>> Now, these are summation formulas that we're going to talk about, and because we're going to be adding a lot of little, a lot of little rectangles actually. A lot of little areas of rectangles, I should say. But just look at the notation here. This is, this notation is upper case Sigma, the Greek letter Sigma, and it stands for the sum of items described using the notation. Now, I is just a counting mechanism, and this says that I goes from 1 to N. So the, a counting mechanism is going from one to N and C is a constant. And it says that the sum as I goes from one to N of this constant is C times the upper limit of this, of this summation. Now, to, to give you an idea of what this means is, is this, that we're adding this number that many times. That's all there is to it, you see, because this counting mechanism says to add according to what we have in this expression according to the number of, the number associated with the counting mechanism. So we're going to add C for the, the first I, and then we're going to add another C to the second I, and when I is equal to 3, we're going to tack on another C and so on. And so we get this. Now, it'll make more sense if I put a number in here. Suppose I runs from 1 to 10, and we're, we're adding 7's, you see. So it's just, well, when I is 1, we, we write a 7. When I is 2, oh, we write another 7. And when I is 3, we write another 7. You see, and so the result of all of this is that we'll have ten 7's up here. So it's going to be nothing more than 7 times 10. So it's 7 times this upper limit business. Now, in this one, we have same notion with this summation notation, but it's a constant times X subscript I. Now, the subscript I means that we're just going to apply the counting mechanism as the subscript on, on X. So the first few items will be C times X subscript 1. See, the counting mechanism is starting at 1, and it's going to N. So it's C times X subscript 1 plus C times X subscript 2 plus C times X subscript 3 and so on all the way through to C times X subscript N. Now, what we notice in this is that we have C as a common factor throughout all of these guys. So we C, then, can be taken out its common factor, and all of this is nothing more than the summation as I runs from 1 to the N of X subscript I. Now, to bottom line the idea, we do this. That when we have a constant involved in this situation of constant times, some expression having to do with that counting mechanism, we can just kind of extract that constant, you see, and write the constant outside like this. OK. Now. Here's another situation. We have the summation of two items that contain a counting mechanism, and we can just kind of sort of distribute the idea of summation. That is, the sum of these two, and this could be plus or minus, by the way, but it's the sum of, of this first one plus or minus the sum of the second. Alright. No big deal. Now, these next three are really important. This one says the sum is I runs from 1 to N of simply I. So we're just adding the accumulation of this counting mechanism, and, and when I is 1, you see, we, we're adding 1. When I is 2, we add a 2. When I is 3, we add 3, and then all the way through to N. Well, this is nothing but more than a, a, an arithmetic series, and we've talked about a formula for adding items in an arithmetic series before, and, and was N over 2 times ace of 1 plus ace of N. What, well, wait a second. Ace of 1 is 1 here, and ace of N, the last item is, is N, and so this is just 1 plus N, and the way it's written here is N times N plus 1. This part is 1 plus N. I can write it as N plus 1, and all of this is over 2. So there's our formula, then, for this situation. Now, there's a formula for the situation of, of adding the I squared, and it's this. Now, I don't expect you to commit this to memory right away, just kind of refer to it a few times, and when you use it a few times, it'll, it'll come to you rather naturally. But, and here we have the summation as I runs from 1 to N of I cubed, and here is the formula associated with that.