For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Geometric Sequences & Series
 A sequence is geometric if every term in the sequence can be given by multiplying the previous term by some constant number r:
a_{n} = r·a_{n−1}.  The formula for the n^{th} term (general term) of a geometric sequence is
a_{n} = r^{n−1} ·a_{1}.  To find the formula for the general term of a geometric sequence, we only need to figure out its first term (a_{1}) and the common ratio (r).
 We can use the following formula to calculate the value of a geometric series. Given any geometric sequence a_{1}, a_{2}, a_{3}, …, the sum of the first n terms (the n^{th} partial sum) is
S_{n} = a_{1} · 1−r^{n}
.  We can find the partial sum (S_{n}) by only knowing the first term (a_{1}), the common ratio (r), and how many terms are being added together (n). [Caution: Be careful to pay attention to how many terms there are in the series. It can be easy to get the value of n confused and accidentally think it is 1 higher or 1 lower than it really is.]
 Unlike an arithmetic series, we can also consider an infinite series when working with a geometric sequence. If r < 1, then
S_{∞} = a_{1} · 1
.
Geometric Sequences & Series

 A sequence is geometric if the ratio between any two consecutive terms is a constant. In other words, there exists some constant value r that we can multiply to get from one term to the following term.
 To check if a sequence is geometric, we simply need to find the ratio between any two consecutive terms, then see if every pair of consecutive terms in the sequence has the same ratio.
 Let's start with the first sequence:
Don't be fooled by the fact that the exponents come as an arithmetic sequence (constant difference), we care about the ratio between the terms. Check to see what the ratio between each consecutive pair of terms is:3^{1}, 3^{3}, 3^{5}, 3^{7}, …
Since the ratio between any two terms is always the same (r=9), we have that the sequence is geometric.3^{3} 3^{1}= 9 3^{5} 3^{3}= 9 3^{7} 3^{5}= 9  Now let's consider the second sequence:
At first glance, it's difficult to see that there is some common ratio (or, equivalently, that there exists some number we can multiply by to get to the next term at each step). While we don't immediately see a value for r, check the ratios between each consecutive pair of terms:−4, 6, −9, 27 2, …
Thus, now that we've actually calculated the ratios, we see that the sequence does have a common ratio of r=−[3/2]. Thus it is a geometric sequence.6 −4= − 3 2−9 6= − 3 2, 27 2−9= − 3 2

 The number given by a_{1} is the first term of the sequence. The value of r is what we multiply by to get each subsequent term. That is r·a_{1} = a_{2}, and so on for later terms.
 The n^{th} term formula a_{n} is a formula where we can plug in n (the number of the term we want) and get out the value of the term for that n^{th} term.
For a geometric sequence, we saw in the video lesson that there is a simple formula if you know the first term and the common ratio. It is simply
a_{n} = r^{n−1} ·a_{1}.  Since we have a formula and already know a_{1} and r, we can just plug in:
a_{n} = r^{n−1} ·a_{1} = 4^{n−1} ·5
[If we want to check our answer, we know the first term is a_{1}=5 and the common ratio is r=4, so we can easily write out the first few terms:
Now that we have the first few terms, check to make sure the formula gives the same values:5, 20, 80, … n=1 ⇒ a_{1} = 4^{1−1} ·5 = 5
Great: our formula for a_{n} checks out, so we know our answer is correct.]n=3 ⇒ a_{3} = 4^{3−1} ·5 = 80

 For a geometric sequence, we saw in the video lesson that there is a simple formula if you know the first term and the common ratio. It is simply
a_{n} = r^{n−1} ·a_{1}.  This means we can easily find the formula if we know the first term (a_{1}) and the common ratio (r). Looking at the sequence, the first term is clearly there, so we know a_{1} = 343. Now we just need to find the common ratio.
To do that, just find the common ratio between two terms (remember, we find the ratio as "later divided by earlier", for example, third over second). To be sure we got it correctly, do it for two different pairs. Below we'll check first and second along with third and fourth:
Thus the common ratio is r=[2/7].98 343= 2 7= 8 28  Now that we know a_{1} = 343 and r=[2/7], we just plug in:
a_{n} = r^{n−1} ·a_{1} = ⎛
⎝2 7⎞
⎠n−1
·343
[If we want to check our answer, we already know the first few terms of the sequence, so we can check that our formula gives the same values:n=1 ⇒ a_{1} = ⎛
⎝2 7⎞
⎠1−1
·343 = 343
Great: our formula for a_{n} checks out, so we know our answer is correct.]n=4 ⇒ a_{4} = ⎛
⎝2 7⎞
⎠4−1
·343 = 8

 We could find the value of the sum by just getting a calculator (or a piece of scratch paper) and adding everything up. It would take a whole lot of time, but it could be done that way. However, instead of that, let's start by noticing that each term in the sum is effectively a term in a geometric sequence. Thus, we're working with a geometric series: a sum where every subsequent term has a common ratio with its preceding term. We see this because every subsequent term is the same as the previous, just multiplied by r=3.
 From the video lesson, we learned that the sum of any geometric sequence is
where a_{1} is the first term, r is the common ratio between terms, and n is the total number of terms in the series.a_{1} · 1−r^{n} 1−r,  Looking at the problem, it's easy to see that the first term is a_{1} = 2. We know it's a geometric series because we noticed that each term is the same as the previous, just multiplied by the common ratio or r=3. The hardest part is figuring out what n ishow many terms there are total. We might be tempted to think there are 10 terms because we have 3^{10} in the last term. That is not the case. Notice that we don't start at 3^{1}, we actually start with no 3's at all: that is a_{1} = 2 ·3^{0} [since 3^{0} = 1, this is equivalent to a_{1} = 2]. Thus, we actually have n=11 terms, because the first term starts at 3^{0}, and we have to count that first term.
 Now that we know all the pertinent values, we can use the formula:
a_{1} · 1−r^{n} 1−r= 2 · 1−3^{11} 1−3= 2 · −177 146 −2= 177 146

 (Note: If you are unfamiliar with using sigma notation (Σ) for compactly showing a sum/series, make sure to check out the lesson Introduction to Series. How to read and use the notation is carefully explained in that lesson, but it will be assumed you already understand it in the below steps.) Begin by noticing that the notation indicates a geometric series. It will have a common ratio of r=1.03 for every term because of the (1.03)^{i} in the sigma notation. That means every subsequent term will be multiplied by another (1.03), so it follows the rules of a geometric series/sequence.
 From the video lesson, we learned that the sum of any geometric sequence is
where a_{1} is the first term, r is the common ratio between terms, and n is the total number of terms in the series. To find the first term of the series, just plug in the lowest value the index can give: i=1.a_{1} · 1−r^{n} 1−r,
We already noticed that the common ratio is r=1.03 since every subsequent term in the series multiplies by an additional (1.03) [this happens because the exponent goes up 1 for every subsequent term]. For this problem, it's not too difficult to figure out how many terms there are (n), because we just need to know how many numbers 1→ 50 is. That's pretty clearly n=50, so we know the number of terms. [Still, be very careful when figuring out the number of terms n in general: it's very easy to make a mistake and over or undershoot by 1.]a_{1} ⇒ i = 1 ⇒ 100 ·(1.03)^{1} = 103  Now that we know all the necessary values for the geometric series formula, we plug in:
a_{1} · 1−r^{n} 1−r= 103 · 1−1.03^{50} 1−1.03= 103 · −3.3839 −0.03≈ 11 618

 (Note: If you are unfamiliar with using sigma notation (Σ) for compactly showing a sum/series, make sure to check out the lesson Introduction to Series. How to read and use the notation is carefully explained in that lesson, but it will be assumed you already understand it in the below steps.) Begin by noticing that the notation indicates a geometric series. It will have a common ratio of r=−[1/2] for every term because of the (−[1/2])^{k} in the sigma notation. That means every subsequent term will be multiplied by another (−[1/2]), so it follows the rules of a geometric series/sequence.
 From the video lesson, we learned that the sum of any geometric sequence is
where a_{1} is the first term, r is the common ratio between terms, and n is the total number of terms in the series. To find the first term of the series, just plug in the lowest value the index can give: k=7.a_{1} · 1−r^{n} 1−r,
We already noticed that the common ratio is r=−[1/2] since every subsequent term in the series multiplies by an additional (−[1/2]) [this happens because the exponent goes up 1 for every subsequent term]. Don't worry about the fact that r is negative: the formula will work as long as r ≠ 1.a_{1} ⇒ k=7 ⇒ −6400 · ⎛
⎝− 1 2⎞
⎠7
= 50  The trickiest part is probably figuring out what the the number of terms (n) is. To do this, notice that the first term has an index of k=7, while the last term has an index of k=22. Thus there are 22−7 = 15 steps between the two terms. However!, we must also remember to include the starting location, since it doesn't get counted as a step. Thus there are a total of n=16 terms (15 steps plus 1 for "home").
Now that we know all the necessary values for the arithmetic series formula, we can just plug in:
a_{1} · 1−r^{n} 1−r= 50 · 1− ⎛
⎝− 1 2⎞
⎠16
1− ⎛
⎝− 1 2⎞
⎠= 50 · 65 535 65 5363 2= 546 125 16 384

 An infinite geometric series is one where we consider what happens as we work towards adding an infinite number of terms together. For this problem, we would sum the below:
Notice that we never stop adding terms. The pattern of adding more and more terms continues forever. We want to know what value the series will converge to as we add ever more terms (what value the series "settles toward" in the long run).1 + 4 5+ 16 25+ 64 125+ …  From the video lesson, we learned that an infinite geometric series converges if r < 1. If that is the case, then the sum converges to
Thus, to find the sum of the infinite geometric series, we only need to know two things: r and a_{1}.a_{1} 1−r.  Looking at the series, we see that with each "step" the next term will be multiplied by an additional [4/5]. This comes from the the ([4/5] )^{i} in the series: since i increases by 1 for every subsequent term, the exponent will increase by 1, causing an additional [4/5] multiplication to occur. Thus
To figure out a_{1}, we need to see what the first term the series produces is. Don't be fooled into thinking that the first term is [4/5]we need to pay attention to what value the lowest index will create. The lowest index for the sigma notation (Σ) is i=0, so use that to find a_{1}.r = 4 5⎛
⎝4 5⎞
⎠i
⇒ i=0 ⇒ a_{1} = ⎛
⎝4 5⎞
⎠0
= 1  Now that we know the values of r and a_{1}, we plug in:
[Notice that we were only able to plug in to the above formula because r < 1. If we had r ≥ 1, then the infinite geometric series would have no sum, because it would never converge to a single value.]a_{1} 1−r= 1 1− 4 5= 1 1 5= 5

 An infinite geometric series is one where we consider what happens as we work towards adding an infinite number of terms together. Notice that we never stop adding terms. The pattern of adding more and more terms continues forever. We want to know what value the series will converge to as we add ever more terms (what value the series "settles toward" in the long run).
 From the video lesson, we learned that an infinite geometric series converges if r < 1. If that is the case, then the sum converges to
Thus, to find the sum of the infinite geometric series, we only need to know two things: r and a_{1}.a_{1} 1−r.  Looking at the terms of the series, notice that the denominator increases by a factor of 2 with each step. This helps us find r, but we also need to pay attention to the signs. We can rewrite the series to make it clearer:
Thus on every "step" the term is also multiplied by a negative. Putting this together with the denominator growing by a factor of 2 each time, we get9 + ⎛
⎝− 9 2⎞
⎠+ 9 4+ ⎛
⎝− 9 8⎞
⎠+ 9 16+ … r = − 1 2.
Finding a_{1} is quite simple for this problem, since a_{1} is just the first term. Looking at the series, we seea_{1} = 9.  Now that we know the values of r and a_{1}, we can plug in. Don't worry about the fact that r is a negative number: the formula works for positive and negative values of r, they just need to have r < 1.
[Notice that we were only able to plug in to the above formula because r < 1. If we had r ≥ 1, then the infinite geometric series would have no sum, because it would never converge to a single value.]a_{1} 1−r= 9 1− ⎛
⎝− 1 2⎞
⎠= 9 3 2= 9·2 3= 6

 Begin by understanding what the question is asking. You have a choice of two different jobs, and you want to know which one will make you more money in total over the course of 30 years. In other words, we want to know how much you'll make for each of those 30 years, then add them all up:
Since we're just adding up a bunch of terms, we're working with a sequence. Let's get a better sense of how the terms are related.⎛
⎝Year one salary ⎞
⎠+ ⎛
⎝Year two salary ⎞
⎠+ … + ⎛
⎝Year thirty salary ⎞
⎠  Consider `Job One' first. During year one, you will make $70 000. The following year, you get a 2% raise. Mathematically, we can express that as
Thus, we see that a 2% raise is equivalent to multiplying the value we previously had by (1.02). [Why? Because the raise increases the salary to 102% of what it had been.] Furthermore, we can now realize that the salary increase will have this affect at every "step": just multiply the previous term by (1.02). This allows us to rewrite each term for `Job One' in this formatJob One, year two: 70 000 + (0.02)·70 000 = (1.02) ·70 000 70 000, (1.02)·70 000, (1.02)^{2}·70 000, (1.02)^{3}·70 000, (1.02)^{4}·70 000, …
By similar logic, we see that for `Job Two' the 5% raise is equivalent to multiplying the previous year's salary by (1.05). This allows us to rewrite each term for `Job Two' as50 000, (1.05)·50 000, (1.05)^{2}·50 000, (1.05)^{3}·50 000, (1.05)^{4}·50 000, …  Notice that we can now see both sums are geometric series (this should come as no surprise, since we're working on a section devoted to that). From the video lesson, we learned that the sum of any geometric sequence is
where a_{1} is the first term, r is the common ratio between terms, and n is the total number of terms in the series. For `Job One', the first salary is a_{1,One} = 70 000 and the common ratio is r_{One}=1.02. For `Job Two', the first salary is a_{1,Two} = 50 000 and the common ratio is r_{Two} = 1.05. For both of the jobs, the number of years we're adding up is n=30.a_{1} · 1−r^{n} 1−r,  Now that we know all the necessary values, we can plug in to the formula to find each sum, then compare them:
Job One: a_{1} · 1−r^{n} 1−r= 70 000 ⎛
⎝1−(1.02)^{30} 1−1.02⎞
⎠≈ 2 839 766
Thus we see that `Job Two' manages to make more money total over the 30 year period.Job Two: a_{1} · 1−r^{n} 1−r= 50 000 ⎛
⎝1−(1.05)^{30} 1−1.05⎞
⎠≈ 3 321 942
Notice that this is a process of creating interior, equilateral triangles and alternating between shading and erasing. If we continue this process forever, what portion of the original, starting triangle will wind up being shaded in?
 We need to know what portion of the previous triangle each subsequent triangle makes up. To figure this out, consider part (2.) of the diagram. Notice that the new (unshaded) triangle must be [1/4] of the original (shaded) triangle. We can show this by symmetry: the new (unshaded) triangle must be the same size as each of the remaining three triangles (all still shaded). This means a total of four, equalsized triangles. Thus, the new (unshaded) triangle must be a quarter of the original triangle. [If you don't see how to prove the symmetry, you can also work out what the area of the new (unshaded) triangle must be since you know each of its vertices are on the midpoints of the sides of the original triangle. This is much harder though, and it's much easier to argue symmetry since the four triangles inside must all be equilateral triangles.]
 By the same logic, we can see that this will happen at every step in the process. The area covered by each subsequent triangle is [1/4] of the previous triangle's coverage.
If we say the first, original triangle had an area of 1, then we can consider the areas of the sequence of subsequent triangles:
Simplifying a bit, we see we can also express this asArea of triangles, in order: 1, 1 4, ⎛
⎝1 4· 1 4⎞
⎠, ⎛
⎝1 4· 1 4· 1 4⎞
⎠, … Area of triangles, in order: 1, ⎛
⎝1 4⎞
⎠, ⎛
⎝1 4⎞
⎠2
, ⎛
⎝1 4⎞
⎠3
, ⎛
⎝1 4⎞
⎠4
, …  Remember, our goal is to find the area of the triangle that is shaded in over the longrun of the process. Notice that the coverage of each triangle alternates in its effect: some shade in, while others erase what they cover.
To express this idea mathematically, let's consider shading area as positive, while erasing shaded area as negative.
With this in mind, we can express the effect each triangle has in terms of it being a positive addition of shaded area or it subtracting from the shaded area:
Effect on shaded area, in order: 1, − ⎛
⎝1 4⎞
⎠, ⎛
⎝1 4⎞
⎠2
, − ⎛
⎝1 4⎞
⎠3
, ⎛
⎝1 4⎞
⎠4
, …  Now that we know what each triangle does to the shaded area, we can find how much will wind up being shaded by adding them all together:
Since the process continues on forever, the sum will go on infinitely. At this point it's probably clear that we're dealing with an infinite geometric series. From the video, we learned that an infinite geometric series converges if r < 1. If that is the case, then the sum converges toShaded area: 1 − ⎛
⎝1 4⎞
⎠+ ⎛
⎝1 4⎞
⎠2
− ⎛
⎝1 4⎞
⎠3
+ ⎛
⎝1 4⎞
⎠4
− …
Thus, to find the sum of the infinite geometric series, we only need to know two things: r and a_{1}.a_{1} 1−r.  Looking at how the series works, we see that the sign alternates every time and so it is multiplied by a factor of r=−[1/4]. Furthermore, the first term in the series is a_{1} = 1, the area of the entire triangle.
Since −[1/4] < 1, the series converges and we can apply the formula:
a_{1} 1−r= 1 1−(− 1 4) = 1 5 4= 4 5
*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
Geometric Sequences & Series
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 0:48
 Form for the nth Term 2:42
 Formula for Geometric Series 5:16
 Infinite Geometric Series 11:48
 Diverges
 Converges
 Formula for Infinite Geometric Series 16:32
 Example 1 20:32
 Example 2 22:02
 Example 3 26:00
 Example 4 30:48
 Example 5 34:28
Precalculus with Limits Online Course
Transcription: Geometric Sequences & Series
Hiwelcome back to Educator.com.0000
Today, we are going to talk about geometric sequences and series.0002
The other specific kind of sequence we will look at in this course is the geometric sequence,0006
a sequence where we multiply by a constant number for each step.0011
In the previous lesson, we looked at the arithmetic sequence, which is where you add by a constant number.0013
Now, we are looking at geometric, where you multiply by a constant number every step.0018
Just like arithmetic sequences, geometric sequences commonly appear in real life.0022
Since geometric sequences are based on ratios, since we are always multiplying by the same thing,0027
and ratios occur a lot in the world, they give us a way to describe a wide variety of things.0030
In this lesson, we will begin by going over what a geometric sequence is, and how we can talk about them in general.0036
Then, we will look into formulas for geometric series to make adding up a bunch of terms really easy and fast; let's go!0041
We will start with a definition: a sequence is geometric if every term in the sequence can be given0047
by multiplying the previous term by some constant number, r.0053
a_{n} is equal to r times a_{n  1}; that is, some term is equal to the previous term, multiplied by r.0058
We call r the common ratio, because we can express it as a_{n} divided by a_{n  1}; that is, some term divided by the previous term.0073
And so, we have a ratio in the way that we are building it.0084
Here are two examples of geometric sequences: 3, 6, 12, 24...continuing on; 4/5, 4/25, 4/125, 4/625...continuing on.0087
They are geometric, because each step multiplies by the same number.0099
For example, in this one, every step we go forward, we are multiplying by 2: 3 to 6times 2; 6 to 12times 2; 12 to 24times 2.0103
And this is going to continue on forever, as long as we keep going with that sequence.0112
Over here, 4/5 and 4/25...it is not quite as simple, but it is basically the same thing, multiplying by 1/5.0116
That is how we get from 4/5 to 4/25, if we are going to multiply.0124
To get to 4/125, once again, we multiply by 1/5; to get from 4/125 to 4/625, we multiply by 1/5.0128
And this is going to keep going, every time we keep stepping.0139
Every step is multiplying by the same number: we are multiplying by the same number each step.0143
The number can be anything: it can positive; it can be larger than one; it can be less than one; it can be negative.0148
It doesn't matter, so long as it is always the same value for every step.0153
The definition of a geometric sequence is based on the recursive relation a_{n} = r(a_{n  1}).0159
That is, every term is equal to the previous term, multiplied by r.0165
How can we turn this into a formula for the general term, where we don't have to know what the previous term is0170
we can just say, "I want to know the n^{th} term," plug it into a formula, and out will come the n^{th} term?0173
Remember: a recursive relation needs an initial term.0179
So, while this relation that defines a geometric sequence is useful, we still need a little bit more.0181
We need this initial term to know where we startwhat our very first term isbecause previous to that...there is nothing previous.0187
So, we just have to state that as one specific term.0193
Since we don't know its value yet, we will just leave it as a_{1}, our first term.0197
From a_{n} = r(a_{n  1}), we see that a_{1} relates to later terms as: a_{2} will be equal to r(a_{1}).0202
The second term will be equal to the first term, times r; this will continue on.0210
The third term, a_{3}, will be equal to r times a_{2}.0215
But we just showed that a_{2} is equal to r(a_{1}), so we can plug that in there;0218
and we will have r times r(a_{1}), so we end up getting r^{2}(a_{1}) = a_{3}.0224
We can continue on here; a_{4} is going to be equal to r times a_{3}.0233
But we just figured out that a_{3} is equal to r^{2} times a_{1}.0237
So, we replace a_{3}, and we end up having r times r^{2} times a_{1};0241
so now we have that r^{3}(a_{1}) = a_{4}.0246
And we will see that this pattern will just keep going like this.0251
So, we have that a_{1} is equal to a_{1} (there is no big surprise there).0254
a_{2} is equal to r times a_{1}; a_{3} is equal to r^{2} times a_{1}.0258
a_{4} is equal to r^{3} times a_{1}; and we see that the pattern is just going to keep going like this.0263
The n^{th} term is n  1 steps away from a_{1}; it is n  1 steps to get from a_{1} to n.0268
If you start on the first steppingstone, and you go to the n^{th} steppingstone, you have to take n  1 steps forward.0277
We start at a_{1}; to get to the a_{n}, we have to go n  1 steps forward.0283
Since every step means multiplying by r, that means we have multiplied by r n  1 times, which is r raised to the n  1 power.0287
Thus, we have that a_{n} = r^{n  1}(a_{1}).0296
So, to find the formula for the general term of a geometric sequence, we only need to figure out what the first term is, and the common ratio.0301
As soon as we figure out a_{1} and our value for r, we have figured out what the general term is, what the a_{n} term ispretty great.0308
What if we want to find the n^{th} partial sum of a geometric sequence (that is, adding up0316
the first n terms of the sequence, a_{1} + a_{2} + a_{3} up until + a_{n})?0320
Well, we could just add it all up by hand for small values of n.0327
If it was n = 2, so it was just a_{1} and then a_{2}, it is probably not that hard to just figure it out by hand.0331
If it was n = 3, we could probably do it by hand; if it was n = 10, n = 100, n = 1000,0336
this gets really, really tiresome, really quickly, as the value of n gets larger and larger.0341
So, how can we create a formulahow can we just have some formula where we can plug some stuff in,0345
and we will immediately know what that n^{th} partial sum is?0350
Well, let's do two things: first, let's give the sum a name.0355
So, we will call our n^{th} partial sum s_{n}, the sum for the n^{th} partial sum.0358
Second, let's use the form for the n^{th} term of a geometric sequence.0364
Remember: we just figured out that the general form for a_{n} is r^{n  1}(a_{1}).0367
We can use this general term to put of the terms in this series into a format that will involve a_{1}.0376
So, we have s_{n} = a_{1} + a_{2} (which is r times a_{1}) + a_{3}0382
(which is now r^{2} times a_{1}), up until we get to + a_{n} (which is now r^{n  1}(a_{1})).0390
Great; but at the moment, we can't do anything with just this.0399
This isn't quite enough information; we can't combine the various r's, because they all have different exponents.0402
r to the 0, what is effectively here; r^{1}; r^{2}; r^{3}; up until r^{n  1}0410
they don't talk the same language, because they don't have the same exponent.0417
So, since they can't really communicate with each other, we can't pull them out all at once.0420
We could pull out all of the a_{1}'s, but then we would still be left with all of these different kinds of r's.0423
So, we don't really have a good thing that we can do right now.0428
What we want, what we are really looking for, is a way to somehow get rid of having so many things to add up.0431
We want fewer things to add up; if we only had a few things to add up and compute, it would be easy for us to calculate these values.0439
So, that is what we want to figure out how to do.0445
This is the really clever part; this is basically the part of the magic trick where suddenly we pull the rabbit out of the hat.0449
And so, you might see this and think, "How would I figure this out?"0455
And it is a little bit confusing at first; but just like, as you study magic more and more (if you were to study magic),0458
you would eventually realize, "Oh, that is how they got the rabbit into the hat," or "this is how the trick works"0463
as you work with math more and more, you will be able to see, "Oh, that is how we can make these sorts of things."0468
So, don't be worried by the fact that you would not be able to think of this immediately.0472
The people who made this proof at first didn't think of it immediately.0477
They thought about it for a while; they figured out different things, and maybe they tried something that did not work;0481
and eventually they stumbled on something and said, "Oh, if I do this, it works,"0486
and they were able to come up with this really easy, cool, clever way to do it.0489
But it is not something that you just have immediately.0492
It is something that you have to think about, until eventually you can "pull your own rabbit out of a hat."0494
But first, you have to get the rabbit into the hat.0498
Anyway, here is the clever part: what is it?0500
We have s_{n} = a_{1} + r(a_{1}) + r^{2}(a_{1}) + ... + r^{n  1} (a_{1}).0503
The really cool trick that we do is say, "What if we multiplied r times s_{n}?"0515
Well, that would end up distributing to everything in here, since we are multiplying it on both sides of the equation.0522
We would have r(a_{1}) + r^{2}(a_{1}) + r^{3}(a_{1})...0528
all the way until we get up to + r^{n}(a_{1}).0534
Now, notice: we now have matching stuff going on.0537
Here is r times a_{1}; here is r times a_{1}; there is a connection here.0540
Here is r^{2} times a_{1}; here is r^{2} times a_{1}; there is a connection there.0544
Here is r^{3} times a_{1}; and somewhere in the next spot in the decimals is r^{3} times a_{1}, as well.0548
Here is r^{n  1} times a_{1}; and in the next back spot in the decimals, here is r^{n  1} times a_{1}.0555
We have all of this matching going on; well, with this idea, since we have all of this matching,0560
we can subtract this equation, the rs_{n} equation, this one right here,0565
from the s_{n} equation by using elimination.0569
From when we talked about systems of linear equations: if we have two equations, we can subtract, and we can add them together.0572
We are just using elimination to do this.0578
So, we have our s_{n} equation here, and then we subtract by minus r times s_{n}.0580
So now, we have this matching pattern going on: r times a_{1} matches to r times a_{1}; they cancel each other out.0586
r^{2} times a_{1} matches to minus r^{2} times a_{1}; they cancel each other out.0592
Everything in the dots here cancels out with all of the negatives in the dots here.0596
We finally get to r^{n  1} times a_{1} in our top equation,0600
which cancels out with minus r^{n  1} times a_{1} in our bottom equation.0603
The only things that end up being left over are minus r^{n} times a_{1} and +a_{1} here.0608
We get s_{n} rs_{n} is equal to a_{1}  r^{n}(a_{1}).0617
So, through this immense cleverness (and once again, this isn't something that you would be expected to just know immediately,0624
and be able to figure out really easilythis is the part that takes the really long thinking.0629
This is the really clever part; this is what takes hours of thought, by just thinking, "I wonder if there is a clever way to do this."0632
And eventually, you end up stumbling on it.0637
So, through immense cleverness, we have shown that s_{n}  rs_{n} = a_{1}  r^{n}(a_{1}).0640
That is pretty great: we have gotten this from what we just figured out.0648
Now, our original goal was to find the value of the n^{th} partial sum, which was s_{n}.0653
Using the above, we can now solve for s_{n}: we just pull out the s_{n},0658
so we have s_{n} times 1 minus r; we can also pull out the a_{1} over here on the right side.0663
We divide both sides by 1  r, and we get that the n^{th} partial sum is equal to0667
the first term, a_{1}, times the fraction 1  r^{n} over 1  r.0672
So, we now have a formula to find the value of any finite geometric series at all, really easily.0678
All that we need to know is the first term, a_{1}; the common ratio, r,0683
which shows up on the top and the bottom; and how many terms are being added together total (the n exponent on our top ratio).0688
It is pretty greata really, really powerful formula that lets us do a lot of what would be very tedious, very slow, difficult addition, just like that.0696
But what if the series was not finite?0705
So far, we have only talked about if we are doing a finite sum of the sequence.0707
But what if we had an infinite sequence, and we wanted to add up infinitely many of the terms, so we kept adding terms forever and ever and ever?0711
To understand this better, let's consider some geometric sequences and what happens as we take partial sums using more and more and more terms.0718
First, we will look at 3, 9, 27, 81, 243...so it is times 3 each time; it is a geometric sequence.0725
So, our first partial sum, s_{1}, would be just the 3.0732
Our next partial sum would be s_{2}, so we add on the 9, as well; we get 12.0737
Our next partial sum would be s_{3}; we add on the 27, as well; we get 39.0741
The next partial sum is s_{4}; we add on the 81; we get 120.0745
The next partial sum: we add on 243; we get 363; and this is just going to keep going on in this pattern.0749
It is going to keep adding more and more and more.0754
We look at this, and we notice that, as the partial sums use more and more terms, it continues to grow at this really fast rate.0756
In fact, the rate is going to get faster and faster and faster as we add more and more terms.0763
We see its rate of growth increasing as it goes to larger sums.0766
So, if we add terms to the series forever, it is not going to really get to anything.0770
It is going to blow out to infinity; it is going to just "blast off" to infinity.0774
There is no stopping this thing; it is not going to give us a single value.0778
It never stops growing; we say that such an infinite series, one that never stops growing, that doesn't go to a single value, "diverges."0781
As we add more and more terms, it continues to change forever.0789
It diverges from giving us a single, nice, clean value, because it instead just blows off to infinity.0792
It keeps moving around on us; it doesn't stay still; it doesn't go to something; it just goes off, so this would be a divergent series.0798
On the other hand, we could consider another partial sum from this below geometric sequence: 1, 1/2, 1/4, 1/8, 1/16...0807
What we are doing each time here is dividing by 2, or multiplying by 1/2; so we see that this is a geometric sequence.0815
Our first partial sum would be s_{1} = 1; we just add in that first term.0821
The next one, s_{2}, would be 1.5, because we added 1/2, so we are at 1.5.0826
The next one, s_{3}: we add in 1/4; that is 0.25 added in; that becomes 1.75.0831
The next one, s_{4}: we add in an 8; that becomes 1.875.0836
The next one, s_{5}: we add in 1/16; that becomes 1.9375.0840
And it would continue in this way; but we notice that it is not really growing the same way.0844
This time, the sums are continuing to grow, but the rate of growth is slowing down with each step.0849
It is not increasing out like the previous one (it was blowing out somewhere; it was becoming really, really big).0856
But with this one, we see it settling down; as we add more and more terms, it is going to a specific value.0861
It is going to 2; the infinite sum, this infinite series, is going to a very specific value; it is working its way towards 2.0868
If you keep adding more, you will see even more, as it gets to 1.99, 1.999, 1.9999999...0876
As you keep adding more and more terms, you will see that it is really just working its way to a single value.0883
It is slowing down as it gets to 2.0887
In this case, we say that such an infinite series converges; it is converging on a specific value.0889
As we add more and more terms, it works its way towards a single value.0895
There is this single value that it is working towards.0899
From the two examples we have seen, we see that whether a series converges or diverges is based on the common ratio of its underlying sequence.0902
If the common ratio is large, it causes the sequence to always grow.0911
It keeps growing, because that ratio keeps multiplying it to get larger and larger and larger, moving around.0917
Specifically, if the absolute value of r is greater than or equal to 1, the partial sums will always be changing,0922
because the size of our terms never shrinks down; so the series will diverge.0927
On the other hand, if the common ratio is small, it causes the sequence to shrink down.0933
If we have a small common ratio, it is going to make it smaller and smaller and smaller with every term we work on.0937
Since it gets smaller and smaller and smaller, we have that if the absolute value of r is less than 1,0944
the rate of change for the series will slowly disappear to nothing, because every time we go to the next term,0948
since r is less than 1, it makes it smaller; and then it makes it smaller; and then it makes it smaller, and makes it smaller,0954
and makes it smaller, and makes it smaller; so every time it is getting smaller.0959
So, every time, the rate of growth is going down to less and less and less.0962
And so, over the long term, over that infinite number of terms, it ends up converging to a single value.0965
So, as a general rule, an infinite geometric series will converge if and only if the absolute value of r is less than 1.0971
So, the absolute value of r being less than 1 means that the infinite geometric series converges.0979
If the infinite geometric series converges, then the absolute value of r must be less than 1; they are equivalent things for a geometric sequence.0983
All right: assuming that the absolute value of r is less than 1 for a geometric sequence, how can we figure out a formula for its corresponding infinite geometric series?0990
Well, we already figured out a formula that is true for any finite geometric series.0999
Remember: s_{n}, the n^{th} partial sum of any finite geometric series,1003
is a_{1} times (1  r^{n})/(1  r).1007
We just figured out that formula; that is pretty cool.1011
Not only that, but we also know that, as we look at partial sums containing more and more terms,1013
as a partial sum has more and more terms, they have to be growing closer and closer to the value that the infinite series will converge to.1018
As we put in more and more terms into our partial sum, it has to be getting closer to this value that it is going to converge to.1027
Think about why that is: if the partial sums were not getting closer to a specific value,1032
if they were moving around away from the specific value,1036
then it couldn't be converging to that, because it would always be changing around.1039
If it is going to converge to a single value, it has to be working its way towards it.1042
If it is working its way towards it, it must always be getting closer to the thing.1046
If it wasn't always getting closer, if it was sometimes jumping away, it wouldn't be working its way towards it; it would be going somewhere else.1049
Since we know that it is converging, we know that it must be working its way, as we have more and more terms in our partial sum.1054
As we put in more terms in our partial sum, we will be closer to the value that we are converging on.1065
As we add many more and more and more terms in our partial sum, we are going to be closer to the thing that we are converging to.1069
What we are asking ourselves is, "As we have really large values for n, what value are we getting close to?"1076
What happens to the formula we figured out, our n^{th} partial sum formula,1082
s_{n} = a_{1} times (1  r^{n})/(1  r), as the number of terms we have, our n, goes off to infinity?1086
As the value for the number of terms we have, our n, becomes infinitely large, what will happen to this formula?1093
Whatever happens to this formula is what we have to be converging to, because of the argument1098
that we just talked about, about how it has to be getting closer as we put in more and more terms.1102
Notice: the only term on the right side affected directly by the n is r^{n}; there is no other term that directly has an n connected to it.1107
We can ask ourselves what happens to r^{n} as our n grows very large.1115
Also, remember: r is less than 1; the absolute value of r has to be less than 1.1119
These two things combine as we ask ourselves...as n goes to infinity, and we are looking at our r^{n} here,1125
since the absolute value of r is less than 1, we have that as n goes to infinity, r^{n} has to go to 0.1133
So, it is going to shrink down to 0 as n grows infinitely large.1140
Why is that the case? Well, since the absolute value of r is less than 1, every step has to make it smaller.1144
For example, if we look at .9 raised to the 100, we get that that is less than .0001.1150
Why is this occurring? Well, since the absolute value of r is less than 1, we know that it is this fractional thing1157
that effectively, every time we iterate it, every time we hit a term with this common ratio, it takes a little bite out of it.1164
Whether it is .1 or 1/2 or 3/5 or 922/1000, it is going to take a bite out of whatever the term that it is being multiplied against is.1170
As it takes infinitely many bites, since it is always shrinking it down, it means that it is always working its way towards this value of 0.1181
Infinitely many bites away gets us to having nothing.1189
Therefore, because r^{n} is going to 0 as n goes to infinity, we have this part right here shrinking down to a 0 in our formula.1194
So, we get the following formula for an infinite geometric series: the infinite sum is equal to a_{1} times 1/(1  r),1202
assuming that the absolute value of r is less than 1.1210
If the absolute value of r is greater than or equal to 1, we couldn't even talk about this in the first place,1212
because our series would be diverging, because it would always be growing and changing around on us.1217
But if we have that the absolute value of r is less than 1, all we need to know is our first term, a_{1}, and the rate that it is growing at.1220
And we work it out through this formula, and we know what it will converge to over the long run.1228
All right, coolwe are ready for some examples.1232
The first one: Show that the sequence below is geometric; then give a formula for the general term (that is, the a_{n}, the n^{th} term).1234
We have 7, 35, 175, 875...so what we want to ask ourselves is, "What number are we multiplying by each time?"1241
How do we get from 7 to 35? Well, we multiply by 5.1247
Let's check and make sure that that works: 35 to 175yes, if we use a calculator (or do it in our heads,1250
or write it out by hand), we realize that, yes, we can multiply by 5 to get from there to there.1256
The same thing: 175 to 875: we multiply by 5; so we see that this is continuing.1259
Yes, it is geometric; that checks out.1263
Now, we want to give a formula for the general term.1266
We talked, in the lesson, about how a_{n} is equal to (let me write it the way we had it last time):1268
r, the rate that we are increasing at, to the n  1, times a_{1}.1276
What is our a_{1}? Well, a_{1} is equal to 7, because it is the first term.1281
What is our r? r is equal to 5, since it is the number we are multiplying each time.1286
So, r = 5; r = 5; a_{1} = 7; so a_{n} is equal to 5^{n  1} times 7.1290
And if we wanted to check this out, we could do a really quick check.1299
We could plug in...let's look at a_{2}; that would be equal to 5^{2  1} times 7,1303
so 5 to the 1 times 7; 5 times 7 is 35; we check that against what our second term was.1310
And indeed, that checks out; so it looks like we have our answerthere is our answer.1316
All right, the next example: Find the value of the finite geometric series below.1320
Notice: this does have an endwe stop at 3072, so it is not an infinite one.1326
If it were infinite, it would go out to infinity, so we wouldn't actually be able to find a value.1330
All right, how do we figure this out?1333
The first thing: what is the rate that we are increasing at?1336
To get from 3 to 6, we multiply by 2; to get from 6 to 12, we multiply by 2; so at this point, we realize that r equals 2.1338
What is the value a_{1}? That is 3.1345
What we are looking for, remember: the formula we are going to do is: the n^{th} partial sum, s_{n},1348
is equal to a_{1} times 1 minus the rate, raised to the n^{th} power, divided by 1 minus the rate.1353
So, the only thing we have to figure out, that is left, is what our n is; n = ?.1359
How many value are we going to be at?1365
Here we are at a_{1}, but this is a_{?}; this is a_{n}, right over here.1367
What would that have to be? Well, we could set this up, using the formula that we talked about before, our general formula for the general term.1374
a_{n} is equal to r^{n  1} times a_{1}.1382
We plug in our a_{n}; that is 3072; 3072 =...our rate is 2, raised to the n  1, times...a_{1} is 3.1388
So, we divide both sides by 3, because we are looking to figure out our n.1398
Divide both sides by 3, and we get 1024 = 2^{n  1}.1401
Now, you might just know immediately that 1024 is the tenth power of 2: 2^{10} = 1024.1407
So, we would see that it is 10 steps to get forward; we would multiply 10 steps forward, so that would mean that our n is equal to 11.1415
We have figured out that to get 2^{10}...that is 10 steps; we multiplied all of them on the 3; we started at 31424
as our first steppingstone; we stepped forward 10 times...so the first steppingstone, plus 10 steps forward,1430
means a total of 11 steppingstones; so we have n = 11.1435
However, alternatively, we could just figure this equation right here out by using the work that we did with logarithms long ago in this class.1439
1024 = 2^{n  1}...well, what we can do is just take the log of both sides: log(1024) = log(2^{n  1}).1448
One of the properties of logs is that we can pull down exponents; that is why this is so useful.1459
We have n  1 times log(2); log(1024) over log(2) equals n  1.1463
log(1024)/log(2) is equal to 10; it equals n  1, which tells us that n equals 11.1475
So, you could work this out just through raw algebra and using logarithms;1486
or you could work it out, if you recognized 2^{10} as equal to 1024, if you just kept dividing 1024 with a calculator1489
until you saw how many steps it was; either way will end up getting us this value, that n equals 11.1495
All right, great; now we have everything that we need for our sum: s_{n}, the n^{th} partial sum1503
(in this case, the 11^{th} partial sum of what this sequence would be) is going to be equal to...1509
a_{1} is 3 (our first term), times 1 minus our rate (it multiplied by 2 on each one of them),1517
raised to the 11^{th} power (because the number of terms we have total is 11), divided by 1 minus the rate (2, once again).1525
We work this out: 3 times 1  2^{11} is 2047; 1  2 is 1, so the 1 cancels out with the negative on top.1534
We have positive 2047 times 3; and we end up getting 6141; that is what we get once we add up all of those terms.1545
Great; the third example: Find the value of the below sum.1554
The sum is in sigma notation: from i = 0 up until 10 of 500 times 1/2 raised to the i minus 3 to the i divided by 64.1558
Previously, when we talked about sigma notation, series notation, we talked about how summations have properties1569
where we can separate some of the things in sigma notation, and we can pull out constants; great.1575
So, let's start by separating this: we can write this as Σ...it still has the same upper limit; the limits will not change...1580
the same index and lower limit of 500, times 1/2 to the i...minus...so what we are doing is separating around this subtraction...1587
minus...the series...same limits...of 3 to the i over 64.1599
So, we can separate, based on addition and subtraction, into two separate series.1607
Furthermore, we can also pull out constants and just multiply the whole thing.1610
We have 500 times the series; 10i equals 0, 1/2 to the i, minus...we pull out the 1/64, since that is just a constant, as well...1613
on the series, 10i equals 0; 3 to the i...cool.1627
So, at this point, we can now use our formulas.1632
Remember: our formula was that the n^{th} sum is equal to the first term, times 1 minus r^{n}, over 1  r.1634
So, each one of these is going to end up having different rates.1643
But they are going to end up having the same n.1646
Our n is going to be based off of going from 0, our starting index, all the way up to 10, our ending index.1648
What is our n if we go from 0 to 10?1656
Remember: we have to count that first step, so 0 up until 10 isn't 10; it is 11.1659
1 to 10 is 10, so 0 to 10 must be one more, 11; so we have n = 11.1664
All right, back into this: we have 500 times...let's use that formula...the series from i = 0 to 10 of 1/2 to the i...1670
well, what would it be for our a_{1}what would be the first term?1678
Well, if we plugged in i = 0 on (1/2)^{i}, (1/2)^{0}, or anything raised to the 0, is just 1; so our a_{1} is 1 there.1681
Times 1 minus...what is the rate? Well, we are multiplying by 1/2 each time, because it is 1/2 with an exponent on it.1690
So, 1/2 is our rate, raised to the n; our n that we figured out was 11, divided by 1 minus the same rate, 1/2.1695
Great; minus...over here, 1/64: we apply that formula again; so, the series from i = 0 to 10 of 3^{i}:1705
what is our a_{1}what is the first term?1714
Well, 3 to the 0, because our starting index is 0 once againthe first term of this series would be 3 to the 0, because it is the first thing that would show up.1716
3 to the 0 is just 1 again; 1 minus...what is our rate? Our rate is 3, because we multiply by 3,1726
successively, for each iteration, because it is an exponent.1732
3 to the...what is our number of terms? 11; n = 11...one minus three.1735
Great; now we can work through a calculator to figure out what these things are.1741
We will save the incredibly boring actual doing of the arithmetic, but it shouldn't be too difficult.1744
This will become 500 times 2047, over 1024, minus 1 over 64, times 88573.1749
We can combine these things; actually, let's distribute our constants first.1765
2047...well, we will distribute just the multiplication; they are all constant numbers now.1769
255 thousand on our left, 875, divided by 256, minus 88573 over 64;1773
we have to put the right side on a common denominator, but once again, I will spare working out every single aspect.1793
This would simplify to 98417 divided by 256; and there is our answer, exactly precise.1799
If we wanted to, we could have approximated to decimals at some point, and we would get 384.44, approximately.1810
Notice: probably back here, when we were working at these steps, where we have these giant fractions,1819
there is a good chance, if you are working with a scientific calculator or a graphing calculator, that it doesn't show it perfectly in fractions.1824
You would end up getting fractional numbers; and for the most part, most teachers and textbooks would be fine with giving an answer in decimals instead.1829
So, either one of these two answersthe specific exact fraction, 98417/256, or approximately 384.44both of them are perfectly fine.1835
The fourth example: Many lessons ago, when we first learned about exponential functions,1847
we had a story where a clever mathematician asked for one grain of rice on the first square of a chessboard,1851
then two on the second square, four on the third square, eight on the fourth square, 16, 32, 64, 128...1857
he keeps asking for doubling amounts of grains of rice.1872
So, if he had actually been paid on all of the squares, just as he was initially promised, how many grains of rice would he have total?1876
So, what we have here is a finite geometric series, because it is doubling each time.1885
What is our first term? Well, a_{1} was on the very first square; he had one grain, so our first term is going to be 1.1891
What is the rate that it increases at each time?1900
It goes 1, 2, 4, 8, 16...so it is doubling each time; so that means that the rate is multiplied by 2.1902
And what is the number of terms total? If we are going from the first square of a chessboard to...1908
a chessboard is 8 x 8, so the total squares on an 8 x 8 chessboard...we have 8 by 8, so that means we have 64 squares total.1913
So, that is going to be 64 times that this is going to happen; we have 64 terms here.1926
If we are going to figure this out, the n^{th} partial sum, the 64^{th} partial sum,1933
is going to be the first term, a_{1}, times 1 minus the rate, 2, raised to the number of terms, 64, over 1 minus the rate, 2.1937
We work this out: we end up getting 1  2^{64} over 1, which is the same thing as negative 1  2^{64},1948
which we will end up seeing comes out to be very positive.1962
This is our precise answer for the number of grains; but let's see what that is approximately.1964
That comes out to be approximately 1.845x10^{19} grains of rice.1968
That is a big, incredibly huge number of grains of rice.1978
It might be a little bit hard to see just how many grains of rice that is: 10^{19}we are not used to working with scientific notation that large.1982
So, let's turn it into some words that we might understand.1989
That would be about the same as 18 quintillion (that is a quintillion: it goes million, billion, trillion, quadrillion, quintillion) grains of rice.1991
And we are, once again, not really used to working with numbers as incredibly large as the quintillion scale.2003
So, let's write that as 18 billion billion grains of rice; that is an incredible number of grains of ricethat is so many grains of rice!2008
How much rice is that? That is about more than the rice that has ever been produced by humanity, over all of humanity's time existing!2022
It might be approximately on the same scale as the amount of rice humanity has ever created.2030
I am talking about all of humanity ever creating this over the whole history of the world.2036
All of the rice that has ever been made...18 billion billion grains...that is probably somewhere on the same scale.2041
It might be a little bit more; it might be a little bit less; but that is the sort of scale.2046
All of the rice that humanity has ever created: we get to very, very large numbers very quickly2049
with these exponential functions when we are working in geometric stuff; that is something to keep in mind.2054
18 billion billion grains is as much rice as humanity has ever madepretty amazing.2058
The final example: a super ball is dropped from a height of 3 meters.2065
On every bounce, it bounces 4/5 of the previous height.2068
If allowed to bounce forever, what is the total backandforth distance that the ball travels over?2071
Notice that it is total, so it is up and down.2076
Let's draw a picture to help ourselves see what is going on here.2079
We start...we have this ball; we drop the ball, and it falls three meters.2081
Then, it bounces; it is a super ball, so it bounces up, and it will go up by how much?2087
It will go up by 3 times 4/5, because it goes up to 4/5 of its previous height on every bounce.2092
Now, it is going to fall down; well, what height did it just fall down from?2099
It is going to fall down from the same thing, 3 times 4/5.2102
Then, it is going to bounce up again; it is going to bounce up by how much?2106
Well, it is going to bounce up to 3 times 4/5 times 4/5, because it is 4/5 of the previous height it came from.2108
3 times 4/5 times 4/5...well, we could just write that as 3 times (4/5)^{2}.2113
Then, it is going to fall down, once again, from that same height; so it is 3 times (4/5)^{2}.2118
And then, it is going to just continue on in this manner of going a little bit less each time:2124
3 times (4/5)^{3}, and then down again: 3 times (4/5)^{3}, and just continuing on in this manner forever.2129
We want to figure out the total backandforth distance that the ball travels over.2140
Notice: what we have here is ups and downs.2144
So, I am going to go with calculating the ups first: how much is from this up plus this up plus this up, going on forever and ever and ever and ever?2147
Notice: what that means we are dealing with is an infinite sum, because we are saying,2158
"If it were to continue bouncing forever, what would be the amount of distance it would travel over?"2162
That is equal to the first value, times 1 over 1 minus the rate; that was what we figured out that the infinite sum comes out to be.2167
It is the first value, times 1 over 1 minus the rate.2176
All right, in this case, for our up bounces, all of the ups are going to end up being...our first value2179
is a_{1}, so our first value is this one right here, 3 times 4/5.2194
We might be tempted to think that it is 3 meters, but remember: this is all of the upswe are looking at the ups first.2198
And we will see why at the endwhy I chose to look at the ups first.2203
So, all ups is going to be...3 times 4/5 is our first value that we end up having occur.2207
And then, that is going to be times 1 over 1 minus...what is the rate? 4/5, because it is 4/5 on every bounce.2213
It goes up to 4/5 of its previous height, so the rate for the next one would be 4/5 of that, and 4/5 of that, and 4/5 of that.2221
Notice: there was one other requirement on being able to use this.2227
The absolute value of our rate must be less than 1.2230
But since it is 4/5, that ends up checking out; OK.2233
Keep going with this: that is 3 times 4/5; let's write it as 12/5; 12/5 times 1 over...1  4/5 is 1/5;2237
well, 1 divided by 1/5 is 12/5 times...1 over 1/5 is 5; so we have 5 times 1/5 on the bottom;2251
that cancels out; so we have 12 meters total of bouncing for our up values.2259
OK, but that is only part of it; now we also have to figure out what all of the down values are2267
what is the down value here? the down value here? the down value here?...going out this way infinitely.2273
Well, we also have this part right here; but notice: all of these green things end up having a matching up value.2279
They each match up: the purple and the green, the purple ups and the green downs; they all match up.2292
The only one who is out of the normal case of matching up is that red first drop down.2297
So, what that means is that we can match all of our downs; they have a matching 12 meters,2303
because it matches all of the ups; but then we just have to add on the initial 3meter drop at the beginning.2311
So, how many downs do we have in total?2319
Well, that 3meter drop, plus the matched value (because it matches to the 12meter value for all of the ups),2321
12 meters, plus 3: we get 15 meters, so the total number is going to be equal to the 12 from our ups,2327
plus the 15 from our downs; 12 + 15 means 27 meters in total.2345
Great; there we arewe finished that one.2354
All right, now we have a pretty good understanding of how sequences work.2355
We have worked through geometric sequences and arithmetic sequences.2358
We understand how series work; we have a pretty good idea of how sequences work; great.2361
All right, we will see you at Educator.com latergoodbye!2365
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