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 1 answerLast reply by: Professor Selhorst-JonesMon Jun 3, 2013 11:17 AMPost by Vanessa Munoz on June 2, 2013on example 3, second part, it could have been the sum, the addition pattern was wrong

### Introduction to Sequences

• A sequence is an ordered list of numbers. We can write out a sequence as
 a1,   a2,  a3,  a4,  …, an, …
We call each of the entries in the sequence a term. Above, a1 is the first term, a2 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.
 a1,  a2, a3, a4, …, ak
We call the number of terms in a finite sequence its length. The length of the above sequence is k.
• If we know a formula for the nth 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 nth term. Similarly, it can be possible to transform an nth 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 nth 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 a1→ a2, then test it against a2→ a3,  a3→ a4, 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 nth 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:
• Mutliplication/Division: multiply by k every term.
• Squares (n2):     1,  4,   9,   16,   25,   36,   49,  …
• Cubes (n3):         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

For each of the sequences below, give the fifth and sixth terms.
 −2,  3,   8,   13,   … ⎢⎢ 98,   14,   2, 2 7 ,   …
• 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:
 −2,  3,   8,   13,   …
Notice that we can get from each term to the following term by adding 5:
 −2+5 = 3     ⇒     3 + 5 = 8     ⇒     8 + 5 = 13
Once we know the pattern, we simply continue to apply it.
 13 + 5 = 18     ⇒     18 + 5 = 23
Thus the fifth and sixth terms are 18 and 23.
• Now consider the right sequence. Check for an addition/subtraction pattern first.
 98,   14,   2, 2 7 ,   …
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 ÷7 = 14     ⇒     14 ÷7 = 2    ⇒     2 ÷7 = 2 7
Sure enough, the pattern works for every step in the sequence. Now that we know the pattern, continue to apply it.
 2 7 ÷7 = 2 49 ⇒ 2 49 ÷7 = 2 343
Thus the fifth and sixth terms are [2/49] and [2/343].
18,  23        |        [2/49],   [2/343]
For each of the sequences below, give the fifth and sixth terms.
 3, 8 3 , 7 3 ,   2,   … ⎢⎢ 7 250 , 7 50 , 7 10 , 7 2 ,   …
• 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:
 3, 8 3 , 7 3 ,   2,   …
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:
 9 3 , 8 3 , 7 3 , 6 3 ,   …
Now it's much easier to see the pattern: we're subtracting [1/3] for each term:
 9 3 − 1 3 = 8 3 ⇒ 8 3 − 1 3 = 7 3 ⇒ 7 3 − 1 3 = 6 3
Once we know the pattern, we simply continue to apply it.
 6 3 − 1 3 = 5 3 ⇒ 5 3 − 1 3 = 4 3
Thus the fifth and sixth terms are [5/3] and [4/3].
• Now consider the right sequence. Check for an addition/subtraction pattern first.
 7 250 , 7 50 , 7 10 , 7 2 ,   …
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 ·5 = 7 50 ⇒ 7 50 ·5 = 7 10 ⇒ 7 10 ·5 = 7 2
Sure enough, the pattern works for every step in the sequence. Now that we know the pattern, continue to apply it.
 7 2 ·5 = 35 2 ⇒ 35 2 ·5 = 175 2
Thus the fifth and sixth terms are [35/2] and [175/2].
[5/3],  [4/3]        |        [35/2],   [175/2]
Given the nth term formulas below, write out the first five terms of each sequence.
 an = 2n−4 ⎢⎢ bn = n2 + 5
• If we have the nth 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:
 n=1     ⇒     2(1) − 4     =     −2
Continue this process for each of the terms we want to find: n=2, 3, 4, and 5.
 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:
 n=1     ⇒     (1)2 + 5     =     6
Continue this process for each of the terms we want to find: n=2, 3, 4, and 5.
 n=2     ⇒     (2)2 + 5     =     9

 n=3     ⇒     (3)2 + 5     =     14

 n=4     ⇒     (4)2 + 5     =     21

 n=5     ⇒     (5)2 + 5     =     30
−2,   0,   2,   4,   6            |            6,   9,   14,   21,   30
Given the sequence below, write an expression for the nth term of the sequence.
 5,   8,   11,   14,   17,   …
• When trying to find the nth 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:
 n=1     ⇒     5 + 3(1)     =     8     BAD
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).
• With this in mind, let's try the formula:
 an = 5 + 3(n−1)
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).
 n=2     ⇒     a2 = 5 + 3(2−1)     =     5 + 3     =     8

 n=5     ⇒     a5 = 5 + 3(5−1)     =     5 + 12     =     17
Great! The formula works, so we're done.
an = 5 + 3(n−1)
Given the sequence below, write an expression for the nth term of the sequence.
 2,   −4,   8,   −16,   32,   …
• When trying to find the nth 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 2n.
• But we can't ignore the negatives! We need some sort of "flip-flopping" positive/negative sign. In the sequence, the sign goes as below:
 +,   −,   +,   −,   +,   …
There's a clever way to create this flip-flopping pattern: raising −1 to a power and having the exponent go up one every step. Notice the below:
 (−1)2,  (−1)3,  (−1)4,  (−1)5,  (−1)6,  …    ⇒     +,   −,   +,   −,   +,   …
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 flip-flopping 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.]
• We can combine these two ideas to describe the sequence:
 an = (−1)(n+1) ·2n
The (−1)(n+1) creates the flip-flopping sign pattern, while the 2n simply ignores the sign and gives the non-negative 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).
 n=1     ⇒     a1 = (−1)(1+1) ·21     =     (−1)2 ·2     =     2

 n=4     ⇒     a4 = (−1)(4+1) ·24     =     (−1)5 ·16     =     −16
an = (−1)(n+1) ·2n [It's also possible to describe the sequence as an = − (−2)n. This is a perfectly acceptable answer, but it's very useful to get used to the idea of the flip-flopping sign patterns created by (−1)n and (−1)(n+1). They are sometimes absolutely necessary, so it's good to keep them in mind.]
Given the sequence below, write an expression for the nth term of the sequence.
 5,   9,   5,   9,   5,   …
• When trying to find the nth 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:
an =

 5,
 when n is odd
 9,
 when n is even
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?
• 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 re-write 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 flip-flopping the sign of the 2. From the lesson and the previous problem, we saw that we can create a flip-flopping sign through the use of (−1)n. If we then multiply that by 2, we've created the flip-flopping ±2 we wanted. Then add that to 7:
 an = 7 + (−1)n ·2
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).
 n=2     ⇒     a2 = 7 + (−1)2 ·2     =     7 +2     =     9

 n=5     ⇒     a2 = 7 + (−1)5 ·2     =     7 −2     =     5
an = 7 + (−1)n ·2
Given the sequence below, write an expression for the nth term of the sequence.
 4, 5 4 , 2 3 , 7 16 , 8 25 , 1 4 ,   …
• When trying to find the nth 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:
 4 1 , 5 4 , 2 3 , 7 16 , 8 25 , 1 4 ,   …
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:
 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 9

 Sixth 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:
 4 1 , 5 4 , 6 9 , 7 16 , 8 25 , 9 36 ,   …
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 (n2).
 an = n+3 n2
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).
 n=1     ⇒     a1 = 1+3 12 = 4 1 =     4

 n=6     ⇒     a1 = 6+3 62 = 9 36 = 1 4
an = [(n+3)/(n2)]
Consider the sequence below. Will the 20th term be greater or less than 1? Explain your reasoning.
 7 2 , 49 4 , 343 12 , 2401 48 , 16807 240 ,   …
• Our knee-jerk response to the question would probably be something like, "Of course the 20th 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 20th 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 20th 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:
 7,   49,   343,   2401,   16807,   …
Checking the basic operations, we see that we can get from each term to the next term through multiplying by 7:
 7 ·7 = 49     ⇒     49 ·7 = 343     ⇒     343 ·7 = 2401     ⇒     2401 ·7 = 16807
Great! This means we can write the numerator as 7n.
• Now let's look at the sequence of denominators:
 2,   4,   12,   48,   240,   …
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.
 n2     ⇒     1,
 4,
 9,
 16,
 25,
 …
 n3     ⇒     1,
 8,
 27,
 64,
 125,
 …
 n!     ⇒     1,
 2,
 6,
 24,
 120,
 …
Ah-ha! 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!.
• Putting both of these ideas together, we now have a formula:
 an = 7n2 ·n!
As always, make sure to check your formula. It's easy to make a mistake, so try it out:
 n=1     ⇒ 712 ·1! = 7 2·1 = 7 2

 n=5     ⇒ 752 ·5! = 16807 2·120 = 16807 240
• Now that we have a formula for the nth term, we can find the 20th term.
 n=20     ⇒ 7202 ·20! ≈     0.016
Therefore the 20th 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 long-run. 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.
Less than 1. To show this, you must find the 20th term, This is most likely done by figuring out the nth term formula then plugging in n=20.
The Lucas numbers form a well-known recursively defined sequence (less well-known than the Fibonacci sequence, though). They are given by the recursion formula and initial terms below. Write out the first 12 terms.
 L1 = 1,       L2=3,              Ln = Ln−1+Ln−2
• 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 Ln = Ln−1+Ln−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     ⇒     L3 = L3−1 + L3−2     =     L2 + L1
• 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:
 n=3     ⇒     L3 = L3−1 + L3−2     =     L2 + L1
The Lucas numbers come with the initial terms of L1=1 and L2 = 3, so plug those in:
 L3 = L2 + L1     =     3 + 1     =     4
• Continue in this method:
 n=4     ⇒     L4 = L3 + L2     =     4 + 3     =     7

 n=5     ⇒     L4 = L4 + L3     =     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:
 1,   3,   4,   7,   11,   …
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+ , 18=
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,   29,   47,   76,   123,   199,   322
Given the initial term and recursion formula below, find an expression for the nth term of the sequence.
 a1 = 625,               an+1 = 2 5 ·an
• 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:
 625· 2 5 = 250     ⇒     250 · 2 5 = 100     ⇒     100 · 2 5 = 40
Thus we have the sequence
 625,   250,   100,   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 = 54. Re-writing with both of these in mind, we see
 54, 54·2 5 , 54·2252 , 54·2353 ,     …
• 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:
 54 ·20,     53 ·21,     52 ·22,     51 ·23,     …
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:
 an = 5(5−n) ·2(n−1)
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:
 n=1     ⇒     a1 = 5(5−1) ·2(1−1)     =     54 ·20     =     625

 n=4     ⇒     a4 = 5(5−4) ·2(4−1)     =     51 ·23     =     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 (a1) a total of n−1 times by the factor. Thus we have
 an = 625 · ⎛⎝ 2 5 ⎞⎠ (n−1)
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 an formula against those terms like we did above. Finally, notice that the above formula and the one figured out in the prior step are equivalent-both ways are perfectly fine ways to solve the problem.
 625 · ⎛⎝ 2 5 ⎞⎠ (n−1) =     54 ·(2 ·5−1 )(n−1)     =     54 ·2(n−1) ·5(1−n)     =     5(5−n) ·2(n−1)
an = 5(5−n) ·2(n−1),    or, equivalently,    an = 625 ·( [2/5])(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.

### 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

### 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 a1, a2, a3, a4...an...0041

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

This would be a1, the first term, because it is the first in the sequence.0050

a2 is the second term, because it is the second in the sequence.0056

a3 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, a4) 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 a1, a2, a3, a4...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 a1, a2, a3, a4, and there is that ...0149

that says to continue in this manner; but then, we actually stop at ak.0154

Notice how there is nothing after the ak; 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 ak, and we just stop at ak, because there is nothing after ak.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 ak here and we have a1 here.0179

So, that means we are counting our first term, our second term, our third term...all the way up until our kth 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 nth 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 nth 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 an equals some stuff, some algebraic formulation,0224

then if we set n equal to 1, we would get the first term, a1.0232

If we set n equal to 17, we get the 17th term, a17.0238

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 an works,0256

for the way this nth 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 an = 7n - 5, then we have the sequence a1, a2, a3, a4...0273

Well, notice here: in a1, 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 a1 is equal to 2.0290

The same thing for a2: we know that, at a2, 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

a3: we have n = 3, so we swap out; we have 7(3) - 5, 21 - 5, 16; so our third entry, a3, is equal to 16.0312

At a4, 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, an = 7n - 5, we are able to find any term.0331

We can find any term if we know this general form an = 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 an = an - 1 + 7, what is that saying?0381

It is that the nth term--that is an, right here--is created by looking at the previous term.0387

Well, what would be one before n? n - 1; so an - 1 is going to be the term just before the nth term.0392

So, an - 1...and then adding 7 to it is that + 7 business right there.0401

an = an - 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 an = an - 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 a1 = 2, with the previous recursion formula an = an - 1 + 7,0484

then our first term is going to be a1, right here.0491

Well, we were just told that a1 = 2; so that means that we have 2 here.0494

Then, from there on, we have an = an - 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 a2 is going to be equal to0521

(a2 is this one right here), then a2 = a2 - 1, or 1, + 7.0527

So, a1 is equal to 2; we figured that out here; then + 7...so we get a2 = 9, which is what we got right here.0535

And the same thing is going on for figuring out a3: a3 is going to be equal to a2 + 7.0547

a4 is going to be equal to a3 + 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 nth 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 nth term.0578

Similarly, it can be possible to transform an nth 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 nth term format.0615

Other ones are really easy to talk about in that nth 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 nth 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 nth 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 an - 1 - 5.0892

And the green one, the second one, with multiplication, times 3...that would be an - 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 nth 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 an = 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 nth 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 a1 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 a2, 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, an = 2(3)n - 1:0992

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 3n - 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 a1 = 2(3)1 - 1, then that would be 2(3)0; 30 is just 1;1023

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

If we did a2, then we would have a2 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 a1 to a2.1068

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

You look at a1 to a2, and then you want to test that hypothesis against the following ones:1078

a2 to a3, a3 to a4, and any other terms that are given.1083

You come up with thinking that the pattern here, the way we get from one stepping-stone1087

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 nth 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 way1112

to formulate that general form, that general term, the nth 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 nth term: that could be tricky--a formula for the nth 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: n2: 12, 22, 32, 42, 52, 62, 72...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

n3 is 13, 23, 33, 43, 53, 63, 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 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 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 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 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 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 relates1685

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 nth term, write the first four terms of each sequence.1831

Assume that each sequence starts at n = 1.1836

So, that nth term, that general term, is an = stuff.1839

In this case, for our first one, we have an = 3n - 2.1843

Our first term, the a1, is going to be when n = 1.1849

We plug in a1 = 3(1) - 2.1855

Similarly, a2 is when n = 2, so we have a2 = 3(2) - 2.1860

a3 = 3(3) - 2; a4 = (I'll write that a little below; there is not quite enough room) 3(4) - 2.1867

We work this out: a1 = 3(1) - 2, 3 - 2, so we get a1 = 1.1880

a2: 3(2) is 6; 6 - 2 is 4; a3 = 3(3), 9, minus 2; 9 - 2 is 7.1886

a4 = 3(4) is 12, minus 2 is 10.1896

So, we have the first four terms here: a1 = 1, a2 = 4, a3 = 7, a4 = 10.1902

Also, I want to point out; notice how a1, to get to a2...we added 3.1909

To get to a3, we added 3; to get to a4, 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: bn = -1n/(n + 3); our first one, a1, is going to be (-1)1/(1 + 3).1935

We swap out all of the n's that occur for whatever we have here, a1.1947

Next, a2 = -1 squared, -1 to the 2, divided by 2, plus 3.1952

a3 is equal to -1 to the 3, over 3 + 3.1959

a4 is going to be equal to -1 to the 4, over 4 + 3.1966

We can work this out here: we have a1 = (-1)1, which is still just -1; 1 + 3 is 4, so we have -1/4.1973

a2: (-1)2, -1 times -1...that cancels to just positive, so we just have a positive 1, divided by 2 + 3, 5.1984

a3 = (-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

a4 = (-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 a1 = -1/4; a2 = +1/5; a3 = -1/6; a4 = 1/7.2008

We found the first four terms.2020

Finally, cn = 47...oh, oops; that whole time shouldn't have been an, because it was bn.2022

So, it actually should have been not a2, not a1...any of these...2030

It should have been b1, b2, b3, b4, 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

b1, b2, b3, b4...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: cn = 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, c1 is going to be equal to 47; c2 is going to be equal to 47;2069

c3 is going to be equal to 47; c4 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 an = an - 1 + an - 2, and a1 = a2, which equals 1.2105

Right away, we know that our first two terms are 1 (a1 = 1), and then a2 also equals 1; so 1, 1.2114

If we want to figure out the next term, we have an = an - 1 + an - 2.2125

So, if we want to figure out what a3 is going to be, then that is going to be equal to a3 - 1,2130

so a2, plus a3 - 2, a1.2135

a3...we don't know what a3 is, but we do know what a1 and a2 are.2140

They are both 1; so we have 1 + 1; a3 = 2.2144

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

If we want to figure out a4, that would be a4 - 1, a3, plus a4 - 2, a2.2155

What is a3? We just figured out a3 = 2, so we have 2 +...a2, once again, is 1; that equals 3.2164

a4 = 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

a5 is equal to...not the general term, but how this pattern is working, on the whole...2186

a5 is equal to a4, the previous term, plus the term previous to that one, a3.2191

So, a5 = a4 + a3: we just figured out that a4 = 3 and a3 = 2.2199

So, we have a5 = 5; there is our next term, 5.2206

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

Well, notice: a5 = a4 + a3; so a5 is equal to the previous term, plus the term previous to that.2215

a4 equaled a3 + a2; a4 is equal to the previous term, plus the term previous to that.2221

a3 = a2 + a1, 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 a1 and a2, 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 nth 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 nth term,2322

the an, 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

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 an = 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

an = 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 a1 = -3; so that means we can have a1 = 4(1) + ?.2405

We know that a1 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 an, 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 a1; that is how we figured it out.2440

Let's check and make sure that a2 ends up working out.2442

a2 = 4(2) - 7, 8 - 7, which equals 1; and that checks out.2445

Next, a3 = 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 an = 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 n2, then that would end up giving is the sequence:2576

1 (12 is 1), then 22 is 4, then 32 is 9; 42 is 16; 52 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 nth term is.2617

It is the formula an = n2 (because it is the number squared), but then we also have to add 1: an = n2 + 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 a1 = 12 + 1, then we get 1 + 1, which equals 2.2639

That checks out; if we plug in for our second term, a2, then we would have 22 + 1.2647

22 is 4; 4 + 1 is 5; that checks out.2654

Our next one: a3 = 32 + 1; 9 + 1 = 10; that checks out; great.2658

At this point, it seems that our n2 + 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 nth term.2702

We have an = -3 times an - 1; that is to say, some term2708

is equal to -3 times the previous term; an is some term; an - 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 a2, the second term,2730

it would be -3 times a1, so a2 is equal to -3 times...a1 is just 2, so times 2; that equals -6.2736

So, a2 = -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 nth term, so how can we figure out what the nth term is?2787

Let's switch colors for this; the nth 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 an = -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, a3, 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 nth term formula; it works out perfectly.2959

The fifth example: Find the nth 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 nth term is.3040

The nth term is going to be equal to an =...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: a1 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 a2 = 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 nth 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 32/2, 33/6, 34/24, 35/120.3135

The first thing that you are probably noticing is that we have 35, 34, 33, 32, 3...3151

Well, we could write this as 31; 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 30/1; 31/1; 32/2; 33/6; 34/24; 35/120, 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 3n - 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 35/5!; we will keep going, so let's work backwards...34/4!;3228

33/3!; 32/2!; 31/1!...that would just get us 1.3240

And here is the most confusing part of all: 30 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: 30, 31, 32, 33, 34, 35.3282

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 an equal to...3335

it is 3n - 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 a1, we would have 31 - 1/(1 - 1)!; so that is 30/0!, 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 a4, we would be at 34 - 1/(4 - 1)!; so we have 33/3!.3386

Everything is in this 33, so we don't have to simplify that to 27; we can just leave it as it is.3400

33/3!...that comes out to be 3 times 2 times 1, or 6; so 33/6 is what we have for our fourth location.3406

1, 2, 3, 4th location: 33/6; 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

nth 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