16 - The Geometric Series - Definition, Meaning & Examples - Part 1 - By Math and Science
Transcript
00:00 | Hello . Welcome back here . We're going to conquer | |
00:02 | the topic of the geometric series . Now of course | |
00:04 | we've already talked about the geometric series in several lessons | |
00:07 | but we never calculated the some of the terms of | |
00:11 | the geometric series . So just like in the last | |
00:13 | lesson we talked about the arithmetic series , we introduce | |
00:16 | the concept of the arithmetic series . We talked about | |
00:18 | the some , we wrote the equation down and I | |
00:20 | proved it to you and then we solve problems . | |
00:22 | We're gonna do the exact same thing for the geometric | |
00:24 | series . I'm gonna introduce it . I'm gonna motivate | |
00:26 | it . I'm gonna write down the equation for the | |
00:28 | some of the terms of the geometric series . Then | |
00:31 | I'm going to prove it to you mathematically . Then | |
00:33 | we'll solve some problems . So let's talk about the | |
00:35 | simplest geometric series that I can think of uh to | |
00:39 | motivate the point here . So this is a geometric | |
00:41 | series . How about one plus two plus four plus | |
00:46 | eight plus 16 plus 32 Plus 64 . So I'm | |
00:52 | just trying to pick something that's easy to understand with | |
00:55 | a finite numbers . How many terms do we have | |
00:57 | ? 1234567 terms . So there's seven terms . Uh | |
01:01 | and how do we know ? It's geometric geometric . | |
01:03 | Just means that to get the next term you take | |
01:05 | the previous term , you multiply by the same number | |
01:08 | by a common ratio . So to go from this | |
01:10 | term to this term , we actually multiply by two | |
01:13 | to go from this term to this term . We | |
01:15 | multiply by two to go from this term to this | |
01:17 | term , multiplied by two . This term to this | |
01:19 | term multiplied by two . And I can keep writing | |
01:21 | it . You can see multiplied by two , multiplied | |
01:23 | by two . So this is a geometric series and | |
01:26 | the common ratio R is equal to two . This | |
01:28 | is the common ratio . All right . So , | |
01:34 | what I want to do is I'm gonna write down | |
01:36 | the some of these terms . And when we've written | |
01:38 | these things down , we've done the summation convention , | |
01:40 | but we haven't actually calculated the sum of any of | |
01:44 | these geometric series like this . So let's go and | |
01:46 | write that down . The sun of the geometric series | |
01:57 | is as follows . The partial sum , which means | |
02:01 | we sum the first in terms is equal to the | |
02:04 | first term multiply by one minus R to the power | |
02:07 | of n . All divided by one minus r . | |
02:12 | Okay , that's what I want you to remember now | |
02:14 | . Is it obvious that this is the case ? | |
02:15 | No , but we're going to prove it in just | |
02:17 | a second . Now , the one thing I want | |
02:19 | you to uh to point out to you here is | |
02:21 | that the only thing you need to calculate the sum | |
02:24 | of these terms . I mean , I know you | |
02:25 | can add them in your calculator . What I'm saying | |
02:26 | ? What if you had 1000 terms that would take | |
02:27 | forever ? How do you calculate the sum ? Well | |
02:30 | you put the number of terms in place . You | |
02:32 | have to go in , you have to know the | |
02:34 | common ratio are , that's what you're multiplying by to | |
02:36 | get these terms are is down here as well and | |
02:39 | you have to know the first term . So you | |
02:40 | actually don't need to know the last term in order | |
02:43 | to do this . Remember for the arithmetic series we | |
02:45 | needed the first term , We needed the last term | |
02:48 | and we need the number of terms here . We | |
02:50 | do not need to know the last term . We | |
02:52 | just need to know what is the common ratio . | |
02:54 | What is the number of terms ? What is the | |
02:56 | first term ? Now , notice in the denominator we | |
03:00 | have one minus R . And because we have that | |
03:02 | in the denominator , what we really need to stay | |
03:04 | out here is that this equation holds for the sum | |
03:07 | . However are cannot be equal to one . Why | |
03:10 | ? Because if I put a one in for the | |
03:12 | common ratio one minus one is zero and we're dividing | |
03:15 | by zero , that's really undefined . Or you could | |
03:17 | say that it goes off to infinity . So either | |
03:19 | way , when you divide by zero it's not going | |
03:21 | to work for a calculation . So you cannot have | |
03:23 | our common ratio equal to one . But when you | |
03:26 | think about that's not a big deal because if I | |
03:28 | had a common ratio of one what would the series | |
03:30 | look like if I take this and multiply by one | |
03:32 | , I'm gonna get one multiplied by one . Again | |
03:34 | I get one multiplied by uh one again I get | |
03:37 | one . So if the common ratio wherever one for | |
03:40 | any series technically it's geometric but it's a boring series | |
03:44 | because every term is just going to be the exact | |
03:47 | same term . So if I wanted to find the | |
03:49 | some of the terms of a geometric series with a | |
03:51 | common ratio of one , I don't need this fancy | |
03:54 | equation . If all of the terms of the same | |
03:56 | , I just take the first term and multiply by | |
03:58 | how many terms I have . If the first term | |
04:00 | is five then every term is five and I just | |
04:03 | say five times however many terms I have and then | |
04:06 | I have the some so it's kind of boring and | |
04:08 | it's not it's not really helpful . And so it | |
04:11 | isn't a big deal that are cannot be won in | |
04:13 | this equation for any other geometric series . This equation | |
04:16 | is going to work now before we actually prove this | |
04:20 | , I want to mathematically prove it to you . | |
04:22 | Before we actually do that , I want to apply | |
04:24 | this to this known geometric series that we have above | |
04:28 | . So we just said that the partial sum is | |
04:32 | the first term times one minus R to the power | |
04:35 | , then over one minus are . That's what we | |
04:38 | said , and we said we have 1234567 terms . | |
04:42 | So for this particular series we know that N is | |
04:45 | equal to seven , We know that the first term | |
04:48 | is equal to one and we know that the common | |
04:51 | ratio , because we've been we just wrote it down | |
04:52 | here . The common ratio is too , so we | |
04:54 | have everything needed . So let's check it out . | |
04:56 | The seventh , partial some seven partial sum is the | |
04:59 | sum of the first seven terms is going to be | |
05:02 | equal to this first term , which is one times | |
05:05 | one minus the common ratio of our two to the | |
05:08 | power of n however many terms we're adding together and | |
05:11 | on the bottom it's one minus are so one minus | |
05:13 | R . Is to all we have to do is | |
05:15 | do this calculation . Okay so the one times this | |
05:20 | the one is going to disappear . So all you're | |
05:21 | going to have on the top is one minus what | |
05:24 | is to do the seventh power you're gonna have 128 | |
05:27 | Wrap it in parentheses if you want on the bottom | |
05:30 | 1 -2 is gonna be negative one . So what | |
05:33 | are you gonna get ? You're gonna have s seven | |
05:36 | you're going to get one minus 1 . 28 is | |
05:38 | negative 1 27 over negative one . So the seventh | |
05:43 | partial sum is positive 127 . So what we're basically | |
05:47 | saying is that if we add up all of these | |
05:49 | terms we should get 127 and I encourage you to | |
05:52 | do that , grab a calculator and say one plus | |
05:55 | two plus four plus eight plus 16 plus 32 plus | |
05:57 | 64 . You're gonna figure out that that's exactly equal | |
06:00 | to 127 . Now for a series with seven terms | |
06:04 | , it's not a big deal to add them up | |
06:06 | . But what if this geometric series had 1000 terms | |
06:08 | ? What have had a million terms or 15 million | |
06:10 | terms and adding them up ? Even with the computer | |
06:13 | would take forever . Right ? I take a long | |
06:14 | time even with the computer . So this equation is | |
06:18 | mathematically calculating what the sum of however many terms of | |
06:22 | the series I want . All I need to know | |
06:24 | is the first term , the common ratio which lets | |
06:27 | me predict all the future terms and however many terms | |
06:30 | I want to consider , it's gonna add them up | |
06:32 | now . The question is , why does it work | |
06:34 | right now , As I said in the last lesson | |
06:36 | , when we derive the arithmetic series formula in math | |
06:39 | , When you prove things , all you care about | |
06:41 | is arriving at this . You want to prove that | |
06:43 | this is true and you can do anything you want | |
06:45 | legally to get there . So what we're gonna do | |
06:48 | is we're gonna write down the terms of the series | |
06:51 | . We're gonna , in this case this is a | |
06:52 | geometric series I wrote down with numbers , but we're | |
06:54 | going to write down a general geometric series . And | |
06:57 | then I'll walk you through the next steps to get | |
06:59 | from point a down to proving that this is the | |
07:02 | equation that describes the son . Mostly I don't want | |
07:05 | you to memorize this , but mostly I just want | |
07:07 | you to know that these equations don't come out of | |
07:09 | thin air , Right ? So how do we do | |
07:10 | it ? Let's take a look at this , proof | |
07:13 | . What I'm going to suggest . What I'm gonna | |
07:14 | say is that the inthe partial sum is gonna be | |
07:17 | the sum of all the terms of the geometric series | |
07:19 | . Every series starts with the first term . So | |
07:22 | we call it T one . What is the next | |
07:24 | term in this series ? We'll for a geometric series | |
07:26 | to get the next term . All you do is | |
07:28 | you start with the first term and you multiply by | |
07:30 | a common ratio to get the next term after that | |
07:32 | . And you multiply by the common ratio , it's | |
07:34 | always the same pattern . So to get the next | |
07:36 | term , it's just gonna be that first term times | |
07:38 | the common ratio . RT one times are . But | |
07:42 | then how do you get the next term after that | |
07:44 | ? It's gonna be this term times are but then | |
07:47 | times are again , so it's going to end up | |
07:48 | being T1 times r squared when you think about it | |
07:52 | , you multiply by two and you multiply by two | |
07:54 | again . So basically you consider it being one times | |
07:57 | four to give me four . So I can predict | |
07:59 | any future term by doing this . So this is | |
08:02 | the pattern like this . And then let me ask | |
08:04 | you another question . Uh Well let me let me | |
08:07 | go and get it down . What would the last | |
08:09 | term be in the series in the last term ? | |
08:11 | Here we had noticed we had seven terms 1234567 There's | |
08:16 | seven terms . How many multiplication is do I need | |
08:18 | to get their one multiplication ? 23456 multiplication . So | |
08:23 | if I start with the first one and do six | |
08:25 | multiplication is I get to the seventh term . So | |
08:28 | in the general form the last term is going to | |
08:31 | be the first term Times are to the power of | |
08:35 | N -1 . So we start with the first term | |
08:37 | to get the next term multiplied by the common ratio | |
08:40 | . Then we multiply by the common ratio again . | |
08:42 | And I'm saying we keep doing that until we get | |
08:43 | to the last term . But the last term is | |
08:46 | R to the n minus one . Why is it | |
08:48 | to the n minus one for the last term ? | |
08:49 | Why isn't it just are to the 10th ? Well | |
08:52 | it's because if they're in terms in the series If | |
08:55 | there's seven terms in this series I do six multiplication | |
08:59 | is to get there . So this is the last | |
09:01 | term , the 10th term . So I do one | |
09:03 | less multiplication is by our to get there . That's | |
09:05 | why it's in -1 . What would the term just | |
09:08 | before that be ? It would be T one Me | |
09:13 | erase This plus make it clear T one R to | |
09:17 | the N -2 . Because again we're multiplying here . | |
09:21 | So this is multiplied by arts and M -1 . | |
09:24 | And this is one less multiplication . Uh going that | |
09:27 | direction . Okay . And then we have dot dot | |
09:30 | dot in the center here . I wish I wouldn't | |
09:32 | smudge that right there . That's okay . We'll get | |
09:35 | there . So this should hopefully make sense . Especially | |
09:38 | when you think about the numbers here , when you | |
09:41 | think about the numbers , you multiply by two by | |
09:42 | two by two and then eventually have a last term | |
09:45 | multiplied In -1 times . And so that's exactly what | |
09:48 | we had here . Last term in -1 times in | |
09:51 | -2 times . And then we fill everything else in | |
09:54 | . All right now , what I want to do | |
09:56 | is I want to multiply this series by are the | |
09:59 | common ratio are and I'm gonna write that down directly | |
10:02 | underneath it and then I'm going to subtract it . | |
10:04 | Why are we doing this ? You can't predict ahead | |
10:06 | of time . This has been proved a long time | |
10:08 | ago by people that played around with it and figured | |
10:10 | it out that it works so we're going to do | |
10:12 | it . But don't worry so much if you don't | |
10:14 | , if you say well I would never would have | |
10:15 | figured that out . That's okay . I didn't wake | |
10:18 | up this morning and decided that I was going to | |
10:20 | do this either . All I care about is that | |
10:21 | you understand what I'm doing . So let's take this | |
10:24 | exact same series and multiply by are the common ratio | |
10:27 | are and I'm gonna change colours to do that . | |
10:30 | The common ratio our is our times as cement . | |
10:35 | So I multiply the left hand side by our I | |
10:37 | must also multiply the right hand side by our The | |
10:40 | first term is this , I'm gonna multiply by our | |
10:43 | but I'm not gonna write it underneath it . I'm | |
10:45 | gonna I'm gonna multiply every term ir but I'm going | |
10:47 | to shift the terms over so I'm gonna multiply by | |
10:49 | this and I'm gonna write it down here . T1 | |
10:52 | times are why am I not writing it down here | |
10:55 | ? Because ultimately I'm going to be adding these terms | |
10:57 | together and these terms go together . I could write | |
10:59 | it here and then I would just have to collect | |
11:01 | terms but it's gonna be easier if I line that | |
11:03 | turns up . So I multiplied by our I write | |
11:05 | it here , I multiply by our here and I | |
11:07 | write it underneath here . T one r squared . | |
11:10 | I multiply this by our and I'm gonna get our | |
11:12 | cube but that's gonna fall into the dot dot dot | |
11:15 | area so I'm not going to have to actually write | |
11:17 | that . It is there , it's just not written | |
11:19 | down , it's in the dot dot dot area . | |
11:21 | And then I'm going to multiply the term right before | |
11:24 | this one by R . C . This will be | |
11:26 | in minus three right here , multiply by . Are | |
11:29 | you add the exponents ? It'll be T one R | |
11:31 | to the n minus two because this one in minus | |
11:34 | three Plus one from the art , you're going to | |
11:37 | get in -2 . Then when I multiply this one | |
11:40 | by our I'm gonna have our multiply all this stuff | |
11:43 | by our but I'm gonna add the exponents , it'll | |
11:45 | be a -1 here , so it'll be T1 are | |
11:49 | in -1 . And then when I multiply by our | |
11:51 | to the last term are to the first power will | |
11:54 | add with this so to become our to the 10th | |
11:56 | power . Um Actually I forgot A . T . | |
12:00 | It will be T . One R . To the | |
12:04 | power . So all I've done is multiplied by this | |
12:06 | by our write it down . Multiply this by our | |
12:09 | write it down here , multiply this by our write | |
12:11 | it down by here , multiply this by our it's | |
12:13 | in the dot dot dots . Multiply the term before | |
12:16 | here by our it's going to be in minus two | |
12:18 | . When you add the exponents multiply this by are | |
12:21 | you subtract add one to this ? It will be | |
12:23 | in minus one , multiply this by our , add | |
12:25 | the exponents . It'll be our to the end . | |
12:27 | So you know it's legal to multiply any equation by | |
12:30 | whatever you want , as long as you do it | |
12:31 | to the left and the right hand side . So | |
12:33 | I've done that now that I've done that . What | |
12:36 | do I want to do ? I'm going to take | |
12:38 | this equation and I'm going to subtract it from the | |
12:41 | one right underneath . I'm gonna do a subtraction . | |
12:44 | Remember when we solve systems of equations ? We can | |
12:47 | add them or subtract them ? That's fine . Okay | |
12:50 | , so what am I going to get sn minus | |
12:52 | ? This is just gonna be S N minus R | |
12:56 | S N . Just this term minus this term . | |
12:58 | No magic here , what's this gonna be ? T | |
13:00 | one minus zero is gonna give me T one . | |
13:04 | What is this term minus this term ? What's exactly | |
13:07 | the same terms ? So it's zero . What does | |
13:09 | this term minus this term ? It's also zero . | |
13:12 | And all the terms in the dot dot dot areas | |
13:13 | are also going to give me zero . So let's | |
13:15 | just kind of do this . This term minus this | |
13:18 | term is exactly the same thing . It's gonna be | |
13:19 | zero . This term minus this term is exactly the | |
13:22 | same thing is going to be zero . Here's a | |
13:24 | zero up here minus this one has to be a | |
13:26 | negative T . One R to the end because it's | |
13:29 | zero minus this because I'm subtracting everything so really everything | |
13:33 | cancelled and the whole thing except for all of this | |
13:35 | stuff . So I have S . N minus R | |
13:38 | . S . And equals T . One . All | |
13:41 | of this is zero minus this T one R to | |
13:45 | the end . So now all I have to do | |
13:48 | is clean up , I have an S in here | |
13:51 | So I pull it out , factor it out . | |
13:52 | I get 1 - are on the left 1 -2 | |
13:55 | . On the right . I pull out a . | |
13:57 | T . One , I'm gonna have one minus R | |
14:00 | . To the N . To verify . You multiply | |
14:01 | through here you get this , you multiply through here | |
14:04 | . You get this . Now we're going to solve | |
14:05 | for the impartial some it's gonna be T . 11 | |
14:09 | minus R . To the power of N . Divided | |
14:11 | by this whole thing one minus R . Which is | |
14:15 | exactly what we said . The partial sum should be | |
14:18 | the first term times one minus R . To the | |
14:20 | power then divided by one minus R . And of | |
14:23 | course because this is in the bottom Our cannot be | |
14:26 | equal to one . We've already discussed why ? That's | |
14:29 | not that big of a deal anyway because if the | |
14:31 | common ratio were won the series would be really simple | |
14:33 | . We could easily add everything together . Now do | |
14:36 | I expect you to be able to recover or do | |
14:38 | a proof like this ? No of course not . | |
14:40 | I don't care about that . I just want you | |
14:42 | to know that . You understand where it comes from | |
14:44 | . Okay it's legal . Every step is legal . | |
14:46 | We wrote the series down . That's legal . We | |
14:49 | multiply this series by our which is a number . | |
14:52 | It's legal to do that . As long as we | |
14:54 | multiply the both sides of the equation by the same | |
14:56 | number . Legal , it's totally legal to subtract two | |
14:59 | equations of one another , you know that ? So | |
15:01 | we subtract it and then when you manipulate terms , | |
15:03 | you're able to get an expression for the sum and | |
15:05 | this is the sum that's presented in the theorems . | |
15:08 | Alright , so um put it in your back pocket | |
15:11 | , make sure you understand where it comes from . | |
15:13 | But mostly we want to focus on from here , | |
15:14 | going on how to apply it . So let's go | |
15:17 | to this next board and just do a couple of | |
15:19 | quick extra problems to apply it . What if I | |
15:23 | have a geometric series within terms We're in his eight | |
15:28 | . And the common ratio between these terms is to | |
15:31 | and the first term is equal to one . And | |
15:34 | I ask you to find the sum . The first | |
15:36 | thing you do is you say well the impartial some | |
15:39 | is you should write this equation down every single time | |
15:41 | one minus are to the end Over 1 -2 . | |
15:45 | Now there's eight terms . So what I want is | |
15:47 | I want the eighth partial sum is the first term | |
15:51 | which is a one Times This 1 - are but | |
15:55 | that's too to the power of in but that's eight | |
15:58 | . There's eight terms here Over 1 -2 . But | |
16:01 | again ours too . So it's 1 -2 . So | |
16:04 | the 8th partial sum is The one is just gonna | |
16:07 | multiply out . So what do you have here ? | |
16:09 | You have 1 -2 to the eighth power . When | |
16:12 | you take two to the power of eight you get | |
16:14 | 256 On the bottom you get -1 . So here | |
16:19 | you have negative 255 over -1 . So The 8th | |
16:24 | partial sum is positive 255 . That is the final | |
16:28 | answer . Now . I always recommend when we do | |
16:32 | these problems not to just plug things in and go | |
16:35 | on a merry way . Let's write the series down | |
16:38 | just so we know where it comes from . Since | |
16:41 | we know the first term , we can write the | |
16:43 | first term right away . It's one and we know | |
16:46 | the common ratio is too and we know there's eight | |
16:48 | terms total . So what we're gonna do is multiply | |
16:50 | by two because that's the common ratio will give it | |
16:52 | to multiply by two . Again we'll get a four | |
16:55 | . Then times two is eight . Times two is | |
16:58 | 16 . Times two is 32 then times two is | |
17:01 | 64 . Times two is 128 . We have 12345678 | |
17:07 | terms , which is what we said , we were | |
17:09 | gonna add up . So this is eight terms . | |
17:13 | So it's basically the same series is what we started | |
17:15 | out with , but we just have one extra term | |
17:17 | and if you grab a calculator and add these numbers | |
17:19 | up , you will find out that you get 255 | |
17:22 | . And so that's what we're gonna do . So | |
17:25 | I only have one more of these guys to illustrate | |
17:27 | how to use the geometric series equation . Mostly I | |
17:31 | wanted to introduce it . I wanted you to understand | |
17:33 | where it comes from , but the proof and then | |
17:36 | solve some problems . What if I ask you solve | |
17:38 | and give me the some of the following series , | |
17:41 | two minus six plus 18 minus 54 plus 160 to | |
17:49 | minus 486 Plus 145 , 8 minus 43 74 plus | |
18:00 | 13 +12 to minus 39 +366 How many terms do | |
18:06 | I have ? +123456789 10 terms . Let me just | |
18:11 | check to minus six plus 18 minus 54 plus 1 | |
18:15 | 60 to minus 46 plus 14 58 plus . Uh | |
18:19 | Oh I missed one . Let's see . 486 Plus | |
18:24 | 1458 . Oh no I didn't minus 4374 plus 1312 | |
18:30 | to minus 39366 That's correct . So what we want | |
18:34 | to do is we want to find the some of | |
18:35 | this first of all , is it a geometric series | |
18:37 | ? That's what we want to figure out . Well | |
18:39 | if you think about it , you grab a calculator | |
18:41 | . If you multiply by negative 32 times negative three | |
18:45 | , you'll get the negative six if you take this | |
18:47 | and also multiplied by negative three . Six times negative | |
18:51 | three , you're gonna get the 18 . If you | |
18:53 | go here and multiply by negative three , you'll get | |
18:55 | this multiplied by negative three , you'll get this multiplied | |
18:57 | by negative three , you'll get this and you go | |
18:58 | all the way through the sequence . So every term | |
19:01 | is just the previous term multiplied by negative three . | |
19:03 | So we then know that it's geometric with the common | |
19:06 | ratio of negative three . We also know that they're | |
19:10 | in equals 10 terms because when you count the terms | |
19:13 | you get 10 and we also know that the first | |
19:15 | term is equal to two . So by giving being | |
19:18 | given the sequence we can pull all of this information | |
19:21 | out and that's all that we need to solve this | |
19:22 | problem . Okay , The 10th partial sum is equal | |
19:28 | to the first term , times one minus R . | |
19:32 | To the power of N divided by one minus are | |
19:35 | . So if we want to find , how many | |
19:36 | times do we have 10 ? The 10th partial sum | |
19:39 | , it's going to be the first term which is | |
19:40 | two times one minus the common ratio . Are are | |
19:44 | we said was negative three . So you put in | |
19:46 | parentheses negative three . Make sure you put in parentheses | |
19:49 | to the power of N , which is 10 . | |
19:52 | And on the bottom it's one minus R . But | |
19:54 | again , Rapidan princes because R is negative , it's | |
19:56 | a negative three . So now we just have to | |
19:59 | crank through this . What do we get ? Two | |
20:00 | times ? Here's one right here minus what is negative | |
20:04 | ? Three to the 10th , negative three to the | |
20:06 | 10th is going to be a positive number 59 oh | |
20:10 | 49 It's a big number 5 59,000 and 49 . | |
20:13 | But it's positive because it's a an even power the | |
20:17 | minus sign comes from the outside here . This evaluates | |
20:20 | to a positive number here . You have won . | |
20:23 | This becomes a plus three . So you have a | |
20:25 | four . All right . So the rest of it | |
20:28 | is just simply math . The 10th partial sum is | |
20:32 | two times when you do one minus this , you'll | |
20:34 | get negative 59 048 over four . So the 10th | |
20:41 | partial sum is negative . 29,000 524 negative . 29 | |
20:46 | 5 24 you might say wait , wait , wait | |
20:48 | . How's it negative ? Well , it's negative because | |
20:50 | the signs are alternating and look at the last number | |
20:53 | . The last number was 39,000 and 366 . That's | |
20:56 | a negative number . This is going to pull the | |
20:58 | whole thing negative because all the other numbers are so | |
20:59 | much smaller . And so you end up with negative | |
21:01 | 29,524 . So in this lesson we have uh used | |
21:08 | the geometric series to uh the equation for the geometric | |
21:11 | series to calculate the sum of any geometric series we | |
21:14 | want and all we have to know to do that | |
21:16 | . We do not need to know the last term | |
21:18 | , but we do need to know the first term | |
21:20 | , whatever that common ratio is and however many terms | |
21:22 | we have in the series we went through the proof | |
21:25 | of it , the proof of it . I hope | |
21:26 | you understand that . I try to make it as | |
21:28 | clear as they can . The proof is important , | |
21:30 | but mostly I want you to know how to use | |
21:32 | it and that's why we did these problems here . | |
21:34 | So make sure you can solve every one of these | |
21:36 | yourself . If one more lesson in geometric series calculations | |
21:39 | that we're gonna do , so follow me on to | |
21:41 | the next lesson , we'll get more practice with the | |
21:43 | geometric series . |
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