WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about arithmetic sequences and series.
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Now that we have an understanding of sequences and series, we are ready to look at specific kinds of sequences.
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The first that we will consider is an arithmetic sequence, a sequence where we add a constant number each step.
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We will add some number, and we keep adding the same number every time we go forward a term.
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Sequences of this form pop up all the time in real life, and we often need to add up their terms.
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We will explore the creation of a formula for arithmetic series that will allow us to quickly and easily add up those terms.
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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.
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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.
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All right, a sequence is **arithmetic** if the difference between any two consecutive terms is constant.
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We can show that with the recursive relationship a<font size="-6">n</font> - a<font size="-6">n - 1</font> = d.
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Notice: the nth term minus the n - 1 term (that is, the term one before the nth term) equals d.
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Some term minus the term before it equals d.
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And d is just some constant number; we call d the **common difference**.
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Here are two examples of arithmetic sequences: they are arithmetic because every step in the sequence has the same change.
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For example, here, 1 to 4, 4 to 7, 7 to 10, 10 to 13...it is + 3, + 3, + 3, + 3.
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It is always the same amount that we change; it is always adding a constant number.
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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.
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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.
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The important thing is that it is always the same; the difference can be positive; the difference can be negative;
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but it always has to be the same for each one--that is what makes it an arithmetic sequence.
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How can we find the nth term? The definition for an arithmetic sequence is based on a recursive relation.
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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.
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So, how can we turn this formula into something for the general term--how can we get a general term formula out of this?
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Remember: a recursive relation needs an initial term--we have to have some starting place.
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There is nothing before our starting place to refer back to, so we actually have to be given the initial term directly.
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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.
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Now, from a<font size="-6">n</font> = a<font size="-6">n - 1</font> + d, we see that a₁ relates to later terms.
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At the most basic level, we have that a₂ is equal to a₁ + d.
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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.
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We can take this out and continue looking at later terms.
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a₃ would be equal to a₂ + d, based on this recurrence relation.
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But we just figured out that a₂ is equal to a₁ + d, so we can swap out for a₂:
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a₁ + d, which now gets us a₁ + 2d when we add this d to that one there.
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So, we have a₁ + 2d for a₃; when we work on a₄,
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well, a₄ is going to be a₃ + d, the previous term, adding d.
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But once again, we have just figured out that a₁ + 2d is what a₃ is.
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So, we can plug in for a₃; we have a₁ + 2d, and now we can add that d onto the 2d.
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So, we end up getting that a₁ + 3d is what a₄ is equal to.
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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.
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So, a₁ = a₁; a₂ = a₁ + d; a₃ = a₁ + 2d;
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a₄ = a₁ + 3d, and this pattern will continue down.
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We see that the nth term is n - 1 steps from a₁.
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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.
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Since it is n - 1 steps away from a₁, we will have added d for each of those steps;
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so we will have added d that many times, or n - 1 times d.
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That means that the a<font size="-6">n</font> term is equal to a₁ plus all of those steps times d.
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So, the nth term, a<font size="-6">n</font>, equals a₁, our first term, plus n - 1 times d.
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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.
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With those two pieces of information, we automatically have the general term; we automatically have that nth term formula--pretty great.
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How about if we want an arithmetic series formula?
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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?
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How could we add that whole thing up? Well, we could do it by brute force, where we would just sit down
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with a piece of paper or a calculator and just punch the whole thing out.
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We could do it by hand; but that is going to take a long time.
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And any time that we end up seeing something that is going to take us forever to do, we want to ask ourselves,
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"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?"
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How are we going to do that? We want to look for some sort of pattern that we can exploit.
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We want to find a pattern that we can exploit--something that will keep happening--
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something that we can rely on, that will keep us from having to add up all of these numbers,
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because we can instead use this pattern to give us a deeper insight to what is going on.
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So, if we look at this for a while, we might start to realize that there is a pattern in the numbers.
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But that doesn't help us, because that is just adding the numbers.
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But is there a way that addition itself has a pattern?
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There is something that we could match up--something that we could create--and this is where we are getting really clever.
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This is the hard part, where you really have to sit down and think about it for a long time.
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And hopefully you just end up getting some "lightning bolt" of insight.
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And hopefully, at some point, we will notice that here is 100; here is 1; if we add them together, we get 101.
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But not only that--if we had 99 (let's use a new color)...if we use 99 and 2, we get 101.
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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...
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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...
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what we can do is pair up each number from 1 to 50 with a number from 51 to 100.
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And we will always be able to make 101 out of it.
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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.
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So, we were able to figure out this pattern; there is something going on.
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Now, we are finding something; now we have something that we can pull into a formula that will make this really easy.
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With this realization in mind, let's look for an easy way to pair up the numbers.
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The first thing: it is nice to give names to things in algebra--it lets us work with them more easily.
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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.
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Now, notice: we can rewrite the order of those numbers, since order of addition doesn't matter.
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1 + 2 + ... + 99 + 100, here, is the same thing as 100 + 99 + ... + 2 + 1.
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We can swap the order, and we still have the same value in the end.
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That is one of the nice things about the real numbers: order of addition doesn't matter.
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Furthermore, we can add two equations together--that is elimination.
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Remember: when we worked on systems of linear equations, if you have an equation, you can just add it
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to another equation, because they are both working equations.
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You can add the left sides and the right sides, and you know that everything works out; there is nothing wrong with doing that.
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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.
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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;
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here we have 2 and 99, so we get 101; here we have 1 and 100, so we get 101.
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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.
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How many terms are there total? Well, we had 1, 2, 3...99, 100; so we had 100 terms, left to right.
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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.
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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.
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We can condense all of that addition with multiplication: 3 + 3 + 3 + 3 is the same thing as 3 times 4, 4 times 3.
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So, if we have 101 appearing 100 times, then we can turn that into 100 times 101.
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Our left side is still just 2s; so we have 2s = 100(101).
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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.
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Divide both sides by 2; 100 divided by 2 gets us 50, so 50 times 101 means we have an answer of 5050.
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So, that probably took about as much time as if we had added up 1 + 2, all the way up to 100.
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If we had done that whole thing by hand, it would have taken a while.
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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!"
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Indeed, that is what will work out.
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We have this method in mind of being able to string all of the things in our arithmetic sequence together,
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and then flip it and add them together and see what happens.
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We can now figure out a general formula for any finite arithmetic series.
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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.
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So, s<font size="-6">n</font> = a₁ + a₂ + a₃ + ... up until we get to + a<font size="-6">n - 1</font>,
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up until, finally, a<font size="-6">n</font> is our end, because we have the nth partial sum; great.
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Earlier, we figured out the general term for any arithmetic sequence is a<font size="-6">n</font> = a₁ + (n - 1)d.
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So, we can swap out a₁ for what it is in the general form, a₂ for what is in the general term,
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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.
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This will get everything in terms of a₁ and d and that n; great.
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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,
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and then a₂ would be a₁ + d (2 minus 1, so 1 times d...a₁ + d).
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We work our way out: a<font size="-6">n - 1</font> would be a₁ + (n - 2)d;
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we plug in n - 1 for the general n term, so n - 1, minus 1...n minus 2 times d.
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And finally, the a<font size="-6">n</font> would be n plus n minus 1 times d.
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Great; so we have this thing where the only thing showing up there is a₁, n, and d.
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We have far fewer things that we have to worry about getting in our way.
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Furthermore, we can write s<font size="-6">n</font> in the opposite order; we are allowed to flip addition order.
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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.
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a₁ + (n - 2)d goes next; and then finally, we work our way down: a₁ + d...a₁...
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so now, we have the equation in its normal order and the equation in its opposite order.
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We can add these two equations for s<font size="-6">n</font> together; they are both equal; they are both fine equations.
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There is nothing wrong with them, so we are allowed to use elimination to be able to add equations together.
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We add them together, and we have our normal way of writing it, s<font size="-6">n</font> = a₁ + ...
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+ up until our a<font size="-6">n</font> term, a₁ + (n - 1)d.
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And then, the opposite order is s<font size="-6">n</font> = a₁ + (n - 1)d + ... up until a₁.
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We add these together; a₁ + a₁ + (n - 1)d ends up getting us 2a₁ + (n - 1)d.
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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.
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And we are going to end up getting the same thing for every term in the middle, as well.
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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.
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They all ended up matching up together; the same thing happens.
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We will always end up having that be the value for each of the additions through our elimination.
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So, notice, at this point, that we can do the following: we can write this 2a₁ + (n - 1)d here: 2a₁ + (n - 1)d.
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Well, that is the same thing: we can split the 2a₁ into two different parts.
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So, we have a₁ plus...and then we can just put parentheses: a₁ + (n - 1)d.
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Well, we already have a way of writing this out: a₁ + (n - 1)d is the a<font size="-6">n</font> term.
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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>.
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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>.
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How many terms are there total? There are n terms here total, because we started at a₁ here,
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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.
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The first term to the nth term--that means that we have a total of n terms.
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So, a₁ + a<font size="-6">n</font> gets added to itself n times (n terms, so n times, since they are all identical).
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At that point, we have 2s<font size="-6">n</font> = n(a₁ + a<font size="-6">n</font>).
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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.
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And we get n/2(a₁ + a<font size="-6">n</font>); great.
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Thus, we now have a formula for the value of any finite arithmetic series.
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Given any arithmetic sequence, a₁, a₂, a₃...the sum of the first n terms is n/2(a₁ + a<font size="-6">n</font>).
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This works for any finite arithmetic sequence, starting at the first term and working up to the nth term.
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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.
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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.
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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.
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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.
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We will see why that is the case in the examples; so just pay really close attention.
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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.
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So, it must be that there are n things there.
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But it can start getting a little bit more confusing if you start at a number that isn't 1--
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if you start at 5 and count your way up to 27, how many things did you just say out loud?
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We will see what we are talking about there as we work through the examples.
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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>.
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First, to show that it is arithmetic, we need to show that it has a constant difference.
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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;
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and we can see that this is going to keep going like this, so it checks out.
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It is an arithmetic sequence, because there is a common difference; its common difference is 0.7.
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To figure out the general term, a<font size="-6">n</font>, we want to figure out what our a₁ is.
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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.
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It is the first term, plus (n - 1) times the common difference.
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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)...
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that is just going in because it is the general term...times our difference of 0.7.
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And there we are; there is our general term; there is the formula for the nth term.
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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).
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Sometimes it is useful to have it in this format; but other times we might want to simplify it.
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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;
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so the 2.6 and the -0.7 interact, and we have 1.9 + 0.7n.
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Alternatively, we could write it like this: either of these two ways is perfectly valid.
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Either one of these two things is a formula for the general term.
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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!