For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
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 a 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.a_{1}, a_{2}, a_{3}, a_{4}, …, a_{n}, …  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 its length. The length of the above sequence is k.a_{1}, a_{2}, a_{3}, a_{4}, …, a_{k}  If we know a formula for the n^{th} term (this is also called the 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^{2}): 1, 4, 9, 16, 25, 36, 49, …
 Cubes (n^{3}): 1, 8, 27, 64, 125, 216, …
 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

 To find the terms of a sequence, we first need to know the pattern it uses. Once you know the pattern, use the pattern to generate the later terms.
 Consider the left sequence first. The first pattern you should look for is addition/subtraction:
Notice that we can get from each term to the following term by adding 5:−2, 3, 8, 13, …
Once we know the pattern, we simply continue to apply it.−2+5 = 3 ⇒ 3 + 5 = 8 ⇒ 8 + 5 = 13
Thus the fifth and sixth terms are 18 and 23.13 + 5 = 18 ⇒ 18 + 5 = 23  Now consider the right sequence. Check for an addition/subtraction pattern first.
Hmmm. Well, clearly it's not adding/subtracting by the same number for each step in the sequence. Next, try looking for multiplication/division. With that in mind, it's probably easiest to see that we can get from 14 to 2 by dividing by 7. Now make sure that pattern works for all the terms we can see:98, 14, 2, 2 7, …
Sure enough, the pattern works for every step in the sequence. Now that we know the pattern, continue to apply it.98 ÷7 = 14 ⇒ 14 ÷7 = 2 ⇒ 2 ÷7 = 2 7
Thus the fifth and sixth terms are [2/49] and [2/343].2 7÷7 = 2 49⇒ 2 49÷7 = 2 343

 To find the terms of a sequence, we first need to know the pattern it uses. Once you know the pattern, use the pattern to generate the later terms.
 Consider the left sequence first. The first pattern you should look for is addition/subtraction:
If you see the pattern immediately, great! If you don't, you can make it easier to see by making each term look similar to the others. For this sequence, some terms are fractions, some are not: let's fix this by making sure everything is over a denominator of 3:3, 8 3, 7 3, 2, …
Now it's much easier to see the pattern: we're subtracting [1/3] for each term:9 3, 8 3, 7 3, 6 3, …
Once we know the pattern, we simply continue to apply it.9 3− 1 3= 8 3⇒ 8 3− 1 3= 7 3⇒ 7 3− 1 3= 6 3
Thus the fifth and sixth terms are [5/3] and [4/3].6 3− 1 3= 5 3⇒ 5 3− 1 3= 4 3  Now consider the right sequence. Check for an addition/subtraction pattern first.
Hmmm. Well, clearly it's not adding/subtracting by the same number for each step in the sequence. Next, try looking for multiplication/division. With that in mind, we might realize that the denominator of the fraction is being reduced by a factor of 5 with every step. How do we reduce a denominator? By multiplying on the fraction. Test out multiplying by 5:7 250, 7 50, 7 10, 7 2, …
Sure enough, the pattern works for every step in the sequence. Now that we know the pattern, continue to apply it.7 250·5 = 7 50⇒ 7 50·5 = 7 10⇒ 7 10·5 = 7 2
Thus the fifth and sixth terms are [35/2] and [175/2].7 2·5 = 35 2⇒ 35 2·5 = 175 2

 If we have the n^{th} term formula for a sequence (also called the general term), we only have to plug in the value of the location we're interested in. For example, if we wanted to know the first term, we'd plug in n=1; if we wanted the third term, we'd plug in n=3; etc.
 We'll start with the left sequence first. To find the first term, we plug in n=1:
Continue this process for each of the terms we want to find: n=2, 3, 4, and 5.n=1 ⇒ 2(1) − 4 = −2 n=2 ⇒ 2(2) − 4 = 0 n=3 ⇒ 2(3) − 4 = 2 n=4 ⇒ 2(4) − 4 = 4 n=5 ⇒ 2(5) − 4 = 6  Now we work on the right sequence. To find the first term, we plug in n=1:
Continue this process for each of the terms we want to find: n=2, 3, 4, and 5.n=1 ⇒ (1)^{2} + 5 = 6 n=2 ⇒ (2)^{2} + 5 = 9 n=3 ⇒ (3)^{2} + 5 = 14 n=4 ⇒ (4)^{2} + 5 = 21 n=5 ⇒ (5)^{2} + 5 = 30

 When trying to find the n^{th} term formula for a sequence, always begin by figuring out what the underlying pattern is. For this problem, it's easy to see: we add by 3 for each step in the sequence.
 We can incorporate the idea of adding by 3 for each step by putting +3n in to the formula. This works, because for every increase to n (every step), it adds 3 more, so we have that stepping pattern we want.
 However, there's an issue with this idea. Notice what happens if we blindly try it in the first term:
Hmmm. Let's think about this: the issue is that 5 does not have a 3 add on to it. There have been no increases, because 5 is the first term. Thus, the number of increases we want is not n, but rather (n−1).n=1 ⇒ 5 + 3(1) = 8 BAD  With this in mind, let's try the formula:
This "starts" at 5, then adds 3 for the number of steps after the first term. Once you think you've found your formula, always check it. It's easy to make a mistake, so try plugging in a a few values for n (use values from various places, not just n=1 and n=2).a_{n} = 5 + 3(n−1) n=2 ⇒ a_{2} = 5 + 3(2−1) = 5 + 3 = 8
Great! The formula works, so we're done.n=5 ⇒ a_{5} = 5 + 3(5−1) = 5 + 12 = 17

 When trying to find the n^{th} term formula for a sequence, always begin by figuring out what the underlying pattern is. For this sequence, it's both easy and difficult. If we imagine this sequence without any of the negatives, it's as simple as multiplying by 2 at each step. This means, if we ignore the negative signs, it would be as simple as 2^{n}.
 But we can't ignore the negatives! We need some sort of "flipflopping" positive/negative sign. In the sequence, the sign goes as below:
There's a clever way to create this flipflopping pattern: raising −1 to a power and having the exponent go up one every step. Notice the below:+, −, +, −, +, …
Great! This means we can create the sign pattern we're looking for by using (−1)^{(n+1)}. [Try to remember this idea of (−1)^{(n+1)}: having to figure out a way to create flipflopping signs comes up often in sequences, and we'll use it again in the next question. Also, if you want to create the opposite sign pattern (starting on negative), notice that you can use (−1)^{n}.](−1)^{2}, (−1)^{3}, (−1)^{4}, (−1)^{5}, (−1)^{6}, … ⇒ +, −, +, −, +, …  We can combine these two ideas to describe the sequence:
The (−1)^{(n+1)} creates the flipflopping sign pattern, while the 2^{n} simply ignores the sign and gives the nonnegative value of the term. Put together, they exactly describe the sequence. Still, always check your formula. It's easy to make a mistake, so try plugging in a a few values for n (use values from various places, not just n=1 and n=2).a_{n} = (−1)^{(n+1)} ·2^{n} n=1 ⇒ a_{1} = (−1)^{(1+1)} ·2^{1} = (−1)^{2} ·2 = 2 n=4 ⇒ a_{4} = (−1)^{(4+1)} ·2^{4} = (−1)^{5} ·16 = −16

 When trying to find the n^{th} term formula for a sequence, always begin by figuring out what the underlying pattern is. For this sequence, the pattern is quite easy to see: the terms bounce back and forth between 5 and 9 with each step. We can describe this easily with words, as seen in the next step: the tricky part is describing it with just math symbols, which we will explore after the next step.
 The pattern is easy to see, and if we can describe it with words, quite easy to explain. It's 5 on every odd term (first, third, fifth, etc.) and 9 on every even term (second, fourth, sixth, etc.). We can write this out using notation similar to what we used for piecewise functions:
Great! The problem is done! ... right? Well, yes. But some teachers won't find this acceptable because it relies on words. Furthermore, and more importantly, some problems and ideas will require us to figure out a way to write the expression in just math, without needing to split it into two different cases. Hmmm. So how can we use just math and keep it as a single case?a_{n} = ⎧
⎪
⎨
⎪
⎩5, when n is odd 9, when n is even  We can see this sequence as bouncing back and forth from 5 to 9. We can expand this viewpoint by noticing that the sequence is bouncing around 7. The number 7 is the "center", and we can see the sequence as alternating up and down from that center by a distance of 2. Mathematically, we could rewrite the sequence as the below:
7−2, 7+2, 7−2, 7 +2, 7−2, …  With this new way of seeing it, we just need to figure out a way to alternate subtracting or adding 2. We can do this by always adding, then just flipflopping the sign of the 2. From the lesson and the previous problem, we saw that we can create a flipflopping sign through the use of (−1)^{n}. If we then multiply that by 2, we've created the flipflopping ±2 we wanted. Then add that to 7:
We can be confident that we figured it out, but still always check your formula. It's easy to make a mistake, so try plugging in a a few values for n (use values from various places, not just n=1 and n=2).a_{n} = 7 + (−1)^{n} ·2 n=2 ⇒ a_{2} = 7 + (−1)^{2} ·2 = 7 +2 = 9 n=5 ⇒ a_{2} = 7 + (−1)^{5} ·2 = 7 −2 = 5

 When trying to find the n^{th} term formula for a sequence, always begin by figuring out what the underlying pattern is. This is an extremely difficult pattern to crack. Start off by looking for basic patterns of addition/subtraction or multiplication/division. Looking at it for a little while, we quickly see that there is no basic operation we can use to move from one term to the next that works for all of them.
 Once we realize that it's not a simple pattern, try to get all the terms to follow a common format. By having them all in a similar pattern, we can more easily notice things. Let's first make them all fractions:
Hmmm. No patterns are jumping out at us yet. Let's try comparing the above to some other common patterns. A good one to check against is squares:4 1, 5 4, 2 3, 7 16, 8 25, 1 4, … 1, 4, 9, 16, 25, 36, …  Okay, now we're starting to get somewhere. Notice that the sequence of squares matches up to some of the denominators. While the pattern does not match up everywhere, it does match the first, second, fourth, and fifth terms. With that in mind, let's convert the third term and sixth term so that the denominators will follow the squares pattern: what numerator will allow the denominator to follow that pattern?
Third term: 2 3= ? 9⇒ 6 9Sixth term: 1 4= ? 36⇒ 9 36  Now that we have alternate ways to write the third and sixth terms which follow the pattern of squares in the denominator, let's swap them in:
Eureka! Now the pattern is easy to spot! The thing that made this problem tough wasn't the pattern itself: it was that the pattern had been obscured by the fractions being simplified. Now we see that the numerator just counts up, starting at four (n+3), while the denominator is the pattern of squares (n^{2}).4 1, 5 4, 6 9, 7 16, 8 25, 9 36, …
Finally, always check your formula: it's easy to make a mistake, so try plugging in a a few values for n (use values from various places, not just n=1 and n=2).a_{n} = n+3 n^{2}n=1 ⇒ a_{1} = 1+3 1^{2}= 4 1= 4 n=6 ⇒ a_{1} = 6+3 6^{2}= 9 36= 1 4

 Our kneejerk response to the question would probably be something like, "Of course the 20^{th} term will be greater than 1! Look at how fast the fraction is growing! Even if the denominator is growing, the numerator is way bigger, so it will always be that way." Still, that is not certainty. In math, you have to show you're right by logic, not just gut intuition. To be sure about the 20^{th} term, we need to know what it's going to look like. Therefore we need to figure out what the pattern is and come up with a way to find the 20^{th} term.
 Let's try to figure out a pattern. Since each term shares a common format, let's try to figure out patterns for the numerator and denominator separately. Consider the numerators first:
Checking the basic operations, we see that we can get from each term to the next term through multiplying by 7:7, 49, 343, 2401, 16807, …
Great! This means we can write the numerator as 7^{n}.7 ·7 = 49 ⇒ 49 ·7 = 343 ⇒ 343 ·7 = 2401 ⇒ 2401 ·7 = 16807  Now let's look at the sequence of denominators:
Hmmm. This one is a bit tougher. Looking at the sequence, we see that the pattern is not based on any basic operation being repeated. If that's the case, we should next compare it to some other common patterns: squares, cubes, and factorials.2, 4, 12, 48, 240, …
Ahha! If we compare the denominators to n!, we see that it's just each of them multiplied by 2. We can write the denominator as 2 ·n!.n^{2} ⇒ 1, 4, 9, 16, 25, … n^{3} ⇒ 1, 8, 27, 64, 125, … n! ⇒ 1, 2, 6, 24, 120, …  Putting both of these ideas together, we now have a formula:
As always, make sure to check your formula. It's easy to make a mistake, so try it out:a_{n} = 7^{n} 2 ·n!n=1 ⇒ 7^{1} 2 ·1!= 7 2·1= 7 2n=5 ⇒ 7^{5} 2 ·5!= 16807 2·120= 16807 240  Now that we have a formula for the n^{th} term, we can find the 20^{th} term.
Therefore the 20^{th} term is less than 1. Looking back on the sequence, we see that while the numerator grows very quickly at first, the denominator eventually "speeds up", outpacing the numerator over the longrun. Because of this, the sequence initially grows to reasonably large numbers, but only at the very start: over the "tail" of the sequence the terms are made extremely small.n=20 ⇒ 7^{20} 2 ·20!≈ 0.016

 A recursion formula allows us to find any given term based on information from the previous term(s). In this case, each term in the Lucas numbers is made by the preceding two terms. We read the recursion formula of L_{n} = L_{n−1}+L_{n−2} as "to find a given term, we add together the two terms that came before it." For example, to find the third term, we add together the second and first terms. Notice how this is what the recursion formula says mathematically:
n=3 ⇒ L_{3} = L_{3−1} + L_{3−2} = L_{2} + L_{1}  Now that we understand how the recursion formula works, let's start finding terms. Since we were already given n=1 and n=2, we start by finding n=3:
The Lucas numbers come with the initial terms of L_{1}=1 and L_{2} = 3, so plug those in:n=3 ⇒ L_{3} = L_{3−1} + L_{3−2} = L_{2} + L_{1} L_{3} = L_{2} + L_{1} = 3 + 1 = 4  Continue in this method:
n=4 ⇒ L_{4} = L_{3} + L_{2} = 4 + 3 = 7 n=5 ⇒ L_{4} = L_{4} + L_{3} = 7 + 4 = 11  If we want to speed up the process by a little, we can write out what we have of the sequence so far:
However, instead of writing in an ellipsis (`...'), like we have above, just leave the space after the last term blank. Notice that we can write the next term by simply adding together the last two terms in the sequence, creating:1, 3, 4, 7, 11, …
We can keep doing this on paper, summing up the last two terms to find the next term. Keep doing so until you have all twelve terms like the problem told you to find.1, 3, 4,
7, 11
+,
18
=

 A recursion formula allows us to find any given term based on information from the previous term(s). In this case, each term in the sequence is found by multiplying the previous term by [2/5].
 If you have difficulty seeing how a given recursion formula works, try writing out the first few terms of the sequence:
Thus we have the sequence625· 2 5= 250 ⇒ 250 · 2 5= 100 ⇒ 100 · 2 5= 40
However, written in this way, it does not particularly help us see the pattern. Often, simplifying the terms in a sequence makes it more difficult to see patterns. Instead, when trying to find patterns, it usually helps to not simplify. Furthermore, notice the terms are based on factors of 5 and 2. With this in mind, it helps to write the first term in a way that shows this: 625 = 5^{4}. Rewriting with both of these in mind, we see625, 250, 100, 40, … 5^{4}, 5^{4}·2 5, 5^{4}·2^{2} 5^{2}, 5^{4}·2^{3} 5^{3}, …  We can make the pattern even clearer by putting the 5's into a single object and clarifying the exponent on the 2 for every term:
Now it's fairly easy to see the pattern: the exponent on the 5 goes down with every step, while the exponent on the 2 goes up. Since we start at n=1, we have to keep that in mind in how we set the n's in the exponents:5^{4} ·2^{0}, 5^{3} ·2^{1}, 5^{2} ·2^{2}, 5^{1} ·2^{3}, …
Lastly, always check your formula: it's easy to make a mistake, so try it out. We check it against the first few terms we figured out:a_{n} = 5^{(5−n)} ·2^{(n−1)} n=1 ⇒ a_{1} = 5^{(5−1)} ·2^{(1−1)} = 5^{4} ·2^{0} = 625 n=4 ⇒ a_{4} = 5^{(5−4)} ·2^{(4−1)} = 5^{1} ·2^{3} = 40  Alternatively, we could figure it out simply from seeing that every time we apply the recursion step, it multiplies by a factor of [2/5]. Thus, if we are at term n, we will have multiplied the first term (a_{1}) a total of n−1 times by the factor. Thus we have
As always, if we were to do it this way, we should still check the formula against the first few terms. We'll have to use the initial recursion formula to build some terms, but then we would check our a_{n} formula against those terms like we did above. Finally, notice that the above formula and the one figured out in the prior step are equivalentboth ways are perfectly fine ways to solve the problem.a_{n} = 625 · ⎛
⎝2 5⎞
⎠(n−1)
625 · ⎛
⎝2 5⎞
⎠(n−1)
= 5^{4} ·(2 ·5^{−1} )^{(n−1)} = 5^{4} ·2^{(n−1)} ·5^{(1−n)} = 5^{(5−n)} ·2^{(n−1)}
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Introduction to Sequences
Lecture Slides are screencaptured 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
Precalculus with Limits Online Course
Transcription: Introduction to Sequences
Hiwelcome 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 thingsan 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 have0190
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} termthat is a_{n}, right hereis 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 persona 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 to0521
(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 pluginanumberforourn...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 fact0728
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 sequencehow 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 thisdoes 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 hypothesisyou 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 steppingstone1087
to the next steppingstone, 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 patternnow you are ready to figure out some way1112
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 trickya 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 equation1159
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 are1295
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 ones1353
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 headjust keep that one in the back of your head1393
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 followingone 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 formatfor example,1532
they are all in fractionstry 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 partsfor 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 numerators1575
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 parts1581
for example, a fraction is a numerator over a denominator, every single timeit 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 sequences1624
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 atare 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 relates1685
to the values inside of the actual term inside of the sequencethe 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 stepbystep 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 letterbecause 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 wellknown, recursivelydefined 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 thathow 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 againit looks like our pattern probably is going to work out.2348
5 to 9we add 4; 9 to 13we add 4; 13 to whatever is nextit 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 methodthat this is going to end up working2460
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 sensewe 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 herewhat 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 upsquares, 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 iswhat 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 term2708
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 latergoodbye!3464
1 answer
Last reply by: Professor SelhorstJones
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