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

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

### 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:

We call r thea _{n}= r·a_{n−1}.*common ratio*(since r = [(a_{n})/(a_{n−1})]). Every "step" in the sequence multiplies by the same number. The number can be anything, so long as it is always the same for each step. - The formula for the n
^{th}term (general term) of a geometric sequence isa _{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) isS _{n}= a_{1}·1−r ^{n}1−r

. - 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, thenS _{∞}= a_{1}·1 1−r

.

### Geometric Sequences & Series

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 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

### Math Analysis Online

### Transcription: Geometric Sequences & Series

*Hi--welcome 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 given*0047

*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 6--times 2; 6 to 12--times 2; 12 to 24--times 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 is--*0170

*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 start--what our very first term is--because 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 stepping-stone, and you go to the n ^{th} stepping-stone, 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 is--pretty great.*0308

*What if we want to find the n ^{th} partial sum of a geometric sequence (that is, adding up*0316

*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 formula--how 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 easily--this 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 to*0667

*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 great--a 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 argument*1098

*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 thing--*1157

*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, cool--we 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 175--yes, 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 answer--there is our answer.*1316

*All right, the next example: Find the value of the finite geometric series below.*1320

*Notice: this does have an end--we 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 3*1424

*as our first stepping-stone; we stepped forward 10 times...so the first stepping-stone, plus 10 steps forward,*1430

*means a total of 11 stepping-stones; 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 calculator*1489

*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 sum*1503

*(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 properties*1569

*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 again--the 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 answers--the specific exact fraction, -98417/256, or approximately -384.44--both 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 rice--that 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 quickly*2049

*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 made--pretty 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 back-and-forth 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 back-and-forth 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 value*2179

*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 ups--we are looking at the ups first.*2198

*And we will see why at the end--why 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 are--*2267

*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 3-meter drop at the beginning.*2311

*So, how many downs do we have in total?*2319

*Well, that 3-meter drop, plus the matched value (because it matches to the 12-meter 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 are--we 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 later--goodbye!*2365

## Start Learning Now

Our free lessons will get you started (Adobe Flash

Sign up for Educator.com^{®}required).Get immediate access to our entire library.

## Membership Overview

Unlimited access to our entire library of courses.Learn at your own pace... anytime, anywhere!