 Alright, let's try to simplify some sigma notation using some formulas we've seen in the past. So if you're, so kind of imagine if you're trying to calculate a derivative or an anti-derivative or something, let's think about derivatives, which we know pretty well by this time in calculus. If you wanted to take the derivative of x times 4x squared minus three, probably the first thing you would want to do is distribute. If you distribute the variable through that kind of avoids some unnecessary product rule here. Sigma is no different. We like things to be distributed out if we can afford it. So if you distribute the i through, you get 4i cubed minus 3i. Take the sum as i goes from one to n. And also if you were taking the derivative of a polynomial, you would break it up into pieces. So like if you broke it up into a difference there, you're going to get the sum of 4i cubed minus the sum of, let me draw that sigma again, the sum of 3i as i goes from one to n, one to n. This is exactly what we do like with derivatives and those constant multiples, right? We have this factor of four, factor it out. We have this factor of three, factor it out of the sum. Because sigma is a linear operation, we are justified in pulling out this factor. So we get four times the sum where i goes from one to n of i cubed minus three times the sum of just i as i goes from one to n, right? And so at this moment, if we're taking derivatives like we're taking the derivative of i cubed and the derivative of i, we would apply the power rule for derivatives. We did the same thing for antiderivatives. Well for sigma, we have to apply basically the sigma power rule. What do you do with powers of i right here? And we talked about these in the previous video. Look at the links below if you want to see what exactly those were. But as a reminder, we saw that if you take the sum of i cubed, this is equal to n squared times n plus one squared over four. And if you take the sum of i, this will be n times n plus one over two. And so we're going to apply those formulas in this context here. We end up with four times n squared times n plus one squared all over four minus three times n times n plus one over two. And so it's going to be helpful if we can combine like terms if possible. Some things to note here, four does divide with four. But in order to add these things together, I do want a common denominator of two. So I'm only going to cancel, I'm going to rewrite four over four as two over two. And so then we can write this sum together as the following. We get two n squared times n plus one minus three times n times n plus one all over two. So we have that common denominator. Let's also factor out any common factors we have. We have a factor of n that's common to both. There's also an n plus one that's common to both. That should be a squared right there. I forgot to write it earlier. And so when we factor that thing out, you can factor out an n and n plus one that leaves behind two n times n plus one minus three. There's not going to be much more hope of factoring that until you distribute the two in through the n plus one there. So if we do that, we have n times n plus one. So then we get two in square plus two in minus three. Like so this all sits above two. And there's really not any other. The two n squared plus two in minus three, there's no factorization that's going to work on that thing right here. So we're going to leave this thing. We're going to say that's done right there. We found a formula for the sum of the original function. And it's going to be a rational function in terms of n, because each of the power rules for sigma give us a rational function in terms of the variable I right there. That is that is it's going to be in the end. Sorry. How about the next one right here? Let's evaluate the sum where I goes from one to n of I to the fourth minus one minus I to the fourth again. Well, I can pause this right here that if we go back to the previous video about this sigma power rule, we don't have a sum for the fourth power. So what do we do with this? Well, we could try to derive the fourth power or we could try to kind of remember where it came from. If we write this in expanded form, we're going to get one to the fourth minus zero to the fourth. The next one, if we do the second term, we're going to get two to the fourth minus one to the fourth. The third term will look like three to the fourth minus two to the fourth, etc. And this will continue on until we get to the end. We're going to get n to the fourth minus n minus one to the fourth. And so this principle of a telescoping sum comes up again. This is what we call a telescoping sum. The principle of the telescoping sum is going to come up here because we get the one to the force cancel, two to the force will cancel the three to the force will cancel, and that everything will cancel except for the n minus one to the fourth here. And so the only thing that said don't cancel is the final into the fourth and also the original zero. So we end up with the sum using this telescoping principle here. Telescoping sum, we never actually finished writing that you get into the fourth minus zero to the fourth, which that of course simplifies just to be into the fourth. So it's important to look for telescoping sums. It turns out they can help us simplify many of these types of sums. Well, this is a calculus class after all. So our goal is going to be taking limits of the sigma operations here. So what if we take the limit as n goes to infinity of the sum where one goes from I goes from one to n of the quantity of the sequence three over n times I squared over n squared plus one. Well, that sounds like a mouthful, right? There's a lot going on there. But why why do we even care about such a thing like that? Well, guess what? We will do that later. These these type of limits will actually be quite natural things in the right context. We're not there right now. We're sort of going to be the karate kid right now. We're learning to wax the wax the car and paint the fence. But we're really learning karate the whole time. So let's not worry about the limit process yet. We'll come back to that later on the problem. Let's try to work with this sigma right here. So we take the limit as n goes to infinity. What can we do here? Well, one thing to note is that let me get rid of these parentheses for a moment that when it comes to the sigma, the variable with respect to this operation is the I. And so with regard to sigma, every other symbol you see is a constant, the three, the plus one, those are constants. But with respect to the sigma, the variables I and therefore this three over n is itself a variable. It's I shouldn't say it's a variable. It's a constant. It doesn't change as I changes. So we can factor it out the same way we factor out a seven. And so if we factor that out, we're going to get the limit as n goes to infinity of three over n. And then we get the sigma of I squared over n squared plus one right here. As in as I excuse me, goes from one to n. So we can factor out the three over n because it doesn't depend on the I. Next, we have a sum inside of our sigma. So we can break it up into two sums using the linearity property of sigma. So we get the limit as n goes to infinity of three over n. We're going to get the one over n squared sigma of I squared, as I goes from one to n. And then we add to that sigma of one as I goes from one to n. You'll notice also over here, I took the liberty of taking the one over n squared out of the sigma, because again, the sigma's variable is I, the one over n squared is constant with respect to sigma. And so for the sum of I squares, we can apply the appropriate formula for that one. And as a reminder of what that is, we saw this in the previous video, that if you take the sum of I squared, as I goes from one to n, this will always turn out to be n times n plus one times two n plus one, all over six. So we're just going to kind of memorize this and use it in this place right here. So the so we get this, the limit of three n times one over n squared times the n times n plus one times two n plus one all over six. And then we have to add up together the sum of the ends, sorry, the sum of one. But this is just one added together in time. So that just becomes an n. So we put that right here. Oh, maybe for the sake of simplicity, I'll try to squeeze it in. So what do we have? So if you look at this all the eyes are now gone, this is the advantage of those formulas we have for sigmas all the eyes are gone. And now we're taking a limit as n goes to infinity of a rational expression in terms of n. This is stuff we did earlier in the semester. Let's try to compute this thing. If we take the limit as n goes to infinity, we're going to end up with the following. We are going to get if we distribute some things through. So we're going to distribute this three over n to both pieces. Doing that, we will get three times n times n plus one times two n plus one all over six in cube. That's the first fraction. For the second fraction, we're going to end up with a three n over n like so. And so we want to take the limit as n goes to infinity. Now you could try to simplify these fractions a little bit. Because after all, there's like a common factor of n right there. There's a three that goes into a six, this n kind of cancels out with this one right here. You can do all those things. I'm gonna I'm gonna wait to do that because as n goes to infinity, when you have a rational function, all that matters is the leading terms. So when you look at this, the first fraction, the the six over n cube to denominator, this is our dominant term. But who's the dominant term on top? Well, you would take the product of all the leading terms. So three n times n times two n on top. And so what we're going to see is that as n goes to infinity, this rational expression is no different than just three n cubed over six in cubed. And then the second term, as you take the three n over n, well, there's those are just monomials already as it is. So you can ignore all the fluff and only look at the dominant terms right here. So as n goes to infinity, this thing can be these things can now simplify down right the n cubes cancel the ends cancel, and we end up with three sixth plus three. Three six, of course, is the same thing as a one half. Oh, I'm sorry. I made a mistake. I got to correct this before anyone punches me in the face or something. When we did this one over here, this wasn't just an end, this was a two n. So we got we get three n times n times two n, which actually makes us a six n cube. Sorry about that. And so that gives us a six right here. Six over six is going to become a one, we get one plus three, which is equal to four. And so the limit of that sum from the very beginning is equal to four. Let's come back to this thing. This thing is equal to four. All right, I see how that calculation works. But what is this thing? Right? What does that even measuring? Well, young grasshopper, we will have to learn that next time before we're ready to battle the Cobra Kai. Stay tuned. For for upcoming lectures, we'll she'll give some answers to the questions I posing at the end of our lecture right here. If you did like the videos you've been watching, feel free to subscribe from further updates, post some comments, like it if you do. If you have any questions, feel free to post them in the comments this time in this video, but all the videos, I'll see your comments and I'll try to respond to them in a quick manner. If you have any questions, let's learn some calculus together. See you next time.