 This video will talk about sequences and series. A sequence can be thought of as a pattern of numbers listed in a prescribed order. So think about the sequence of events that you do. You have to do them in a certain order. So that's what we're going to deal with with our numbers. And it's a pattern of numbers which continues to do that over and over and over again. And a series is just the sum of all the terms in a sequence. A finite series is the sum of an indicated number of terms. So say I want to add up the first five terms and I would say s of five down here. So terms of a sequence, the general term is going to be a sub n. That's going to be very important. A sub n. n is going to be equal to a natural number. So when we're talking about term numbers, we're not going to talk about integers. There won't ever be a negative term number and there won't be any decimal term numbers. There are going to be natural numbers. One, two, three, four, the counting numbers. N being the number of term. So if we don't know yet how many terms we have a sub n. And then the initial terms would be a sub 1, a sub 2, a sub 3, a sub 4 all separated with commas. So then we have types of sequences. Well, we have a finite sequence which as it sounds, it has a beginning and an end first and last term. And an infinite sequence gives you a first term but it just goes on forever. So we want to find the first four terms of a sequence and then we want to find the eighth and twelfth terms. So a sub 1, the first term would be 2 times the 1 which is my term number, plus 3. So 2 plus 3 would be 5. So the first term of my sequence is 5. A sub 2 is going to be 2 times 2 plus 3 or 7. A sub 3, 2 times 3 plus 3 or 9. A sub 4, which is our last one, 2 times 4 plus 3 or it's going to give us 11. So our sequence is 5, 7, 9 and 11. So they want to know the eighth and the twelfth. So a sub 8, just like we did with all the others, 2 times 8 would be 16 plus 3 more would be 19. And a sub 12 is going to be 2 times 12 which is 24 plus 3 more which would be 27. So now we want to turn to our technology and see if it can do some of these things for us. When we do use this, we're going to have to tell it four things once we figure out how to set it up. We're going to tell it what the actual sequence is, then we're going to tell us what variable we're using and we're just going to use x. And then with the starting value or what term number that is, like the first, second, what term number I want to start with and then the last term number that I end with. It's not the value, it's the term number. So it is started. We need our calculator and we're going to compute the sequence so it says take second and then stat and then you'll notice over here we see ops. So you want to arrow over to ops and then down here at 5 we see sequence which is exactly what we want. So we put in 5 and now we're going to tell it those four things. Let's go back and use the one that we had before which was 2n plus 3 and so we put 2x plus 3 and then a comma and then we used x so we'll use say x and then comma we want to do the first four terms. So 1 and then we want to do the fourth term too, add from the first up to the fourth term and then we can close our parentheses and if we have that we have the 57911. We want to try doing the same thing with our technology but we want to talk about the first five terms of 2n. So try again, get back to your calculator, second, stat, over to ops, operations, what operation would I like to do? I'd like to have a sequence and I'm using 2x. My variable is an x comma my term 1 comma 5 and close my parentheses. So if I want to know a sub 1 through a sub 5 it's going to be 2, 4, 6, 8, 10. And if I specifically wanted to know what a sub 5 was, maybe I had just done the first four, then we're going to have the sequence is going to be 2x and our variable is going to be x but our first term number is going to be 5 and our second term number will also be 5 because we only want to know one particular. So coming back in here, proving that it's 10 because we already know it's 10. Second, stat, over to ops, choose 5, 2x. Now you can see why I picked a nice easy sequence, comma x and then comma 5, comma 5, close the parentheses, what is the fifth term? It is 10. So if you want to know the whole sequence or you just want to know a given term, you don't have any problems. Now we can also talk about a sum. We can sum up all the terms of the sequence and to do that we have one extra step and I think we talked about it on your paper and what we want to do is take the sum and then we want to have of our sequence. This we know with second op. But sum is going to be second, oh stat, sorry, second stat and then op and then 5. This is going to be second and it's also going to be stat. But we want to go all the way over to math first and choose 5. So let's try, clear all this out. Second stat, go past the ops over to math and we want the sum, 5. What do we want to sum? We want to sum the sequence, so second stat over to ops and then we want again the sequence is 5 and we're going to use our same sequence, so 2x. And again it has the same variable of x. And we want to sum up the first 5 terms, so we want to start with the first term and we want to end with the last term or the fifth term. And then we press enter and we find out that is 30. We can verify that it's 30 over here. 2 plus 4 would be 6, 6 plus 6 would be 12, 12 plus 8 would be 20 and 10 more would be 30. So we really did find that s sub 5 was equal to 30. All right, sometimes we have these recursive sequences which these sequences rely on the term before it. And you have to have what they call a seed element that's usually going to, they have to give you a sub 1 or some element so that you can refer back to it since it relies on a term before. So we want to find the first four terms. Well a sub 1 obviously is negative 2. A sub 2 is going to be a sub 2 minus 1 or a sub 1 minus 16. So that in reality is negative 2 minus 16 or negative 18. So a sub 3 is going to be a sub 3 minus 1 which is really a sub 2 minus 16. So again, a sub 2 was negative 18 and we want to subtract 16 from that. And we're going to have a negative 34 and our last term is going to be a sub 4 minus 1 which is really just a sub 3 minus 16. And a sub 3 was negative 34 and a sub 4 we're going to subtract 16 from that. And if we do that, we end up with 50. So our sequence would be negative 2, negative 18, negative 34, negative 50. Okay, here's a new idea for us. It's a factorial. Some factorials are just the product of the number and every number before it. Just the natural numbers. So 3 times 2 times 1 would be 3 factorial and if you multiply all that, you get 6. So we want to write the first four terms of this sequence where we're going to use our calculator. We have a sub 1 is going to be 1 factorial over 1 plus 3 factorial. Or it's going to be 1 factorial over 4 factorial. And we're just going to let the calculator do that. And to do that, we need to come over to our calculator, hit math. And get some new things in here, go all the way over to prb. I'm not sure what that stands for. But you see 4 right here has that exclamation point. So we can say that now we can go back to our home screen and start putting it in 1 math over to prob, perb, whatever that is, number 4. So 1 factorial divided by 4 factorial. And these are actually simple enough that we probably could just do them by hand. But if we do 1 factorial divided by 4 factorial, we get this number and I want to see if I can make it a fraction. So just press enter and enter. And it's equal to 1 over 24. 1 factorial is just going to be 1. And 4 factorial is going to be 4 times 3 times 2 times 1. 4 times 3 is 12, times 2 is 24, times 1 is 24. The second term is going to be a sub 2 is going to be 2 factorial over 2 plus 3 or 5 factorial. Well, we can simplify by saying that this is 2 factorial. And if I start with seeing the 5, I got 5 times 4 times 3 times, well, 2 factorial is 2 times 1. So the 2 factorial and the 2 factorial cancel and I have a 1 up here. Maybe I should have done that with a different color. Those cancel to give me 1 up here. So I really just have 1 over 4 times 5 would be 20 times 3 would be 60. So we can do them either way. Let's go with for a sub 3. That's going to be 3 factorial over 3 plus 3 or 6 factorial. And we would like to make this as simple as possible. So 3 factorial over 6 times 5 times 4 times 3 factorial. So those cancel each other out and I love to the 1 on top. And then I have this. So 6 times 5 would be 30 times 4 would be 120. And then if I do a sub 4, I'm going to have 4 factorial over 4 plus 3, which would be 7 factorial. And I have on the bottom 7 times 6 times 5 and then 4 factorial cancels. So this is equal to 7 times 6 is 42. And I'm not sure what that is 42 times 5. I could have done it quicker, but it's 210. So we have 1 because the 4 factorials cancel each other out 210. So what is our sequence then? Our sequence is 1 over 24 and 1 over 60 and 1 over 120 and 1 over 210. All right, summation notation. When the general term is the sequence is known, then a summation notation can be used to write the related series in a formula. So we have this nice looking summation. It's a sigma and then we have i is equal to 1 and 2k. So this is telling us that we want to sum from the first term to the k term. First k term, even though it may not be th. And our sequence is obviously a sub n. These letters, this i down here could be j, k, l, and those are the ones that are usually used. And we're just going to go the simple route here. Given a sub n, we want to write the sigma notation for s of 6. This means we want to sum the first six terms. So that means that we're going to have our sigma. And we're going to start with 1 because it said the first six terms. And on top, we're going to have 6 because that tells me the last term that I want to include. And we actually just write 2n minus 1. This a sub n is our formula. So now that it says we went to expand and then verify using our calculator. So we just need to plug and chug, basically. So we are summing here. So the first term would be 2 times 1, which is 2 minus 3, or negative 1. And then if we put 2 in here, we have 2 times 4 minus 3 would be 1. And if we put 3 in here, we have 2 times 3, which is 6 minus 3, which would be plus 3. And if we put 4 in here, we're going to have 5. And then we have, if we put 5 in here, 2 times 5 is 10 minus 3 is going to be plus 7. And if we sum all that up, 1 plus negative 1 is going to cancel each other out. This gives me 10 plus 5, so I have that is equal to 15. And if we come back and practicing our technology again, second stat, all we care about is the sum. So that means we have to go all the way over to math and choose sum. And then I have to tell it the sequence. So second math over to op, 5. And my sequence is 2x minus 3. And then I use the variable x. And I want to go from the first term to the fifth term. It's going to come out right there. So I can insert that. And now I have my, I can come down here to the bend and close my parentheses because I've got it all in there. And it tells me it's 15. And finally, we want to use, write using segment notation. Now we have to find the formula. So what are we doing? It looks like we have, we start with 5, and then we added 5. And we add 5 every time. So we could say, okay, how about we take 5 times n. Isn't this difference, sound like slope? Difference is 5, and we started with 5. Now we have to double check it though because if we just have n here, if I use 1, I'm going to have 10 instead of 5. So it would be better if we came in here and said, okay, I really need to have 1 less because if I have 1 less, now I will have 1 minus 1, which is 0. And 5 plus 0 is 5. So summation, we're going to go from the first term and count how many there are, 1, 2, 3, 4, 5, 6 terms. And we want to simplify this just a little bit because we don't want a whole bunch in here. So 5n minus 5 and negative 5 and the positive 5 cancel out. So we just have the summation of 1 to 6 of 5n.