How To Derive The Sum Formula of a Geometric Series - By The Organic Chemistry Tutor
Transcript
00:00 | in this video , we're going to talk about how | |
00:02 | to prove the formula that will help us to calculate | |
00:06 | the sum of a geometric series and there's two of | |
00:09 | them . There's a finite geometric series and the infinite | |
00:12 | geometric series . We're going to talk about how to | |
00:14 | prove the formula to calculate the sum of both of | |
00:17 | those . So let's start with a geometric sequence . | |
00:22 | Let's say we have a sequence three six 12 24 | |
00:30 | 48 96 and so forth . three is the first | |
00:36 | term . Six is the second term . 12 is | |
00:39 | the third term . The common ratio is too you | |
00:46 | need to multiply the first term by two to get | |
00:48 | the second term three times to a 66 times two | |
00:52 | is 12 , 12 times two is 24 . So | |
00:57 | we have a common ratio of two . So no | |
01:01 | to at the second term is the first term times | |
01:05 | are six is 3 times two . The third term | |
01:11 | is the first term times r squared 12 , His | |
01:17 | three times 2 squared What three times for . Likewise | |
01:22 | , if you want to find 1/4 term , it's | |
01:24 | the first term times our cube 4 -1 History . | |
01:33 | Make sure you understand that . Because we're going to | |
01:35 | use that later to prove the formula . Now , | |
01:39 | what we have here is a sequence to convert it | |
01:43 | into a series . We need to use the addition | |
01:46 | sign . So this is a geometric series . So | |
01:54 | let's add up the first six terms . So this | |
01:57 | is going to be S6 , the partial some of | |
02:00 | the first six terms , three plus six plus 12 | |
02:07 | plus 24 plus 48 plus 96 . That's 189 . | |
02:18 | Now , let's use the formula to calculate that son | |
02:22 | . This is the formula that helps us to calculate | |
02:24 | the sum of a finite geometric series . It's a | |
02:28 | sub one times one minus R a c n over | |
02:33 | one minus R . So to calculate as sub six | |
02:37 | is going to be a sub one the first term | |
02:39 | which is three Times 1 -2 . Ours too . | |
02:45 | It's race to the end And is basically the number | |
02:48 | of terms which is six Over 1 -2 . So | |
02:53 | this is going to be three one minus two races | |
02:57 | . Six power . If you multiply 26 times You | |
03:01 | get 64 , 1 - She was -1 , 1 | |
03:06 | -64 is -63 . The two negative signs will cancel | |
03:10 | becoming positive . So it's three times 63 . three | |
03:15 | times 6 is 18 , three times 3 is nine | |
03:18 | . So we get 189 which is the same as | |
03:21 | what we see here . So this is the formula | |
03:24 | that helps us to calculate to some of a finite | |
03:29 | geometric series . Now , what if we were to | |
03:34 | have an infinite geometric series ? So let's say we | |
03:39 | have the series eight plus four Plus two Plus 1 | |
03:44 | plus one half plus 1/4 and then plus dot dot | |
03:50 | dot . So this series here it has the beginning | |
03:55 | and it has an end . So it's a finite | |
03:58 | geometric series . This series doesn't have an end , | |
04:02 | it goes on forever . So it's an infinite geometric | |
04:05 | series . The common ratio , if you take the | |
04:10 | second term divided by the first four divided by eight | |
04:13 | is one half . The combination is 1/2 for all | |
04:16 | terms . Now because the common ratio or rather the | |
04:21 | absolute value of art Because it's less than one . | |
04:24 | The series converges which means that the sum is finite | |
04:30 | . If the absolute value of our if it was | |
04:32 | greater than one , this series will diverge . And | |
04:35 | you're you wouldn't be able to calculate the sum because | |
04:38 | it can increase towards positive or negative infinity . So | |
04:42 | it's important that the common ratio , the absolute value | |
04:45 | of income ratio be less than one for this to | |
04:47 | work . And the sum of this infinite geometric series | |
04:53 | , we can write s sub infinity . It's the | |
04:58 | first term divided by 1 -1 . So in this | |
05:01 | example it's 8/1 minus a half . 1 -1 , | |
05:09 | eight divided by a half . If you multiply the | |
05:12 | top and bottom by two , This becomes 16/1 , | |
05:16 | which is 16 . So if you were to add | |
05:19 | these numbers eight plus four is 12 plus two , | |
05:23 | that's 14 plus 1 , 15 Plus 1 15.5 Plus | |
05:30 | 14 , If you keep doing this you're gonna get | |
05:33 | closer to and closer to 16 but you're not going | |
05:35 | to pass 16 . So that's the some of this | |
05:40 | infinite geometric series . Now let's talk about how to | |
05:43 | prove this formula as well as the other one . | |
05:50 | So let's start with the formula that describes the sum | |
05:53 | of an finance geometric series . So we're gonna have | |
06:00 | S . S . N . Is a one plus | |
06:05 | a two plus a three plus a four . And | |
06:13 | then the second to last term is a sub N | |
06:15 | -1 And the last term will be a seven . | |
06:24 | Now , a sub two . Recall that a sub | |
06:27 | two is a 71 times are and a sub three | |
06:32 | . We said that it's a 71 times R squared | |
06:36 | A four is a sub one times our Cube . | |
06:42 | By the way , the formula that describes the f | |
06:45 | term of a geometric series Or a geometric sequence is | |
06:50 | a sub one . Our race to the N -1 | |
06:55 | . So in our first example where we had the | |
06:56 | series or the sequence three , 12 24 48 let's | |
07:04 | say . If we want to calculate the fifth term | |
07:06 | , It would be a sub five Is equal to | |
07:09 | a sub one which is three times are ours too | |
07:12 | . Race the end -1 Or 5 -1 . 5 | |
07:16 | -1 is four . 2 to the 4th power two | |
07:20 | times two times two times two at 16 , three | |
07:23 | times 16 is 48 . So this formula here gives | |
07:27 | you the f term of a geometric sequence . So | |
07:40 | listening about that , if a sub N is a | |
07:43 | sub one times are Race to the N -1 , | |
07:47 | What is a sub and -1 ? A 7 -1 | |
07:52 | . It's gonna be a sub one times are but | |
07:55 | we're going to plug in and -1 into that expression | |
07:59 | . So replace an end With N -1 . We | |
08:03 | get a sub one . Are race to the End | |
08:06 | monastery And that's what we're going to replace this thing | |
08:09 | with . It's going to be a someone are To | |
08:14 | the end -2 and then a sub N . We | |
08:17 | already have that here . That's just a one To | |
08:21 | the end -1 . Now for the next line , | |
08:30 | what we're gonna do is we're going to take this | |
08:32 | equation and we're gonna multiply it by our so the | |
08:36 | speed , this is gonna be our sn and then | |
08:41 | this is going to be a one times are this | |
08:47 | becomes if you multiply by our it's going to be | |
08:50 | a one R squared and then plus a one . | |
08:54 | Our cube plus a one Are to the 4th and | |
09:02 | then multiply this by our The exponent is going to | |
09:05 | increase by one . So N -2 plus one that's | |
09:11 | going to be and minus one multiplying this by our | |
09:16 | the export will increase by one as well . So | |
09:19 | n minus one plus one Becomes simply and -1 and | |
09:23 | positive one will cancel . Now what we're gonna do | |
09:28 | is we're gonna multiply this equation by -1 . So | |
09:32 | this is going to be negative . And then every | |
09:35 | term within the bracket , once we distribute the negative | |
09:40 | sign will be negative . So now we're gonna add | |
09:45 | these two equations . So we're gonna have S seven | |
09:50 | N plus negative R . S . F . N | |
09:54 | . Or simply ss men minus R . S . | |
09:58 | Event . And then we'll have a one . Now | |
10:07 | A one AR -1 R . They're going to cancel | |
10:14 | a one R squared minus a one R squared . | |
10:20 | They will cancel as well . A one Our Cube | |
10:23 | is going to cancel A one arch . The fourth | |
10:27 | will cancel with something that's here and then A one | |
10:35 | RN -2 will cancel with something that's there . And | |
10:39 | then a one RN -1 will cancel as well . | |
10:45 | This one will not cancel so but it has a | |
10:47 | negative sign . So it's gonna be negative . A | |
10:51 | one are raised to end . Now what we're gonna | |
10:54 | do is we're going to factor out s event S | |
10:59 | . M.N . by itself is one negative R . | |
11:03 | S . F . N . Divided by S . | |
11:04 | F . N . Is going to be negative . | |
11:06 | Are Here we're going to factor out the first term | |
11:09 | a sub one . A someone divided by itself is | |
11:13 | one negative A one R . C N divided by | |
11:17 | a one is negative . Our race to the end | |
11:22 | And then we're gonna divide by 1 -2 . So | |
11:29 | this gives us the formula for the partial sum of | |
11:31 | a finite geometric series . It's the first term times | |
11:36 | one minus the common ratio race to the end over | |
11:39 | one minus R . So that's how we can derive | |
11:45 | the son of a finite geometric series . Now , | |
11:50 | let's talk about how we can get this formula in | |
11:56 | order to get that formula , we need to realize | |
11:59 | that if the the absolute value of our is less | |
12:04 | than one , and as N goes to infinity , | |
12:10 | our race to the end Goes to an Eagles0 , | |
12:14 | not infinity . Now , let's talk about that . | |
12:18 | So if we have .9 , which is less than | |
12:21 | one , raise the first power , that's 0.9 . | |
12:25 | But what happens if we increase the exponent from 1-10 | |
12:30 | ? Will this number get bigger or lower .9 ? | |
12:35 | Race to the 10th Power Is a smaller number . | |
12:39 | It's .348 678 Let's use the approximate symbol . Now | |
12:53 | , let's increase it to 100 .9 . Race to | |
12:58 | 100 . It's even smaller , it's point 00002656 . | |
13:09 | So we can see that the limit as N goes | |
13:15 | to infinity of our sub . And assuming that our | |
13:20 | is less than one or the absolute value of our | |
13:22 | is less than one . This Becomes a zero when | |
13:27 | an approaches infinity . So the sum of an infinite | |
13:35 | geometric series where The absolute value of ours less than | |
13:39 | one . This is going to be the limit as | |
13:44 | N goes to infinity of the the finite geometric series | |
13:48 | formula , which is a sub one times one minus | |
13:52 | our race of the N Over 1 -2 . Now | |
13:57 | this is the only part of the equation that has | |
13:59 | ended And we know that are to the end will | |
14:02 | go to zero once we apply This limit , so | |
14:06 | it becomes a sub one times 1 -0/1 -2 . | |
14:11 | And 1 0 is simply one . A sub one | |
14:17 | times 1 is simply a sub one . So the | |
14:20 | sum of an infinite geometric series Boy , the athlete | |
14:24 | valley of ours , less than one Is this equation | |
14:27 | a sub 1/1 -2 . So that's how you could | |
14:31 | derive it from this formula . It's by realizing that | |
14:36 | as N goes to infinity are certain goes to zero | |
00:0-1 | . |
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