How To Derive The Formula For The Sum of an Arithmetic Series - By The Organic Chemistry Tutor
Transcript
| 00:00 | in this lesson we're going to talk about how to | |
| 00:03 | prove the formula that gives you the sum of an | |
| 00:07 | arithmetic series . Now before we do that let's talk | |
| 00:10 | about an arithmetic sequence versus an arithmetic series . Here's | |
| 00:15 | an example of and everyone takes sequence . Their number | |
| 00:19 | is 5811 14 17 and so forth . The first | |
| 00:26 | term a sub one is 5 . The second term | |
| 00:30 | a sub two is 8 . The common difference between | |
| 00:34 | them . History To go from the first term to | |
| 00:37 | the second term you've got to add three and then | |
| 00:40 | to get the next term you gotta add three again | |
| 00:42 | . Eight plus three is 11 . 11 plus three | |
| 00:44 | is 14 . So that is known as the common | |
| 00:48 | difference . The sequence is simply a list of numbers | |
| 00:55 | . When you add that list of numbers you get | |
| 00:57 | a series . So let's say we want to calculate | |
| 01:03 | the sum Of the 1st 6 terms . This would | |
| 01:06 | be s sub six . So it's gonna be five | |
| 01:12 | plus eight plus 11 Plus 14 Plus 17 . So | |
| 01:18 | that gives us a some 55 . Now we can | |
| 01:22 | also calculate that some . Using this formula It's the | |
| 01:26 | sum of the first and the last terms divided by | |
| 01:28 | two times in . So basically this part here means | |
| 01:33 | that it's the average of the first and the last | |
| 01:35 | time . The first term is five . The last | |
| 01:44 | term in the sequence a six . Actually that's I | |
| 01:48 | take that back . That's a set of five . | |
| 01:49 | Rather We're adding five terms . Not six mistakes happen | |
| 01:55 | but it's going to catch him . So the fifth | |
| 01:58 | term that's 17 , divide by two And then times | |
| 02:07 | n . n . is five . Now the average | |
| 02:11 | of five and 17 That's going to give you the | |
| 02:14 | middle number 11 five plus 17 is 20 to 22 | |
| 02:18 | divided by two is 11 And then 11 times five | |
| 02:22 | is 55 . And we can see that these two | |
| 02:26 | answers match . So this formula gives you the partial | |
| 02:31 | sum of an arithmetic series . Now let's talk about | |
| 02:36 | how we can derive that formula . So S . | |
| 02:47 | F . N . Is the sum Of the terms | |
| 02:51 | in an inauthentic series . It's a sub one , | |
| 02:54 | the first term plus the second term . Ace up | |
| 02:56 | to Plus III term , a sub three . And | |
| 03:01 | this can keep on going until we get the last | |
| 03:05 | time in the series . Now the second term we | |
| 03:15 | know it's the first term plus the common difference in | |
| 03:20 | the previous problem . The first time was five . | |
| 03:23 | If you add a common difference of three , you | |
| 03:25 | get the next term eight . So a sub two | |
| 03:29 | , we can write that as a sev one plus | |
| 03:32 | D . Now if we want to get the third | |
| 03:35 | term from the first term we need to add to | |
| 03:39 | common differences . 3 -1 is two , so the | |
| 03:46 | third term is going to be A sub one plus | |
| 03:50 | 2 d . And then this pattern will continue to | |
| 03:54 | repeat until we get to the last term . So | |
| 04:01 | this expression is based on writing to some from left | |
| 04:06 | to right . Now we're gonna write this some formula | |
| 04:08 | from right to left in the next line . So | |
| 04:15 | we're gonna start with the last term first . So | |
| 04:18 | S seven is also a seven plus . Now the | |
| 04:23 | second to last term It's going to be this but | |
| 04:27 | -1 common difference . So it's a sub n minus | |
| 04:33 | D . Now the third from the last term , | |
| 04:38 | It's going to be a sub N -2 d . | |
| 04:41 | And then that pattern will continue to repeat as we | |
| 04:44 | go all the way to the first term . A | |
| 04:46 | sub one . Now we're going to add these two | |
| 04:53 | equations . So let's put a plus sign S . | |
| 04:58 | F N plus S F N . That's gonna be | |
| 05:02 | too times S F N . Here we have a | |
| 05:06 | sub one plus a seven . I'm going to put | |
| 05:10 | it in parentheses . Then if we add these two | |
| 05:15 | notice that D cancels with negativity and we get a | |
| 05:19 | sub one plus N . Now if we add these | |
| 05:31 | two , two D and -2 D will cancel . | |
| 05:37 | And so we left with a someone plus a seven | |
| 05:40 | and this pattern will continue until we add the last | |
| 05:44 | two terms , which is also A sub one plus | |
| 05:48 | A seven . Now we need to factor out the | |
| 05:53 | G c f on the right side . The G | |
| 05:55 | c f , the greatest common factor is clearly A | |
| 05:58 | sub one plus ace event . The question is , | |
| 06:02 | how many ace of one plus asset ? And do | |
| 06:04 | we have , Would you say it's 345 , 20 | |
| 06:09 | , 36 ? The answer is we have In terms | |
| 06:14 | of these things . So if we factor out a | |
| 06:17 | sublime plus , a seven will be left with and | |
| 06:21 | Next we need to divide both sides by two . | |
| 06:25 | And so that's how we can derive the formula for | |
| 06:29 | the some of the different IQ series . It's the | |
| 06:32 | first term plus the last term Divided by two times | |
| 06:36 | and |
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