 Welcome back to our lecture series math 1210 calculus one for students at Southern Utah University as usual I am your professor today. Dr. Andrew miss a nine In lecture 43 of this series. We're actually going to break it up over two sections the The first part I actually want to continue about the sigma notation Continue talking about the signal patient in the previous lecture So this idea of rectangles being our friend actually has nothing to do with anything in this video But it'll be in the next part Which is which is the majority of this lecture content here So if we continue on an appendix e inside of Jane Stewart's calculus textbook I want to talk a little bit more about this sigma notation and we had talked about last time about This sigma operator adding together We're adding together terms in a sequence and how we can find formulas for sums of terms in a sequence We talked about things like the sum of I the sum of I squared The sum of I cubed and we gave some hints on how one could actually find formulas for larger powers But we're gonna stop at just three because that already is quite complicated enough So that that'd be kind of like us on day one of derivatives. We learned the power rule Well, also on day one of derivatives. We learned about taking derivatives of Exponentials and so when it comes to the sigma operation, there is a counterpart of such a thing and this is called a geometric sequence It's kind of like a an exponential growth of some kind So if we have a sequence a sub n like right here We say that the sequence of real numbers is geometric if it has the form what you see right here a n equals A times r to the n minus 1 where you have some initial value a so like the first term of the sequence We're gonna call it a and then there's this constant ratio This term is formed recursively by the second term The second term a 2 will just be the first term a 1 times it by r, which is just a r The second or the third term in the sequence a 3 will look like a 2 r Which has a 2 is just a r times r you end up with a r squared The fourth term of the sequence You're gonna get the third term times by r Which gives us a r squared times r again you end up with a r cubed and then if we did this another time the fifth term You're gonna get the fourth term times r Which is a r cubed times r which gives you a r to the fourth And so that's where this general formula is coming from this a n is equal to the initial term a times this common ratio r And you take that in minus one time So the power of r is one less than where you are in the sequence And the reason behind that is this the reason why we call r the common ratio is the following If we take two consecutive terms in a geometric sequence take a n over a n minus one You'll end up with a r to the a n minus one over a r to the n minus two power And if you simplify this thing the a's will cancel and we can take away powers of r and the end You'll be just with an r and so if you take the quotient the ratio of consecutive terms in a geometric sequence You always get the same number over and over and over again, and that's what characterizes a geometric sequence So as an example of that the sequence 2 6 18 54 162 is geometric You'll notice if you take 6 divided by 2 that's equal to 3 Likewise 18 divided by 6 is equal to 3 if you take 54 and divide that by 18 that's equal to 3 162 divided by 54 is equal to 3 and so the terms in the sequence are always just They're they're the ratio is always 3 and that's also this is another way of defining Geometric sequence is that to get through the sequence 2 times 2 times 6 2 times 3 is 6 times 3 is 18 times 3 is 54 times 3 is 162 if you want the next term of the sequence you'll take 162 times 3 Which gives you 486 and then the next term of the sequence you'll times that by 3 and you'll get 1458 and you can keep on doing this just by multiplying by 3 over and over and over again This works with fractions as well You can take the sequence the geometric sequence 1 half 1 4th 1 8 1 16th 1 32nd 1 64th keep on going It would be like 1 over 128 1 over 256 Etc. Etc. What we do is as we go from one term to the other we just multiply by 1 half our common ratio here is 1 half The number a of course is the first term in the sequence which here is likewise 1 half in the previous example The first term was a 2 So recognizing the geometric sequences that's all that's involved in that And so what I want to talk about is the sum of a geometric sequence. So let's take sn here To represent the sum of a geometric sequence where you go from i equals 1 To n and you add up this geometric sequence ai. Well, this would look like a 1 plus a 2 plus a 3 Up to an that's just what the sum means and as each of these terms is geometric you're going to get a plus ar plus ar squared All the way up to ar to the n minus 1 So let's remember this right here s Is equal to this expression right here And so what we're going to do this time is we're going to take this sn and we're going to times it by this common ratio r Well, if you do that, you're going to take r times all the things we had before a ar plus ar squared All the way up to ar n minus 1 and if you distribute the r through onto all these terms You end up with ar plus ar squared plus ar cubed All the way up to ar in my or in just in right Because a times r will give you ar a times our r times r Sorry, sorry that again r times ar right there gives you the ar squared r times ar squared gives you ar cubed Etc and so you end up with this ar to the n right there And so if we put these things together If we take the sn minus r sn Well, this will look like the a plus the ar Plus the ar squared all the way up to ar n minus 1 and we're going to subtract from this the r sn which we see above which looks like ar ar squared ar cubed All the way up to ar to the n like so And so we're going to try to simplify things on on both sides of the equation in terms of the left hand side You'll notice that uh sn and r sn have a common factor of sn if you factor that out You're going to get sn times 1 minus r On the right hand side, we have another telescoping sum going on here. You get ar cancelled to the ar ar squared cancels the ar cubes cancelled and the ar and everything cancels up up until ar minus 1 The things that don't cancel are going to be ar n and an a And so if we record that down here, we're going to get a minus ar to the n Which if you factor out the common factor of a you get 1 minus r n left over And so dividing both sides by the 1 minus r So we can solve for So we can solve for the s Over there. We end up with the with the famous formula here sn Equals a times 1 minus r to the n over 1 minus r And so this gives us a formula for the sum of a geometric sequence or what we call a geometric sum Where a is the initial term of the sum and r is that common ratio We're going to use this to try to calculate a geometric sum right here So if you look at this right thing right here, you take the sum from what i equals 1 to n 1 to 8, excuse me This is the sum one half plus one fourth Plus one eighth All the way up to um one over two to the sixth What is that 164th? Right, we're trying to add all these things together which this is like a ratio phobics nightmare You have to add up eight fractions all of which have a different denominator. Oh, no, what are you going to do? Well recognizing this as a geometric sum we can apply the geometric sum formula So for example, we we identify the first term of the sequence the first term is here a And like we saw before with this example the common ratio is one half One half times one half is one fourth one fourth times one half is one eighth one eight times one half is one 16th the next term in the sum And so if we treat this as a geometric sum We end up with this using the formula we had before a times 1 minus r to the eighth over 1 minus r We end up with one half for a We're going to get one minus one over two to the eighth And then one minus one half right here In the denominator one minus one half that is itself one half So we get one half over one minus one over 64 over one half One minus one half in the bottom and you'll notice that the one halves cancel And so you're left with just one over one over 64 Which if you want to write this together, you can take 64 minus one over 64 Ending up the sum's going to be 63 over 64 Which does provide a much simpler way of computing Uh this sum even though there's eight different fractions We can handle the geometric sum pretty quickly And I want to make mention that if we were to change this to the take the sum from i equals 1 to n This does not fundamentally change the problem at the end. We end up with a 2 to the n right here We have a 2 to the n right here Final form this is going to look like one Minus one over two to the n and that's the sum right then and there And so we can compute these geometric sums very quickly with this formula