Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Sigma notation is the summation with the index starting at K equal 1 and the index ending at K equal N of some a sub K The formula for the KTH term K starts at K equal 1. In this case, K doesn't have to equal 1, it could be any number. So if we have the summation of K equal 1 to north of a sub K, what it really means is a sub 1 + a sub 2 + a sub three plus... plus a sub n -, 2 + a sub n - 1 + a sub N. So if we have the summation of a sub K + b sub K, that would really be A1 plus B1 plus A2 plus B2 plus a 3 + b three... Plus AN plus BN. So grouping all those A's together A1A2 plus... plus AN grouping all the BS together B1 plus B2 plus... plus BN. We know that when we combine all the A's, that's really just the summation of K equal 1 to north of A sub K plus the summation of K equal 1 to north of B sub K. We can do the same thing with subtraction. If we thought about doing the summation of A sub K -, B sub KA 1 -, B one plus A 2 -, B two plus all the way out to AN minus BN, grouping all the A's together, factoring out a negative to make all the B's turn positive, and then grouping all the B's together, we can see that that's really just the summation of K equal 1 to north of a sub K minus the summation of K equal 1 to north of B sub K. We have a couple other ones. If we thought about C times ace of K, if we had the summation of K equal 1 to north of CA sub K, so we'd have CA sub one plus CA sub 2 plus... dot plus CA sub K or say a sub N Sorry, if we then factored out AC from all those, we'd get a sub 1 + a sub 2 plus... Dot dot plus a sub N and that portion is really just the summation of K equal 1 to north of a sub K. We also have the summation of K equal 1 to north of C and when K is one we have C When K is 2, we have C when K is 3, we have C, when K is 4, we have C when K is five, etcetera. When K is N we have C. So how many of these do we have total? We have N of them or n * C when we look at summations of K equal 1 to north of K if we write it in ascending order 1 + 2 + 3 plus... all the way out to north. And then if we turn around and write the exact same summation but in descending order NN -1 N -2 + 4 + 3 + 2 + 1. If we thought about adding those two together and we looked at it term by term, we would see that the one and the n + 1 would go together to give me an n + 1 and the next term the two and the n -, 1 will go together to give me n + 1. And with the next term, the three and the n -, 2 would go together to give me n + 1. And if we looked all the way down at the end, the N and the one would give me this n + 1. The n -, 1 and two would give me the next n + 1. So we'd actually have NN plus ones to get the summation by itself. We would then divide each side by two. Because if we add the left side together and we add the right side together, if I have one of them plus another one of them, I have two of them. But I wanted to solve for the summation, so I'm going to divide each side by two. So we're going to have the summation of K equal 1 to north of K equaling N times north plus 1 / 2. So the summation K equal 1 to north is is that quantity. We have two more. We have the K equal 1 to north of K ^2, which is n * n + 1 * 2 M plus one all over six. It's a little more complex than I'm going to show at this point. K equal 1 to north of K ^3. It's the quantity of the n * n + 1 / 2 ^2. If we look at our K and our K ^3, they're almost identical, we just square 1. So if we look at the idea of a Riemann sum next, a Riemann sum takes partitions and the norm of a partition. Our new notation is going to be things that look like absolute values, but two of them. That P is the largest of all the sub interval widths. So now we're going to do the rectangular concept to find the area under the curve. But we can make the rectangles different widths now. So we're going to take the largest one and force the largest one to go to 0. If the largest goes to 0, we know that all the rest have to also go to 0. So a norm of our partition is the largest of all the subinterval widths where we're trying to force that particular piece to go to 0. If we look at some examples, if we thought about the summation of K equal 1 to three of K - 1 / K, we stick in one, then we stick in two, then we stick in three. So 1 - 1 / 1 is 0 + 2 - 1 / 2 is 1/2 + 3 - 1 / 3 is 2/3, We'd get 7 sixths. If we went K equal 1 to three of the quantity 1 - 1 / K, we'd have K equal 1 to three of 1 - K equal 1 to three of 1 / K So the first one 1 + 1 + 1 would give us 3, the second one 1 / K So 1 / 1 + 1 / 2 + 1 / 3 would get 3 - 11 six or seven six. Continuing on with another example, if we have 1 - 2 + 4 - 8 + 16 - 30, two 1 - 2 to the first plus 2 ^2 - 2 ^3 + 2 to the 4th -2 to the 5th, we could think of one as two to the 0. So 2 to the 0 -, 2 to the first plus 2 ^2 - 2 ^3, etcetera. Now, if we thought about counting these numbers off, if I thought about letting this be the number one, then two, then 345 and six. RK would start at one and go to six. The -1 has to be positive for the first term, so I'd need it to an even power. 1 + 1 would give me two, so an even power would make the first term +2. I need it to the 0 power, so if I put in one for K, 1 -, 1 would give me 0. So that's a possibility. There are lots of possibilities. If I thought about starting with 0 here and counting 1234 and 5K would start at 0 and go to five. And now we need the first term to be positive. So -1 to the zero power, anything to the 0 powers, one except for 0 to the zero, which is undefined. 2 to the zero power is 1. So our first term would be that one. If we thought about our next term to the first power, we'd get a negative from the -1 to the 1st, and two to the first would be two if we wanted to start counting with something like a -2. There are truly infinitely many possibilities here. If we started with -2 and counted, we'd go from -2 to 3. We need the first term to be positive, so I need the -1 to be to an even power -1 to the -2. The negative just means to take the reciprocal, and two -1 ^2 is still going to be a +2. I need the first exponent to be 0, so -2 + 2 would give me zero. We could also think about pulling the -1 inside if the exponents were the same. So -1 to the K * 2 to the K is really the same thing as -2 to the K. If we look at just one or two more examples, say we wanted to come up with K equal one to seven of -2 K, we know that that -2 is a constant, so it could come in or out. So we have K equal one to seven of K. So that would be -2. Now we're going to use our formula n * n + 1 all over 2. So the twos will cancel, leaving us a negative, and we're going to get 7 * 8. So -56 there. If we wanted to look at that, we could write it out. We could say when K is one, we get -2 when K is 2, we're going to get +4. When K is three, negative 6K is 4. These are all going to be negatives. I'm sorry -2 negative four -6 negative 8 -10 negative 12 and -14 we add those up, we should get -56 there's negative 20304456. So -56 Thank you and have a wonderful day.