For more information, please see full course syllabus of Math Analysis

For more information, please see full course syllabus of Math Analysis

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## Transcription

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### Introduction to Sequences

- A
*sequence*is an ordered list of numbers. We can write out a sequence as

We call each of the entries in the sequence aa _{1}, a_{2}, a_{3}, a_{4}, …, a_{n}, …*term*. Above, a_{1}is the first term, a_{2}is the second term, and so on. Any symbol can be used to denote the sequence, while the subscript (small number to the right) tells us which term it is in the sequence. - If a sequence goes on forever without stopping, it is called an
*infinite sequence*. Most of the sequences we will work with will be infinite sequences. If a sequence does stop, it is called a*finite sequence*.

We call the number of terms in a finite sequence itsa _{1}, a_{2}, a_{3}, a_{4}, …, a_{k}*length*. The length of the above sequence is k. - If we know a formula for the
*n*(this is also called the^{th}term*general term*), we can easily find any term. Plug the appropriate value for n into the formula, then work out what that term is. For example, if we want to find the seventh term, we would plug in n=7. - A sequence can also be defined
*recursively*: each term is based on what came before. The sequence is built on a*recursion formula*that shows how each term is built from preceding terms. To use a recursion formula, we need a "starting" place before we can make a sequence. This is called the*initial term*(or terms, if multiple are needed). - Given a recursion formula and initial term(s), it can be possible to find a formula for the n
^{th}term. Similarly, it can be possible to transform an n^{th}term formula into a recursion formula and initial term(s). Still, there is no guarantee we can do this. Sometimes it will be easy, sometimes hard, and sometimes impossible. - Very, very often you will be given the first few terms of a sequence and told to either give more terms, or figure out a formula for the n
^{th}term. To do this, you will have to find some__pattern__in the sequence, then exploit it. - When trying to recognize a pattern in a sequence, try to think in terms of how to get from one term to the next. Establish a hypothesis by looking at a
_{1}→ a_{2}, then test it against a_{2}→ a_{3}, a_{3}→ a_{4}, and any other terms given. Once you've figured out the pattern, it's easy to find further terms in the sequence. Finding a formula for the n^{th}term can be tricky, though. Think carefully about how you can put the pattern in an equation, then make sure to check some terms after you create the formula. - When trying to find the pattern in a sequence, there are a variety of common pattern types that appear. Here are some important ones to keep in mind:
__Addition/Subtraction:__add k every term.__Mutliplication/Division:__multiply by k every term.__Squares (n__1, 4, 9, 16, 25, 36, 49, …^{2}):__Cubes (n__1, 8, 27, 64, 125, 216, …^{3}):__Factorials (n!):__1, 2, 6, 24, 120, 720, …-
__Alternating Signs:__Alternating signs can created by (−1)^{n+1}or (−1)^{n}.

- If most of the terms in a sequence are presented in a certain format, like fractions, try to figure out a way to put all the terms in that format. It can be easier to see patterns if everything is in the same format. Furthermore, if the format can be clearly broken into multiple parts (in a fraction
^{[¯]}/^{[¯]}, we can break it into numerator and denominator), it can help to figure out patterns for each part separately. - It can sometimes help to write the number of the term above or below each term (n=1, n=2, etc). This helps you keep track of numerical location, which often makes it easier to identify patterns.
- In the end, there is no one way to identify all patterns. Try to take a broad view of the sequence and look for repetitions or similarities to other patterns you've seen. If you still can't figure it out, see if there's an alternative way to write the terms out. Persevere, and be creative.

### Introduction to Sequences

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

- Intro 0:00
- Introduction 0:06
- Definition: Sequence 0:28
- Infinite Sequence
- Finite Sequence
- Length
- Formula for the nth Term 3:22
- Defining a Sequence Recursively 5:54
- Initial Term
- Sequences and Patterns 10:40
- First, Identify a Pattern
- How to Get From One Term to the Next
- Tips for Finding Patterns 19:52
- More Tips for Finding Patterns
- Even More Tips
- Example 1 30:32
- Example 2 34:54
- Fibonacci Sequence
- Example 3 38:40
- Example 4 45:02
- Example 5 49:26
- Example 6 51:54

### Math Analysis Online

### Transcription: Introduction to Sequences

*Hi--welcome back to Educator.com.*0000

*Today, we are going to talk about an introduction to sequences.*0002

*For the most part, a sequence is simply an ordered list of numbers.*0005

*While the idea is simple, there are a huge variety of uses for sequences.*0009

*They come up in a wide variety of fields, and they are an extremely important tool in advanced mathematics.*0013

*In this lesson, we will learn what a sequence is, various ways to describe them, and how to find patterns that they may be based on.*0018

*Let's go: the definition of a sequence: a sequence, in math, means pretty much the same thing as it does in English.*0027

*It is an order of things; specifically, in this case, it is an ordered list of numbers.*0034

*We could write a sequence as a _{1}, a_{2}, a_{3}, a_{4}...a_{n}...*0041

*We call each of the entries in the sequence a term.*0048

*This would be a _{1}, the first term, because it is the first in the sequence.*0050

*a _{2} is the second term, because it is the second in the sequence.*0056

*a _{3} would be the third term, because it is the third in the sequence.*0061

*So, any symbol can be used to denote the sequence.*0065

*In this case, we are just the using the lowercase letter a, but we could use any letter, or any symbol, that we wanted.*0069

*a is a common, convenient one, though.*0074

*The subscript (that is the small number to the right: here, this little number 4, a _{4}) tells us which term it is in the sequence.*0076

*This 4 here tells us that it is the fourth term in the sequence.*0087

*Here are some examples: we could have 1, 2, 3...(and the ... just says "keep going in this manner; it continues on").*0094

*2, 9, 16...; √x, √2x, √3x...a sequence is just some ordered list of things--an ordered list of numbers, in this case.*0103

*So, even here, it ends up being an ordered list of numbers.*0116

*It is using a variable, but once we set x as some number, it is going to end up just being an ordered list of numbers there, as well.*0119

*All right, if a sequence goes on forever without stopping, it is called an infinite sequence.*0127

*Most of the sequences that we are going to work with are infinite sequences.*0132

*That is something where it is a _{1}, a_{2}, a_{3}, a_{4}...*0135

*and then there is nothing after those dots; it just says that it keeps going, and there is no stop to this thing.*0139

*On the other hand, we could have a finite sequence, if the sequence does stop.*0145

*In that case, we have a _{1}, a_{2}, a_{3}, a_{4}, and there is that ...*0149

*that says to continue in this manner; but then, we actually stop at a _{k}.*0154

*Notice how there is nothing after the a _{k}; it is just blank after that.*0158

*That says that we have reached the endpoint; there is not ... to tell us to keep going in this manner.*0162

*We get to a _{k}, and we just stop at a_{k}, because there is nothing after a_{k}.*0167

*So, it says that this is the end of our sequence.*0171

*We call the number of terms in a finite sequence its length.*0174

*In this case, in the length of the above sequence, we would have k, because we have a _{k} here and we have a_{1} here.*0179

*So, that means we are counting our first term, our second term, our third term...all the way up until our k ^{th} term.*0185

*1 up until k...if we count from 1 to k, whatever k is, that means we are going to have*0190

*a total of k things, so we have a length of k for that finite sequence.*0195

*We can often talk about some formula that allows us to find the n ^{th} term, also called the general term.*0201

*If we know such a formula, we can easily find any term.*0208

*By plugging in different values for n, since we know what the n ^{th} term is going to be,*0211

*well, if we say some value for n, we can find the term that is that value.*0216

*If we plug in n = 3, we can find the third locations.*0221

*So, as long as we know that a _{n} equals some stuff, some algebraic formulation,*0224

*then if we set n equal to 1, we would get the first term, a _{1}.*0232

*If we set n equal to 17, we get the 17 ^{th} term, a_{17}.*0238

*Notice how the n = 17 replaces where the n would have been; it is a subscript n;*0245

*but since we swapped it out for 17, we now have a, subscript 17.*0251

*So, because we have some algebraic formulation for the way a _{n} works,*0256

*for the way this n ^{th} term works, the way this general term works,*0260

*we can plug in our value for n, use this algebraic formulation, and churn out some number to know what a for any term is going to be.*0264

*For example, if we know that a _{n} = 7n - 5, then we have the sequence a_{1}, a_{2}, a_{3}, a_{4}...*0273

*Well, notice here: in a _{1}, that means n = 1; so we swap out the n in 7n - 5 for 7(1) - 5.*0281

*7 - 5 gets us 2, so we now know that a _{1} is equal to 2.*0290

*The same thing for a _{2}: we know that, at a_{2}, we have n = 2.*0296

*So, we swap out 7n - 5 to 7(2) - 5; 7(2) is 14; 14 - 5 is 9.*0299

*So now, we have that the second entry, the second term, in our sequence is 9.*0307

*a _{3}: we have n = 3, so we swap out; we have 7(3) - 5, 21 - 5, 16; so our third entry, a_{3}, is equal to 16.*0312

*At a _{4}, we have n = 4; 7(4) - 5...7 times 4 is 28; 28 - 5...so we have 23 for our fourth term, as well.*0322

*So, by knowing the general term, a _{n} = 7n - 5, we are able to find any term.*0331

*We can find any term if we know this general form a _{n} = some algebraic format, like 7n - 5.*0340

*We can also define a sequence based on terms that came previously.*0351

*We just figured out a way to just say that the absolute thing is going to be this; this will be this, based on this formula.*0354

*We have this definite, general term.*0360

*But we can also define it based on terms that came previously.*0363

*This is called defining a sequence recursively.*0367

*In this, the sequence is built on a recursion formula that shows how each term is based on preceding terms.*0371

*Recursive: we are looking backwards to something that came previously.*0377

*For example, if we have the recursion formula a _{n} = a_{n - 1} + 7, what is that saying?*0381

*It is that the n ^{th} term--that is a_{n}, right here--is created by looking at the previous term.*0387

*Well, what would be one before n? n - 1; so a _{n - 1} is going to be the term just before the n^{th} term.*0392

*So, a _{n - 1}...and then adding 7 to it is that + 7 business right there.*0401

*a _{n} = a_{n - 1} + 7: some term is equal to the previous term, plus 7.*0407

*Since this is true for any n at all, the recursion formula tells us that every term is equal to the term before it, plus 7.*0414

*We didn't say n has to be some specific value; we haven't nailed down what the value of it is going to be.*0422

*So, since this is true for any n, the recursion formula tells us that every term (because it is true for any n,*0427

*so a _{n} = a_{n - 1} + 7 for any value of n) will be equal to the term before, plus 7.*0433

*However, there is one special term that doesn't have a term before it.*0442

*Our recursion formula was based on looking at the one behind you and adding 7.*0446

*But in this case, there is one number that isn't going to have anything behind it.*0450

*The person who is the first in line (not really a person--a number)--whatever term is first, our first term: there is nothing behind it.*0454

*There is nothing to look at behind that term.*0461

*So, if that is the case, we need something to start from.*0464

*A recursion formula on its own is not enough to obtain a sequence; we need some sort of starting place before we can make a sequence.*0466

*We need to know what that first term is, what that seed is that our recursion formula will grow off of.*0473

*This is called the initial term (or terms, if we need multiple of them).*0479

*So, using the initial term a _{1} = 2, with the previous recursion formula a_{n} = a_{n - 1} + 7,*0484

*then our first term is going to be a _{1}, right here.*0491

*Well, we were just told that a _{1} = 2; so that means that we have 2 here.*0494

*Then, from there on, we have a _{n} = a_{n - 1} + 7.*0499

*So, that says that to get a term, you take the previous term, and you add 7.*0505

*So, to get from 2 to the next term, we get + 7, so 2 + 7 gets us 9.*0509

*To get to the next one, we have + 7; so that gets us 16.*0515

*To get to the next term, we have + 7; that gets us 23.*0518

*Writing out exactly what happens: if we want to know what a _{2} is going to be equal to*0521

*(a _{2} is this one right here), then a_{2} = a_{2 - 1}, or 1, + 7.*0527

*So, a _{1} is equal to 2; we figured that out here; then + 7...so we get a_{2} = 9, which is what we got right here.*0535

*And the same thing is going on for figuring out a _{3}: a_{3} is going to be equal to a_{2} + 7.*0547

*a _{4} is going to be equal to a_{3} + 7, because our recursion formula is telling us to go that way.*0552

*Given a recursion formula and initial terms, it can be possible to find a formula for the n ^{th} term.*0559

*That absolute plug-in-a-number-for-our-n...using the general term, it just puts out what the value is.*0564

*There are sometimes ways to be able to do this; if you have a recursion formula and initial terms,*0571

*you can sometimes transform it into a formula for the n ^{th} term.*0578

*Similarly, it can be possible to transform an n ^{th} term formula into a recursion formula and initial terms.*0582

*So, if we know the general term, we can go to the recursion formula.*0589

*If we go to the recursion formula, we can go to the general form.*0592

*However, we can't always end up doing this.*0596

*There is no guarantee that we can do this; we very often can, especially at this level in math.*0598

*But sometimes it is not going to be so easy to do; sometimes it is going to be really hard.*0603

*Sometimes, it will be totally easy; but sometimes it is going to be hard, and sometimes it is going to be impossible.*0607

*It will depend on the specific sequence that we are working with.*0612

*Some sequences are really easy to talk about in a recursion formula, but basically impossible to talk about as a general term, an n ^{th} term format.*0615

*Other ones are really easy to talk about in that n ^{th} term format, that general term, but really hard,*0623

*practically impossible, to talk about in a recursion formula.*0628

*So, we won't necessarily be able to switch between the two, but we will often be able to switch between the two.*0630

*And if a problem ever asks us to switch between the two, it will certainly be possible.*0635

*Very, very often, you will be given the first few terms of a sequence and told to either give more terms or figure out a formula for the n ^{th} term.*0640

*To do this, you will have to figure out some pattern in the sequence, and then exploit it.*0647

*You will have to look at the pattern, look at the sequence, and ask how these things are connected.*0651

*What is a pattern in here that I can use to create a formula?*0656

*However, before we learn how to do this, before we learn how to find patterns and sequences,*0660

*I want to point out that there is technically no guarantee that a sequence must have a pattern.*0665

*There is no guarantee that we will have patterns in our sequences.*0671

*For example, the below is a perfectly legitimate sequence with no pattern that we are going to be able to find in it.*0676

*47, then -3, then .0012, then π raised to the negative fifth power, then 17, then 1, then 1 again, then 800, and then a whole bunch of other numbers.*0681

*There is no pattern going on here; there is no rhyme; there is no reason.*0693

*There is nothing that we are going to be able to figure out to create some formula here.*0696

*So, there is no guarantee that a sequence has to have some pattern that we are going to be able to create a formula from.*0700

*Things can be really confusing with sequences, and we won't be able to figure out a way to generate a formula.*0706

*But good news: all of the sequences at this level in math will have patterns.*0712

*Technically, a sequence is not required to have a pattern; but at this level in math, all of the sequences that we see are definitely going to have patterns.*0717

*We will always be able to find patterns in the sequences that we are working with.*0725

*So, we don't have to worry about problems being unsolvable, because we can always rely on the fact*0728

*that there is going to be some pattern in there somewhere; there will always be a pattern that we can find, if we look hard enough.*0733

*If the problem is about finding patterns, there is definitely going to be a pattern for us to find.*0740

*We just have to look carefully and be really creative.*0743

*It won't necessarily be easy to find the pattern; but it will be in there somewhere.*0746

*It is not going to be something like this, where it is really, really hard for us to be able to see a pattern,*0749

*because there is simply no pattern; there are going to be cases...*0755

*in all of the things that we are looking at, it is always going to be the case that we are going to be able to find a pattern somehow.*0758

*We don't have to worry about the stuff where there is just no pattern whatsoever.*0763

*All of the problems that we will be asked to do, we will be able to do.*0767

*To find more terms in a sequence, or figure out a formula for the n ^{th} term,*0771

*the first thing that we have to do is identify a pattern in the sequence.*0774

*If we want to find a formula, if we want to be able to talk about more terms,*0777

*the first thing that we have to figure out is what goes on in the sequence--how does the sequence work?*0780

*Consider these two sequences: 17, 12, 7, 2... and 2, 6, 18, 54...continuing on with both of them.*0785

*Let's look at the first one: 17, 12, 7, 2: what we can do is say, "Well, what is the connection here between 17 and 12,*0794

*and how is that related to 12, 7; and how is that related to 7 to 2?"*0801

*Well, looking at this for a little while, we probably realize that what we are doing,*0804

*to get from 17 to 12, is subtracting by 5; what we are doing to get from 12 to 7 is subtracting by 5;*0808

*what we are doing to get from 7 to 2 is subtracting by 5.*0813

*So, we know that this pattern of subtract by 5, subtract by 5, subtract by 5...it is going to continue on,*0816

*because it showed up everywhere in the sequence so far.*0820

*Similarly, over here, how do we get from 2 to 6, from 6 to 18, from 18 to 54?*0823

*Looking at it for a while, we probably realize that what we are doing is multiplying by 3.*0828

*2 times 3 gets us 6; 6 times 3 gets us 18; 18 times 3 gets us 54; so this pattern will continue on,*0832

*throughout the rest of the sequence, because it showed up in all of the sequence that we saw.*0840

*We have figured out what the patterns are.*0843

*We notice that the sequence on the left subtracts by 5 every term; the one on the right multiplies by 3.*0846

*At this point, it would be easy to find more terms.*0850

*If we wanted to find the next term in the sequence 17, 12, 7, 2, we would just subtract 5 again; 2 - 5 would get us -3.*0853

*And we would be able to keep going, if we wanted to.*0862

*Similarly, over here, if we wanted to figure out the next term in 2, 6, 18, 54, we would just have to multiply by 3 again.*0864

*So, what would come after that? 54 times 2 is 150 + 4(3) is 12, so 162.*0872

*So, we would get 162 as the next one, and we would be able to continue on in that manner, if we wanted to find any more terms in the sequence.*0880

*It is easy to find more terms, and it would also be easy to find a recursion formula.*0887

*The red one that we were doing, the minus 5 one, is just going to be a _{n - 1} - 5.*0892

*And the green one, the second one, with multiplication, times 3...that would be a _{n - 1} times 3.*0899

*So, it wouldn't be very hard to figure out recursion formulas, because they are really deeply connected to the pattern we saw.*0904

*However, if we want a formula for the n ^{th} term, it is going to take a little more thought,*0909

*because we have to be able to figure out how this works for all of them.*0913

*It is not just describing the pattern of how we get from one to the next, or how we get from this one to the next one.*0916

*It is describing how we get to any of them, without being able to have any sort of reference points.*0923

*We can't just say - 5, - 5, - 5; we have to figure out a way of collecting all of the - 5s that happened, or all of the times 3s that happened.*0928

*So, we think about it for a while; and we will be able to figure this out.*0934

*We would be able to get a _{n} = 22 - 5n; we would be able to realize that,*0937

*since what we are doing is subtracting by 5 a bunch of times, it is going to be - 5 times n;*0942

*and then we need to figure out what number we are subtracting 5 from.*0947

*And we want to make sure to check the first few terms.*0949

*Always check the first few terms, once you think you have figured out a formula for the general term.*0953

*Once you think you know what the n ^{th} term is going to be, make sure you check that what you figured out is right,*0957

*because it is easy to make a mistake and be off by a little bit, to be off by one number in the sequence.*0963

*So, just make sure that you try and check.*0968

*For example, if we have figured out what a _{1} is going to be, well, that would be 22 - 5(1).*0970

*22 - 5(1) is 22 - 5, or 17; that checks out with what we have here.*0975

*If we did a _{2}, then that would be equal to 22 - 5(2); 22 - 5(2) is 22 - 10; 22 - 10 is 12; that checks out with what we have here.*0980

*So, it looks like it ends up working out.*0990

*Similarly, for the green one, our multiplication one, a _{n} = 2(3)^{n - 1}:*0992

*well, let's think about this--does this end up working out?*0999

*We realize that it has to be something about 3 to the some exponent, because every step, we are multiplying by some 3.*1001

*So, if we stack all of those 3s together, it is going to be 3 to the some sort of exponent.*1009

*The question is what that exponent should be: it is 3 ^{n - 1}, because this very first one hasn't been affected by the 3 at all.*1014

*Let's check and make sure that that is the case.*1021

*If we plug in a _{1} = 2(3)^{1 - 1}, then that would be 2(3)^{0}; 3^{0} is just 1;*1023

*any number raised to the 0 is just 1; so we get 2; 2 checks out.*1032

*If we did a _{2}, then we would have a_{2} is 2(3)^{2 - 1}, or to the 1; so it equals 6; that checks out.*1036

*We can see that this is going to continue to work; so our general formula works out.*1045

*But make sure you check it and think about it; it can be a little bit difficult,*1049

*but as long as you check it, you can be sure that what you have is going to work.*1052

*When trying to recognize a pattern in a sequence, try to think in terms of how to get from one term to the next.*1056

*How do you get from this first term to the next term?*1062

*How do you get from the first term to the second term?*1066

*Establish a hypothesis, some guess at what you think the pattern is, by looking a _{1} to a_{2}.*1068

*You want to start with a hypothesis--you think that this is probably how the pattern works.*1074

*You look at a _{1} to a_{2}, and then you want to test that hypothesis against the following ones:*1078

*a _{2} to a_{3}, a_{3} to a_{4}, and any other terms that are given.*1083

*You come up with thinking that the pattern here, the way we get from one stepping-stone*1087

*to the next stepping-stone, is that we do some operation.*1091

*Add some number, multiply by some number...it is doing some sort of thing; how is it working out?*1094

*Figure out what you think is going on for the way that the pattern works.*1099

*And then, test and make sure that that works on the way that we get to the next one and the way that we get to the next one,*1101

*or that it just works for the n ^{th} number location.*1106

*Once you have some hypothesis, test it against all of the other ones.*1109

*If it works, great; you just figured out the pattern--now you are ready to figure out some way*1112

*to formulate that general form, that general term, the n ^{th} term.*1116

*If it doesn't work, go back to the beginning: figure out a new hypothesis, and then try again.*1120

*It is all about figuring out something that you think might work, and then testing it against the information you have.*1128

*Keep testing until you get something that actually ends up working out,*1134

*at which point, you have found the pattern; now you are ready to start working your way towards a general term formula.*1137

*Once you figure out the pattern, it is easy to find further terms in the sequence.*1143

*That is the easiest part; you just continue with that pattern to generate any more terms that they tell you to generate.*1146

*Finding a formula for the n ^{th} term: that could be tricky--a formula for the n^{th} term can be a little bit tricky.*1153

*What you want to do is think carefully about how you can put the pattern into an equation*1159

*and make sure to check some terms after you create the formula.*1164

*That checking is really, really important; think carefully about how you can put that pattern into an equation,*1166

*and then check after you have come up with some sort of equation that you think will probably work.*1173

*It is really important to check, because it is really easy to make mistakes, especially the first couple of times you are doing it.*1177

*We will also see this a whole bunch of times in the examples.*1182

*We are going to work with a whole bunch of examples here.*1184

*So, that will really help to cement our understanding of how to do this.*1186

*We have lots of examples to make this clear.*1188

*All right, how do we find patterns?--that can sometimes be a tricky thing.*1191

*When trying to find the pattern in a sequence, the two most common pattern types that appear are addition/subtraction,*1195

*where we just add some k every term (k could be a positive number; k could be a negative number;*1202

*that allows us to add or subtract, depending on what the k is; but we are just adding the same k, adding some constant number,*1207

*every time we do a step); and then the other one is multiplication/division, where we multiply by some k every term.*1213

*If it is multiplication, it is just some constant number k; and we can also effectively divide.*1221

*If it is a fraction as our k, we are effectively doing division there, as well.*1225

*So, we are just multiplying by some constant number every term.*1229

*Every step is either going to be addition or multiplication.*1231

*This is the largest portion of the patterns; many, many of the patterns that we are going to work with,*1235

*at this level, and really at any level, are going to be connected to addition/subtraction or multiplication and division.*1240

*So, these are the first two that you want to keep in your head.*1245

*A large number of patterns can be figured out just by keeping these two types in mind.*1248

*Always think first in terms of addition/subtraction, multiplication/division.*1254

*Check to see if you see those first.*1257

*However, those are not the only kinds of patterns that you end up seeing.*1259

*These two pattern types are not enough to figure out the pattern for all sequences.*1262

*In that case, it can help to keep various other patterns in mind.*1267

*A good one to keep in mind is the squares: n ^{2}: 1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2}, 7^{2}...*1270

*But often, you won't see them as a number squared, because then it would be really easy to recognize the pattern.*1278

*Instead, you see it as 1, 4, 9, 16, 25, 36, 49...and continuing on, if they still have even more terms.*1283

*So, it just helps to keep that structure of numbers in the back of your head.*1291

*It is really useful to be able to remember what all of those perfect squares are, what all those squares are*1295

*that you are used to working with, that you have seen in previous algebra classes.*1299

*Just keep them in your mind, and look for things that look somewhat like that, or along those lines.*1303

*Another one that often shows up is cubes; this one is less often than squares, but it does show up.*1309

*n ^{3} is 1^{3}, 2^{3}, 3^{3}, 4^{3}, 5^{3}, 6^{3}, etc.*1314

*Once again, you very often won't see it as a number cubed, because then it would be easy to see that the pattern is cubed.*1319

*And it is supposed to be a little more challenging than that.*1324

*So instead, it is normally 1, and then 8, and then 27, and 64, then 125, and 216, and so on, and so on.*1326

*So, you are probably less used to using cubes; it is really important to just pay attention to at least these first four.*1334

*Memorize 1, 8, 27, and maybe 64, maybe 125; keep at least those first few terms in mind,*1340

*because you want to be prepared to say, "Well, I am not used to seeing this pattern; I am not used to seeing something like this;*1348

*but oh, maybe it is cubes, because I see that the first three are like that," and then you can check the other ones*1353

*by hand, and make sure that that does work out.*1358

*You don't have to memorize the whole thing, but you do have to be ready,*1360

*when you see the pattern, to be able to think that maybe that is going to be something.*1363

*You have to be prepared to recognize it; you don't have to know the whole pattern, but you have to be prepared to recognize it.*1366

*Finally, factorials: n!: 1!, 2!, 3!, 4!, 5!, 6!, etc., etc.: once again, you aren't normally going to see that as factorials written out.*1371

*You will instead often see it as something like 1, then 2, then 6, 24, 120, 720...a really good one to memorize is 1, 2, 6, 24, 120.*1383

*That is 1!, then 2!, 3!, 4!, 5!; so if you can keep that pattern in your head--just keep that one in the back of your head--*1393

*then that will end up showing up a lot, and if you are not prepared to recognize that pattern,*1402

*that kind of problem would be really, really hard, because you won't be able to see that pattern when it shows up.*1405

*And in case you forgot how a factorial works, factorials multiply the number that is factorial by every integer below the number.*1410

*So, for example, 5! would be 5 times 4 times 3 times 2 times 1, which works out to 120.*1419

*And we define...we just specifically define 0! = 1 for ease.*1426

*It helps things out; it helps a lot of other things in math work out.*1431

*So, 0! = 1; we just set it that way and try not to get worried about how that doesn't make sense, compared to how the other one works out.*1434

*It actually sort of makes sense; but it is better to just remember and memorize 0! = 1.*1443

*That one will come up occasionally.*1449

*All right, sometimes the sign will change with each term; it will flip between positive and negative, positive/negative, positive/negative.*1451

*You will see a positive on one, and the next one will be negative, and the next one will be positive, and the next will be negative.*1458

*And the next one will be positive, and then negative, and so on, and so forth.*1463

*And you will see this flipping pattern; if that is the case, it might be one of these two following--one of these two methods, these two types of patterns.*1466

*-1 raised to the n + 1: notice that, if we have n at 1, then we will have -1 squared, which is going to come out to be a positive 1.*1474

*And then, the next one would be -1 to the 2 + 1; n = 2, so -1 to the 3 is going to end up being -1.*1482

*And then, we get +1 and -1 and +1 and -1, because -1 raised to some integer is just going to multiply by -1 that many times.*1489

*So, it will flip between positive and negative, positive/negative, as our n's step up, one at a time.*1496

*Similarly, -1 raised to the n gives us the exact same pattern, flipping and turning.*1501

*It just starts, instead of starting at +1, at -1; and then, it will be +1 and -1, then +1, then -1, then +1.*1506

*So, if you see a flipping sign pattern, and you don't see something else to be able to cause that to happen,*1513

*in the sequence that you are working with, whatever pattern you are working with,*1520

*these two right here are a really good thing to keep in the back of your mind for a way to just cause a flipping sign to appear.*1523

*If most of the terms in the sequence are presented in a certain format--for example,*1532

*they are all in fractions--try to figure out a way to put all of the terms in that format.*1535

*So, if you see a certain format in your sequence, put all of the terms into that format.*1539

*If most of your terms are in fractions, make all of your terms in fractions.*1545

*It can be a lot easier to see patterns if everything is in the same format.*1549

*So, if you get all of your sequence terms in the same format, it normally causes patterns to appear more readily.*1555

*Furthermore, if the format can clearly be broken into multiple parts--for example,*1562

*if we have a fraction, _/_, we can break it into the numerator (the part on top) and the denominator (the part on the bottom)--*1565

*in that case, we can clearly talk about how all of our numerators*1575

*behave by some pattern, and all of our denominators behave by some pattern.*1577

*If you can break the thing into multiple parts, if a format can be broken into multiple parts--*1581

*for example, a fraction is a numerator over a denominator, every single time--it can help to figure out patterns for each part separately.*1587

*Figure out the numerator pattern on its own; figure out the denominator pattern on its own.*1596

*Sometimes, that will really help clarify things; you don't have to worry about trying to figure out the whole fraction.*1599

*You can instead break it apart piecemeal and then just put them back together once you recognize each pattern on its own.*1603

*It is important to note that all of these different pattern types that we have talked about so far-- they don't necessarily occur in isolation.*1609

*While that will sometimes happen (you will sometimes just have addition; you will sometimes just have multiplication;*1619

*you will sometimes just have switching signs), we are often going to end up working with sequences*1624

*that use multiple pattern types at once, so it will be up to us to figure out that it is using this pattern and this pattern and this pattern,*1629

*and then figure out some way to merge all of those three patterns together, once we are trying to describe the whole thing.*1637

*They might even end up involving patterns that you haven't seen before.*1642

*They might do something that you don't immediately recognize; we have to come up with a new way to describe it.*1645

*So, keep a lookout for something that looks weird, that is totally new to the way that you are doing things.*1650

*And you might end up having to learn a new method of describing things, which will just take longer to think through.*1654

*It can sometimes help to write the number of the term above or below each term,*1660

*to write n = 1 and then n = 2, and so on and so forth.*1664

*By writing the number of the term above or below, you are able to keep track of numerical location.*1669

*By being able to see what number we are at--are we at the first term? Are we at the fifth term?--*1676

*by being able to have this clear reference point of "this is term #1; this is term #5,"*1680

*you will be able to see how the number of the term relates*1685

*to the values inside of the actual term inside of the sequence--the values of that term.*1688

*The location of the term will normally be related to the values inside of the term.*1694

*That will often make it easier to identify patterns, by being able to see that this has number location 5;*1698

*and because of that, we see that the general form is working in this certain way.*1704

*So, writing the numbers above or below can really help you see that sort of thing.*1708

*In the end, there is no one way to identify all patterns.*1712

*I want you to try to take a broad view of the sequence and look for repetitions or similarities to other patterns that you have seen.*1716

*Don't try to focus on it always being the same thing, because it won't.*1723

*Each sequence is probably going to have its own special pattern.*1726

*You will start getting used to certain ways that patterns interact, or certain types of patterns.*1729

*And it will be easier to recognize them on future ones.*1733

*And a lot of this stuff will show up in a whole bunch of different places, like standardized tests,*1736

*later in different math classes, in science classes...so what you are learning in this class will definitely be applicable for a long time to come.*1741

*But it is not necessarily always going to be the same thing.*1748

*So, just take a broad view of what is going on.*1751

*Don't think that it is definitely going to work in one way, because you don't really know until you have figured out what the pattern is.*1754

*So, look at the thing carefully; think, "How does one term interact with the next term?"*1760

*How are these two terms related to each other?*1766

*Is there some sort of general pattern that is occurring as I look through all of the numbers, all of my terms, at once?*1768

*Try to think in really large, broad strokes before you try to come up with a very specific pattern showing up.*1774

*And if you still can't figure it out, if you are looking at it for a long time and you can't figure it out,*1780

*see if there is an alternative way to write the terms out.*1784

*That was what I was talking about with...if most of them are in fractions, put all of them in fractions.*1787

*Figure out if there is some way to write the terms in something, so that they all have this new alternative way,*1791

*because maybe that will help you see what is going on better.*1796

*And just in general, persevere; be creative.*1799

*Figuring out patterns is not something that is just a step-by-step method, and you have gotten to the answer.*1802

*It is something where you have to look at it and sort of think for a while.*1807

*Just be clever; have a little bit of luck; and just work at it.*1809

*If you really can't figure it out for a long time, go to the next problem and come back to that problem later.*1812

*Sometimes just a little bit of time will cause it to "bounce around" in your head,*1817

*and you will be able to see the pattern easily, when earlier it was really, really difficult.*1821

*So, just stick with it, and as you work with it more and more, it will make more and more sense.*1825

*All right, we are ready for some examples.*1829

*Given the n ^{th} term, write the first four terms of each sequence.*1831

*Assume that each sequence starts at n = 1.*1836

*So, that n ^{th} term, that general term, is a_{n} = stuff.*1839

*In this case, for our first one, we have a _{n} = 3n - 2.*1843

*Our first term, the a _{1}, is going to be when n = 1.*1849

*We plug in a _{1} = 3(1) - 2.*1855

*Similarly, a _{2} is when n = 2, so we have a_{2} = 3(2) - 2.*1860

*a _{3} = 3(3) - 2; a_{4} = (I'll write that a little below; there is not quite enough room) 3(4) - 2.*1867

*We work this out: a _{1} = 3(1) - 2, 3 - 2, so we get a_{1} = 1.*1880

*a _{2}: 3(2) is 6; 6 - 2 is 4; a_{3} = 3(3), 9, minus 2; 9 - 2 is 7.*1886

*a _{4} = 3(4) is 12, minus 2 is 10.*1896

*So, we have the first four terms here: a _{1} = 1, a_{2} = 4, a_{3} = 7, a_{4} = 10.*1902

*Also, I want to point out; notice how a _{1}, to get to a_{2}...we added 3.*1909

*To get to a _{3}, we added 3; to get to a_{4}, we added 3.*1913

*Each term here has this + 3 step, which we are going to end up seeing from this 3 times n,*1916

*because the 3 times n...the n is what term location we are at, so as we go up more term locations,*1922

*we are going to end up seeing more times that we have ended up adding on this number 3 to the thing.*1928

*All right, the next one: b _{n} = -1^{n}/(n + 3); our first one, a_{1}, is going to be (-1)^{1}/(1 + 3).*1935

*We swap out all of the n's that occur for whatever we have here, a _{1}.*1947

*Next, a _{2} = -1 squared, -1 to the 2, divided by 2, plus 3.*1952

*a _{3} is equal to -1 to the 3, over 3 + 3.*1959

*a _{4} is going to be equal to -1 to the 4, over 4 + 3.*1966

*We can work this out here: we have a _{1} = (-1)^{1}, which is still just -1; 1 + 3 is 4, so we have -1/4.*1973

*a _{2}: (-1)^{2}, -1 times -1...that cancels to just positive, so we just have a positive 1, divided by 2 + 3, 5.*1984

*a _{3} = (-1)^{3}; -1 to an odd exponent is going to end up leaving a negative after.*1992

*So, we have -1 over 3 + 3, 6.*1998

*a _{4} = (-1)^{4}, to an even exponent; it is going to cancel out; we are going to have a positive.*2002

*So, we have 1/(4 + 3) is 7: so a _{1} = -1/4; a_{2} = +1/5; a_{3} = -1/6; a_{4} = 1/7.*2008

*We found the first four terms.*2020

*Finally, c _{n} = 47...oh, oops; that whole time shouldn't have been a_{n}, because it was b_{n}.*2022

*So, it actually should have been not a _{2}, not a_{1}...any of these...*2030

*It should have been b _{1}, b_{2}, b_{3}, b_{4}, because it has a different name than the sequence at the top.*2036

*That is why we are using a different letter--because it is a different sequence for this problem.*2043

*b _{1}, b_{2}, b_{3}, b_{4}...it is easy to end up forgetting that we are changing symbols sometimes.*2048

*So, pay attention to the symbol of the sequence that you are working with.*2055

*All right, the last one: c _{n} = 47; the thing to notice here is...does this side, the right side, end up involving n at all?*2058

*It doesn't; as the n changes, the right side doesn't notice the n change.*2065

*So, c _{1} is going to be equal to 47; c_{2} is going to be equal to 47;*2069

*c _{3} is going to be equal to 47; c_{4} is going to be equal to 47.*2074

*So, whatever value we end up using for n, it is always going to end up coming out to 47.*2080

*So, the first term is 47; the second term is 47; the third term is 47; the fourth term is 47.*2084

*We always end up getting 47, because it is just a constant sequence.*2089

*All right, the second example: the Fibonacci sequence is a well-known, recursively-defined sequence.*2094

*It is given by the recursion formula and the initial terms below; write out the first 12 terms.*2100

*Its recursion formula is a _{n} = a_{n - 1} + a_{n - 2}, and a_{1} = a_{2}, which equals 1.*2105

*Right away, we know that our first two terms are 1 (a _{1} = 1), and then a_{2} also equals 1; so 1, 1.*2114

*If we want to figure out the next term, we have a _{n} = a_{n - 1} + a_{n - 2}.*2125

*So, if we want to figure out what a _{3} is going to be, then that is going to be equal to a_{3 - 1},*2130

*so a _{2}, plus a_{3 - 2}, a_{1}.*2135

*a _{3}...we don't know what a_{3} is, but we do know what a_{1} and a_{2} are.*2140

*They are both 1; so we have 1 + 1; a _{3} = 2.*2144

*So, our next term is 2; what comes after that?*2150

*If we want to figure out a _{4}, that would be a_{4 - 1}, a_{3}, plus a_{4 - 2}, a_{2}.*2155

*What is a _{3}? We just figured out a_{3} = 2, so we have 2 +...a_{2}, once again, is 1; that equals 3.*2164

*a _{4} = 3, so the next thing is going to be a 3.*2175

*Let's do one more of these, and then we will see what the general pattern here, going on, is.*2181

*a _{5} is equal to...not the general term, but how this pattern is working, on the whole...*2186

*a _{5} is equal to a_{4}, the previous term, plus the term previous to that one, a_{3}.*2191

*So, a _{5} = a_{4} + a_{3}: we just figured out that a_{4} = 3 and a_{3} = 2.*2199

*So, we have a _{5} = 5; there is our next term, 5.*2206

*What comes after that--how do we get this?*2213

*Well, notice: a _{5} = a_{4} + a_{3}; so a_{5} is equal to the previous term, plus the term previous to that.*2215

*a _{4} equaled a_{3} + a_{2}; a_{4} is equal to the previous term, plus the term previous to that.*2221

*a _{3} = a_{2} + a_{1}, so the previous term, plus the term previous to that.*2227

*What we are doing to make the next term: it is looking at the previous term and the previous previous term.*2232

*To make whatever comes after the 5, it is going to be: add 3 and 5 together, and that will make the next thing.*2237

*So, 3 + 5 gets us 8; then, the next one is going to, once again, be: take the 5 and the 8; add them together.*2243

*5 + 8 gets us 13; we see that this relationship here is that any term is equal to the previous term and the previous previous term, added together.*2251

*So, that is why we needed a _{1} and a_{2}, because we needed 2 terms,*2263

*so that we could talk in terms of the previous previous term.*2267

*We needed that larger start; we needed two initial terms before we would be able to get started.*2269

*So, at this point, we just add things together: 8 + 13 is 21; 21 + 13 is 34; 21 + 34 is 55; 34 + 55 is 89; 55 + 89 is 144.*2274

*And it will continue on in this manner.*2294

*And there we are; there is the Fibonacci sequence; there are the first 12 terms of the Fibonacci sequence.*2296

*Cool; the Fibonacci sequence has some other interesting properties.*2304

*You might end up studying it more in the class that you are currently in, or in a future class.*2306

*It is a pretty cool thing; but it is enough for us now just to understand how the thing works out.*2310

*All right, the third example: Find the n ^{th} term for each sequence below, and assume that the sequence starts at n = 1.*2315

*The first thing that we always have to do, if we are looking at a sequence, and we want to figure out what the n ^{th} term,*2322

*the a _{n}, the general term, is (it equals some formulaic algebra expression), then it is going to be...*2326

*we have to figure out the pattern, and then we use that pattern to come up with an equation.*2335

*So, what is the pattern in this first sequence? -3, 1, 5, 9, 13.*2338

*Well, notice: how do we get from -3 to 1? We can just add 4.*2344

*How do we get from 1 to 5? We add 4 again--it looks like our pattern probably is going to work out.*2348

*5 to 9--we add 4; 9 to 13--we add 4; 13 to whatever is next--it is probably going to be add 4, add 4, add 4.*2353

*Since the pattern worked for everything that we have seen so far in the sequence,*2361

*we can assume that the pattern is definitely just "add 4."*2365

*So, if that is going to be the case, we know that it has to be something of the form a _{n} = 4n +...we don't know yet.*2369

*We don't know what it is going to be, so we will just leave it as a question mark.*2377

*We could also use a variable like normal, but in this case, the variable I have decided to use is ?.*2379

*a _{n} = 4n...that represents this step, step, step, step: +4, +4, +4, +4.*2384

*Every step brings +4; every term you move on to brings this adding by 4.*2391

*And so, 4 times n will allow us to represent how many steps we have taken, multiplied by 4; and that brings that many 4s to the table.*2396

*So now, we just need to figure out what the question mark is.*2403

*Well, notice: we know that a _{1} = -3; so that means we can have a_{1} = 4(1) + ?.*2405

*We know that a _{1} equals -3; so -3 = 4 (4 times 1 is just 4) + ?; so we have -7 (solving for question mark) = ?.*2414

*So, at this point, plugging that in, we have that a _{n}, the general term, is equal to 4n - 7.*2426

*It is always a good idea to check this sort of thing out; that probably will end up working out, but let's make sure.*2436

*We already checked a _{1}; that is how we figured it out.*2440

*Let's check and make sure that a _{2} ends up working out.*2442

*a _{2} = 4(2) - 7, 8 - 7, which equals 1; and that checks out.*2445

*Next, a _{3} = 4(3) - 7 = 12 - 7 = 5; and that checks out.*2452

*And we can see, going along, this method--that this is going to end up working--*2460

*because we have this 4n here, so every time we go forward another term, we are going to end up adding another 4,*2464

*which also represents our pattern of + 4 each time; so it makes sense--we have our answer.*2470

*The general term for that sequence is a _{n} = 4n - 7.*2476

*All right, the next one: 2, 5, 10, 17, 26.*2480

*The first thing we want to do is figure out what the pattern is.*2485

*If it is addition, then we would have something like + 3.*2488

*All right, to get from 2 to 5, it is + 3; to get from 5 to 10, it is + 5; to get from 10 to 17, it is + 7...that is not going to end up working out.*2492

*It can't be addition as our pattern; we see that pretty quickly.*2502

*So, let's try another really common one, which is multiplying.*2506

*Well, to get from 2 to 5, you have to multiply by 5/2; to get from 5 to 10, we multiply by...that is not going to work.*2508

*We see very quickly that multiplication is just not friendly here; it is not going to work out in a very good way.*2517

*So, multiplication is out; at that point, we see that addition fails; multiplication fails.*2522

*So now, this is really where we get creative, and we start thinking.*2531

*What is going to be able to get us the answer here?*2536

*What is going on here--what is the pattern 2, 5, 10, 17, 26...?*2539

*And this is the part where we lean back, and we just think for a while.*2545

*We think, "What does this look like? What have I seen that looks even vaguely similar to the way this grows...*2549

*the fifth term is 26...how does this work?*2555

*We might try coming up with some pattern the first time that doesn't end up working.*2557

*That is OK; the important thing is just to keep looking and keep going, pondering and thinking: how is this related to something else?*2560

*Remember: we talked about some of the other patterns that are likely to show up--squares, cubes, factorials...*2566

*Those are good ones to start trying out if you can't figure it out yet.*2571

*So, let's look at squares: what are the squares?*2574

*Well, the squares would end up going...if we had simply n ^{2}, then that would end up giving is the sequence:*2576

*1 (1 ^{2} is 1), then 2^{2} is 4, then 3^{2} is 9; 4^{2} is 16; 5^{2} is 25...*2582

*How does 1, 4, 9, 16, 25 relate to 2, 5, 10, 17, 26?*2594

*Oh, they are very similar; we are just adding 1.*2601

*If we add 1, +1 would get us 2; +1 would get us 5; +1 would get us 10; +1 would get us 17; +1 would get us 26.*2605

*We have figured out what the general term here is--what the n ^{th} term is.*2617

*It is the formula a _{n} = n^{2} (because it is the number squared), but then we also have to add 1: a_{n} = n^{2} + 1.*2621

*That seems to give us the formula for our general term for this sequence.*2632

*Let's check and make sure that that is, indeed, the case.*2637

*We do a quick check; if we plug in a _{1} = 1^{2} + 1, then we get 1 + 1, which equals 2.*2639

*That checks out; if we plug in for our second term, a _{2}, then we would have 2^{2} + 1.*2647

*2 ^{2} is 4; 4 + 1 is 5; that checks out.*2654

*Our next one: a _{3} = 3^{2} + 1; 9 + 1 = 10; that checks out; great.*2658

*At this point, it seems that our n ^{2} + 1 ends up working out; it ends up following this sort of like the squares method,*2666

*but a little bit more, just adding one each time; we see that the pattern that we figure out ends up being the same.*2673

*Notice: this pattern isn't really so much a pattern about a recursive thing going on.*2679

*It is not really so much about adding the same number each time, multiplying by some number each time...*2683

*We could figure it out as a recursion formula, but it is easier to think of it just in terms of this absolute:*2690

*here is the general term; here is how any given location ends up working.*2695

*It is the number of the location, squared, plus 1.*2698

*All right, the fourth example: Given the recursion relationship below, write the first five terms; then give the n ^{th} term.*2702

*We have a _{n} = -3 times a_{n - 1}; that is to say, some term*2708

*is equal to -3 times the previous term; a _{n} is some term; a_{n - 1} is the term 1 back, going backwards by 1.*2713

*So, it is -3 times the previous term; and we have to have some starting place to begin with;*2722

*otherwise we won't be able to always look at previous terms.*2727

*So, we start at 2; then, the next one...if we want to talk about a _{2}, the second term,*2730

*it would be -3 times a _{1}, so a_{2} is equal to -3 times...a_{1} is just 2, so times 2; that equals -6.*2736

*So, a _{2} = -6; notice: all that we really did there was just multiply by -3.*2745

*So, we can probably just end up doing this in our head.*2751

*2, then -6; what would come after that?*2754

*2, then, -6; then we would multiply by -3 again, -3 times the previous term, to get our next term.*2758

*-6 times -3 gets us +18; to get the next term, 18 times -3 is going to end up being -54, because it is -3.*2765

*The next term is going to end up being...times another -3...positive 162.*2777

*And it will continue on in this matter; so we have now figured out the first 5 terms.*2782

*Great; but we also have to give the n ^{th} term, so how can we figure out what the n^{th} term is?*2787

*Let's switch colors for this; the n ^{th} term...notice: 2, -6, 18, -54, 162...it is kind of hard to see a pattern there really obviously.*2792

*We see that every time, it is multiplying by -3, because we were told very explicitly that the recursion formula says to multiply by -3.*2801

*So maybe we could make that -3 show up so that we could see that a little more easily.*2808

*If we do that, we could write its 2 first; we have 2 show up at the beginning.*2812

*It doesn't have any -3s multiplying by it yet.*2818

*But next is going to be (-3) ^{1} times...well, we wouldn't be able to figure out immediately that it is going to be to the 1.*2820

*So, -3 times 2; then the next one would be -3 times it again, so now we have (-3) ^{2} times 2.*2827

*And the next one would be (-3) ^{3} times 2, and the next one would be (-3)^{4} times 2.*2837

*And it would just continue on in this manner.*2843

*So remember: we want to get the whole thing to look like a similar format.*2846

*Everything has these -3's on it, raised to some exponent, for the most part.*2849

*Most of them have the exponents, and all of them end up having the -3 business here,*2854

*with the exception of this one, which has neither exponents nor -3.*2861

*So, we need to get them to all have this similar format.*2864

*We have -3 to the what?--what number can we multiply by?--what can we always multiply by?*2867

*We can always multiply by -3 to the 0; it is (-3) ^{0} times 2, because that is just 1.*2872

*Over here, we have -3 to the 1, times 2; -3 to the 2, times 2; -3 to the 3, times 2; -3 to the 4, times 2; and so on.*2879

*So, if that is the case, we can match this up to n =...1 is here; n = 2 is here; n = 3 is here; n = 4 is here; n = 5 is here...*2893

*What changes each time? The only thing that ends up changing is 0, 1, 2, 3, 4...everything else is the same.*2903

*n = 1 gets us 0; n = 2 gets us 1; so what are we doing to the n? We are subtracting 1 each time.*2910

*So, we see that the general term can be given as a _{n} = -3 raised to the n - 1 times 2.*2916

*And there we go; if we want to check that, let's just check for the third term.*2928

*Just randomly, to make sure that everything ends up working out: the third term, a _{3}, is equal to -3, raised to the 3 - 1 times 2.*2934

*That is -3...3 - 1 is squared, times 2; -3 squared equals 9, times 2...9 times 2 is 18.*2944

*And that checks out with what we already figured out is the third term.*2956

*Great; so we figured out our general term, our n ^{th} term formula; it works out perfectly.*2959

*The fifth example: Find the n ^{th} term for the sequence below; assume that the sequence starts at n = 1.*2965

*The first thing: most of it ends up being in this fraction, fraction, fraction, fraction...not a fraction.*2971

*We want to get everything in terms of these fractions, so let's get everything into fraction format.*2979

*1 + 2 over 1 is how we will replace that; and then, 2 + 3 over 2, and so on and so on.*2984

*The rest of them are now in fractions.*2991

*We can write this as...here is our n = 1; here is our n = 2; n = 3; n = 4; n = 5; n = 6.*2994

*Notice: everything ends up changing; all of the numbers in each of these terms end up being different from the previous and the next term.*3003

*But we end up seeing some connections here: 1 matches to the 1's here; 2 matches to the 2's here; 3 matches to the 3's here;*3012

*4 matches to the 4's here; 5 matches to the 5's here; 6 matches to the 6's here.*3022

*And we see that this other number is just + 1 each time.*3028

*1 + 1 gets us 2; 2 + 1 gets us 3; 3 + 1 is 4; 4 + 1 is 5; 5 + 1 is 6; 6 + 1 is 7.*3032

*So now, we have an easy way to figure out what the n ^{th} term is.*3040

*The n ^{th} term is going to be equal to a_{n} =...well, it seems to be...*3043

*this one is just our value of n; plus some fraction...the bottom is also the value of n, and the top is n + 1.*3049

*And there we go: do a quick check, because it is always a good idea to do a check.*3060

*Let's check it out: a _{1} would be equal to 1 + 2, over 1; so we get 1 + 2...that checks out with what we already had as the first term.*3065

*If we wanted to try another one, like a _{2} = 1 + 2 + 1, over 1, which would be 1 + 3, over 1...*3078

*oops, sorry: not over 1; not over 1; it is divided by n as well; sorry about that mistake; divided by 2; 1 + 3 over 2.*3087

*Oh, I did it on both of them; it is important to end up using your formula in the check.*3096

*It is also an n here; 2 + 2 + 1 over 2; 2 + 3 over 2; and that does check out with what we had here.*3100

*Great; the last example: Find the n ^{th} term for the sequence below; assume the sequence starts at n = 1.*3108

*We see, right away, that this thing kind of changes its format a fair bit in these two.*3115

*It is totally different in these two.*3122

*Before we even really start looking for patterns, we want to get everything into the same format.*3124

*That will just make it easier to see patterns.*3127

*So, how can we get them into the same format?*3130

*We first certainly need to have them as fractions.*3132

*So, as fractions, we have 1; we can always divide by 1, so 1/1; then 3/1, and then 3 ^{2}/2, 3^{3}/6, 3^{4}/24, 3^{5}/120.*3135

*The first thing that you are probably noticing is that we have 3 ^{5}, 3^{4}, 3^{3}, 3^{2}, 3...*3151

*Well, we could write this as 3 ^{1}; how can we write 1 out?*3158

*Well, remember: 3 to the 0 equals 1; any number raised to the 0 comes out to be 1.*3161

*So, we could rewrite the top as 3 ^{0}/1; 3^{1}/1; 3^{2}/2; 3^{3}/6; 3^{4}/24; 3^{5}/120, and so on.*3166

*At the top part, we now see pretty clearly that there is a pattern.*3192

*It is that the number increases by 1 each time from the exponent on the 3; it starts at 0, so that would be 3 ^{n - 1}.*3194

*But what about this bottom part, 120, 24, 6, 2, and then we have 1 and 1 here...?*3201

*1, 1, 2, 6, 24, 120...well, if we worked backwards, we might recognize factorials.*3208

*Remember: we talked about factorials earlier in the lesson; that looks like factorials.*3215

*So, 5 times 4 times 3 times 2 times 1 is 120; 4 times 3 times 2 times 1 is 24; so we have 5! on the far right...*3220

*We can write this as 3 ^{5}/5!; we will keep going, so let's work backwards...3^{4}/4!;*3228

*3 ^{3}/3!; 3^{2}/2!; 3^{1}/1!...that would just get us 1.*3240

*And here is the most confusing part of all: 3 ^{0} over...well, what can we do that will end up having factorials involved, and still get us 1?*3254

*Now we have to go back and remember: there is a very specific thing about factorials.*3261

*The way that factorials work is that 0! is just defined to equal 1.*3265

*So, that means we could also write this as 0!; so we have maintained a pattern.*3270

*We are going to end up seeing patterns, because all of these problems are going to be based on patterns.*3275

*So, we know that it is a factorial pattern; it is no surprise that it is going to keep going.*3279

*We have on the bottom 0!, 1!, 2!, 3!, 4!, 5!...on the top: 3 ^{0}, 3^{1}, 3^{2}, 3^{3}, 3^{4}, 3^{5}.*3282

*Let's compare that to the numbers n = 1, n = 2, n = 3, n = 4, n = 5, n = 6, our first location, our second location, third location, etc.*3291

*With that in mind, we see that the top exponent is always equal to the value of the location.*3302

*At n = 5, we get a 4 exponent, so it is always minus 1.*3311

*Similarly, the 0!, 1!, 2!, 3!, 4!, 5!...it is always the number of the location, minus 1, to get to the number in the factorial.*3315

*In the third location, at n = 3, it is a 2! (3 minus 1); in the sixth location, it is a 5!, 6 minus 1.*3327

*So, we see that they are both based off of this n - 1 business; so that means we can finally set our a _{n} equal to...*3335

*it is 3 ^{n - 1}, over (n - 1)!; and there is our answer.*3343

*There is the general term; and it is always, always a good idea to check your work with this sort of thing;*3354

*So, let's just do a quick check to make sure that this ends up coming out.*3359

*At a _{1}, we would have 3^{1 - 1}/(1 - 1)!; so that is 3^{0}/0!, so we get 1/1,*3365

*which equals 1, which checks out with what we initially had.*3378

*Let's try jumping forward to a slightly larger number, so we can check against something else.*3382

*At a _{4}, we would be at 3^{4 - 1}/(4 - 1)!; so we have 3^{3}/3!.*3386

*Everything is in this 3 ^{3}, so we don't have to simplify that to 27; we can just leave it as it is.*3400

*3 ^{3}/3!...that comes out to be 3 times 2 times 1, or 6; so 3^{3}/6 is what we have for our fourth location.*3406

*1, 2, 3, 4 ^{th} location: 3^{3}/6; that ends up checking out.*3414

*We end up seeing that our general term makes sense.*3420

*All right, we have a really good understanding of how sequences work, and how we can get general term,*3423

*n ^{th} term, formulas from looking at them for a while.*3427

*Recognizing patterns is a really useful skill; it will show up in a whole bunch of different things in math.*3430

*Even if you end up thinking that this is a little bit difficult now, trust me: it is going to end up paying dividends later on.*3435

*You are going to end up using this stuff a lot, finding patterns in a variety of stuff,*3440

*whether it is in science class, math class, economics...whatever you end up studying.*3443

*You are going to end up having to find patterns of some sort.*3448

*Even in English, patterns are really important things; you are going to talk about themes in a book, so patterns really matter.*3451

*This sort of thing is really important.*3456

*We have a good understanding for sequences; now we are ready for the rest of this section.*3458

*We will have a whole bunch of ideas, working from this base of sequences.*3461

*All right, we will see you at Educator.com later--goodbye!*3464

1 answer

Last reply by: Professor Selhorst-Jones

Mon Jun 3, 2013 11:17 AM

Post by Vanessa Munoz on June 2, 2013

on example 3, second part, it could have been the sum, the addition pattern was wrong