Transcript for:
Deriving the Formula for the Sum of an Arithmetic Series

[Music] hey guys in this video I'm going to show you how to derive the formula for the sum of an arithmetic series so firstly what is in a dramatic series well this is the sum of a sequence of terms or each consecutive term has got a common difference so this is an example where the common difference is 1 because we're adding on one every time first time here is one also and there's a finite number of terms which is the hundred so I'm gonna show you how to do this using numbers first and then we'll generally isn't using letters so the trick we use is by writing this series out backwards so s equals 100 plus 99 plus 98 Plus cetera all the way down to 3 plus 2 plus 1 now you can see here both of these series they're gonna have the same son so we can add these two together and we'll do this by pairing them up like this so first time first time second time seconds down etc so I'm left I'm sorry I'm just gonna get to lots of us now we add these two together so one plus 100 gives us Andrea 1 2 plus 99 engine 1 3 plus 98 also hand in 1 so you can see here the pattern just goes on and every term is gonna be 101 and we can take advantage of this trick to find the sum of the series so how many terms are they gonna be now well we defined it as 1 to 100 said there's going to be a hundred times so we can simplify this as 100 times 101 so if we saw for us by dividing my to say 100 over T is 50 50 times 101 so the sum of this series is going to be 5050 and they get if you add up all the numbers from 1 to 1 you get 5050 so now let's try to generalize and firstly we need to define some terms so let's let a equal the first time and D being a common difference and then n being the number of terms so what's our general arithmetic series gonna look like this is gonna be Estevan equals this notation represents the series with n number of terms so S sub N equals our first time is going to be a second term is going to be a plus T better a plus 2 D etc all the way up to the last time a plus a plus and this is our last time here a plus n minus 1 D the reason we've got n minus 1 is because we introduced D and the second terms is a lag if you like there's always one last even the terminal so we'll use the same trick as before s of n equals by writing our backwards for s of N equals a plus n minus 1 D plus a plus n minus 2 D I'll just write a couple times a plus D and so just as before we're gonna add these two series together so left hand side gets to lots of s of N and now we need to pair up these terms just was before first so my first term second term second term all the way to last time I said let's at least get now so yeah to a plus and minus one D that's the first time let's look at this one now it's it still today and we have n minus 2 D plus D so we get n minus one thing just so you can see just as before every times gonna be the same all the way up to the last time so how many terms are there gonna be or from our definition there's gonna be n number of terms so we can simplify this as excessive N equals n times 2 plus n minus 1 D now solving for n by just dividing by 2 we get the formula say n over 2 times 2 a + n - 1 D and they have it how to drive the son of an arithmetic series