600 So, that should actually read as a k, or the thing on the inside should read as an i, for this problem--I'm sorry about that. 00:24:34.600 --> 00:24:50.900 53 - 4 times 4 is our first one; plus 53 - 4 times 5 (our next step up--our index goes up by one) plus 53 - 4 times 6 00:24:50.900 --> 00:25:02.700 (our index goes up one again); and it keeps doing this, until we get to our last upper limit for our sum, 53 minus 4 times 25; cool. 00:25:02.700 --> 00:25:05.500 Now, at this point, we think, "OK, how can we add this up?" 00:25:05.500 --> 00:25:12.800 Well...oh, this is an arithmetic sequence; it is 4 times some steadily-increasing, one-by-one thing. 00:25:12.800 --> 00:25:16.700 So, it is an arithmetic sequence, an arithmetic series, that is appearing here. 00:25:16.700 --> 00:25:18.200 If that is the case, what do we need? 00:25:18.200 --> 00:25:25.200 We need to know the first term, the last term, and the number of terms that there are total. 00:25:25.200 --> 00:25:28.900 If that is the case, all we really care about is this first term and this last term. 00:25:28.900 --> 00:25:31.800 All of the stuff in the middle--we don't really need to work with it. 00:25:31.800 --> 00:25:53.800 53 - 4 times 4, so 53 - 16, plus...up until our last term of 53 - 4 times 25 is 100; 53 - 16 comes out to be 37, plus...plus...53 - 100 comes out to be -47. 00:25:53.800 --> 00:26:05.700 Our first term is 37; our last term is -47; the only real question that we have now is what is the value for n. 00:26:05.700 --> 00:26:10.300 How many terms are there total? We are counting from 4 up to 25. 00:26:10.300 --> 00:26:13.400 So, from 4 up to 25, how many steps do we have to take there? 00:26:13.400 --> 00:26:24.600 25 - 4 means 21 steps; but notice, it is steps; there are 21 steps, but we also have to count the k = 4. 00:26:24.600 --> 00:26:32.500 It is 21 steps above 4, so we also have to count the step at 4; so 21 + 1 counts where we start. 00:26:32.500 --> 00:26:37.700 It is not just how many steps you take forward, but how many stones there are total, so to speak. 00:26:37.700 --> 00:26:45.800 21 + 1 = 22 for our value of n; so we get n = 22; great. 00:26:45.800 --> 00:26:51.400 n = 22; we know what the first one is; we know what the last one is; we are ready to work this out. 00:26:51.400 --> 00:27:02.000 22 is our n, divided by 2; n/2 times the first term, 37, plus the last term, -47...we work this out. 00:27:02.000 --> 00:27:14.800 22/2 is 11, times 37 + -47 (is -10)...we get -110; that is what that whole series ends up working out to be; cool. 00:27:14.800 --> 00:27:17.100 All right, and we are ready for our fifth and final example. 00:27:17.100 --> 00:27:21.500 An amphitheater has 24 seats in the third row, 26 in the fourth. 00:27:21.500 --> 00:27:29.900 If this pattern of seat increase between rows is the same for any two consecutive rows, and there are 27 rows total, how many seats are there in total? 00:27:29.900 --> 00:27:32.400 The first thing we want to do is understand how this is working. 00:27:32.400 --> 00:27:37.100 Well, it is an amphitheater; we can see this picture here to help illustrate what is going on. 00:27:37.100 --> 00:27:40.600 As we get farther and farther from the stage, it curves out more and more. 00:27:40.600 --> 00:27:44.900 It is pretty small near the stage; as it gets farther and farther, it expands out and out. 00:27:44.900 --> 00:27:50.100 So, that means there are more seats in every row, the farther back in the row we go. 00:27:50.100 --> 00:27:52.800 Later rows will end up having more seats than earlier rows. 00:27:52.800 --> 00:27:56.600 That is why we have 24 in the third, but 26 in the fourth. 00:27:56.600 --> 00:28:02.700 We can see this: the early rows have fewer seats than the later rows, from how far they are from the stage. 00:28:02.700 --> 00:28:09.400 OK, what we are looking for is how many seats there are total. 00:28:09.400 --> 00:28:17.900 What we can do is talk about the third row having 24 seats; the fourth row has 26 seats; so we could think of this as a sequence, 00:28:17.900 --> 00:28:23.300 where you know that it has the constant increase; the pattern of seat increase between rows is always the same. 00:28:23.300 --> 00:28:27.900 What we have here is an arithmetic sequence; it makes sense, since that is what the lesson is about. 00:28:27.900 --> 00:28:36.100 We can write 24 seats in the third row as a₃ = 24. 00:28:36.100 --> 00:28:45.800 We also know that it is 26 in the fourth row; so a₄ = 26. 00:28:45.800 --> 00:28:54.200 OK, so if that is the case, and the pattern of seat increase is always the same for two consecutive rows, 00:28:54.200 --> 00:29:00.700 that means...to get from 24 to 26, we added +2; we have a common difference of positive 2. 00:29:00.700 --> 00:29:04.000 So, if that is the case, what would the second row have to be? 00:29:04.000 --> 00:29:10.000 Well, it would have to be -2 from the third row, so it would be at 22; 22 + 2 gets us to 24. 00:29:10.000 --> 00:29:17.800 The same logic works for the first row, so the first row must be at 12 20, 22, 24, 26...that is the number of seats. 00:29:17.800 --> 00:29:20.000 We see that we have a nice arithmetic sequence here. 00:29:20.000 --> 00:29:23.200 What we are really looking to do is take a finite arithmetic series. 00:29:23.200 --> 00:29:28.200 We are looking to figure out what is the 27th partial sum, because what we want to do 00:29:28.200 --> 00:29:34.100 is add the number of seats in the first, second, third, fourth...up until the twenty-seventh row. 00:29:34.100 --> 00:29:35.800 And we will be able to figure out all of those. 00:29:35.800 --> 00:29:40.900 So, what we need to use the formula that we figured out: we need to know how many seats there are in the first row, 00:29:40.900 --> 00:29:44.600 how many seats there are in the last row, and the total number of rows. 00:29:44.600 --> 00:29:47.100 How many seats are there in the last row? 00:29:47.100 --> 00:29:51.100 a<font size="-6">27</font> is going to be our last row, because there are 27 rows total. 00:29:51.100 --> 00:30:03.300 a<font size="-6">27</font> is going to be the number in the first row, plus...27 - 1 (n - 1 is 27 - 1) times our difference (our difference is 2, so times 2). 00:30:03.300 --> 00:30:05.900 This makes sense, because what we have here is that the 27th row 00:30:05.900 --> 00:30:12.200 is going to be equal to our first row, 20, plus...how many steps is it to get from the first to the twenty-seventh row? 00:30:12.200 --> 00:30:18.000 That is going to be 26 steps, times an increase of 2 for every row we go forward. 00:30:18.000 --> 00:30:24.800 We work this out; that means that our 27th row is equal to 20 plus 26 times 2 is 52; 00:30:24.800 --> 00:30:32.700 a<font size="-6">27</font>, our 27th row...the number of seats in our 27th row, 20 + 52, is 72 seats total. 00:30:32.700 --> 00:30:39.000 So, at this point, we have 72 seats for our final row, 20 seats for our first row... 00:30:39.000 --> 00:30:43.000 how many total rows are there? Well, that is going to be...if we are going from the first row, 00:30:43.000 --> 00:30:46.800 up until the 27th row, then we can just count: 1, 2, 3, 4...counting up to 27. 00:30:46.800 --> 00:30:51.600 That is easy; that is 27, so n = 27. 00:30:51.600 --> 00:30:58.900 So, our formula is the number of terms total, divided by 2, times the first term plus the last term. 00:30:58.900 --> 00:31:04.400 So, our number of terms total (number of rows total) is 27, divided by 2, times... 00:31:04.400 --> 00:31:12.400 what is the first term, the first number of seats? 20, plus...what is the last number of seats, our last term? 72. 00:31:12.400 --> 00:31:25.000 27/2 times 92...we work that out with a calculator, and we end up getting 1242 seats total in the amphitheater. 00:31:25.000 --> 00:31:26.300 Great; there we are with the answer. 00:31:26.300 --> 00:31:30.200 All right, in the next lesson, we will end up looking at geometric sequences and series, 00:31:30.200 --> 00:31:33.700 which give us a way to look at this through multiplying instead of just adding. 00:31:33.700 --> 00:31:36.000 All right, we will see you at Educator.com later--goodbye!