Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Sequences, numbers that have something of a pattern. So a sub one, a sub two, a sub three, a sub four..., a sub N... Finite sequence means that the sequence terminates or it has a last term. Infinite sequence goes on, so a sub 1A sub two... all the way out to Infinity. A sub N is the general term or the NTH term. N factorial means starting with the number N we're going to subtract 1 and multiply by every single term until we get down to 1. So n * n - 1 * n - 2 * n - 3 dot dot .321. So if I had five factorial, that would be 5 * 4 * 3 * 2 * 1 or 120. If I had two factorial, that would be two times one or two. One factorial is just one, and we're going to need to introduce 0 factorial and that's going to be by definition A1. Also, we have summation notation, and a summation notation is used when we're adding the terms of a sequence. That's called a series. And the way that a series looks is we're going to add the terms. In this case the I is called our index. The one is our lower limit, the N is our upper limit. So a sub I is going to be a sub 1 + a sub 2 + a sub three plus... Out to a sub N, if we have the summation of I equal 1 to north of Cai, we're going to come up with what that's actually going to be equivalent to. So if we thought about writing out our terms for this, we'd have CA sub one plus CA sub two plus CA sub three plus all the way out to CA sub N. This is going to be finite because we have an ending spot. We can see that all of these terms have AC in common. So I'm going to pull out AC and I'm going to get a sub 1 + a sub 2 + a sub three plus... plus a sub N. And a moment ago we said that this is really just the summation of I equaling one to north of a sub I. So one of our properties is going to be if I goes from 1:00 to north of California sub I, we can pull that constant in and out of the summation. So if we look at an example, if I have the summation of I equal 1 to six of five, I, I'm going to pull out the five. I'm going to get I equal 1 to six of I. So we're going to have 5 * 1 + 2 + 3 + 4 + 5 + 6, and that's going to give us five times 21 or 105. If we look at the summation of I equal 1 to north of a sub I + B sub I, that's going to be a sub 1 + b sub one plus a sub 2 + b sub two plus... plus a sub n + b sub N. Now we know that we can add in any order we want. So I could group all those A's together and I could also then group all those B's together. Well by our definition that a sub one through a sub N is just I equaling one to north of a sub I and the second one is just our summation of I equal 1 to north of B sub I. So we just have developed another property that says the summation of I equal 1 to N of a quantity AI plus BI could be thought of as the summation of each of those terms separately, doing a similar type of thing for the next one. If we have a sub 1 -, B sub 1 + a sub 2 -, B sub 2 plus a sub n -, b sub N, we can group all the A's together, and then we're going to group all the B's together. But we're going to also factor out a -1. If I take out a negative in front of all those subtractions, it turns those all into addition. So now we have the summation of I equal 1 to north of a sub I minus the summation I equal 1 to north of B sub I. So this is I equal 1 to north of the quantity a sub I -, b sub I just equals those two individually subtracted. If we have a constant, our first term is going to be C, Our second term is going to be C our third term is going to be C, our NTH term is going to be C How many of these do we have total? We have N of them, so we're going to have n * C If you don't see that, you could think about factoring out AC from each of these terms, which would give us N ones added together. So if we have the summation of I equal 1 to north of C, that's just C * n OK. If we have a sub N of three n + 2, we're going to ask to be found. We're going to ask to find a sub one. We literally just stick one in for our N. So our A sub one would be 5. If you were asked to find a sub two, we're going to stick 2IN for our N. So a sub 2 is going to be 8A sub three we would stick 3IN for our N and get 11 this next one. A sub one would be -3 to the 1st or -3. A sub two would be -3 ^2 or +9. A sub three would be -3 ^3 or -27. The last example of this type, if we wanted a sub one, we're going to put a -1 to the 1 + 1 over 2 to the 1 - 1. 1 + 1 would give us -1 ^2, which would give us a +1 - 1 is 0. Anything other than 0 to the zero is 1, so our a sub one is just one. If we were asked to find a sub two negative one 2 + 1 / 2 to the 2 - 1 so -1 to the third power is going to be a negative and two to the 2 - 1 is going to be 1/2. So those are some examples of finding the first terms in a sequence recursive formula. A recursive formula means we have to know the previous term to get to the next term. So in this example, I know a sub one equal 5. So if I wanted a sub 2A sub two would be 3 * a sub 2 - 1 - 1. Well, that's really 3A sub 1 - 1 or 3 * 5 - 1 or 14. If I wanted a sub three, I'd have three times a 3 - 1 - 1 or a sub three would equal 3A sub 2 - 1. So now we had to know what our A sub 2 was and that was 14. So we're going to get 42 - 1 or 41. If I wanted a sub four, I'd have 3A sub 4 - 1 - 1. A sub four would equal 3A sub 3 - 1. The A sub three from a moment ago was 41, so 123 -, 122 factorials factorial. Remember we start with the number and we count down. So if I have 20 factorial, that's 20 * 19 * 18 * 17 * 16 * 15... Dot dot all the way down to 3 * 2 * 1. In the denominator we have 3 * 2 * 1 * 17 * 16 * 15 * 14 times... dot .321. We can see here that the 17 factorial on top and the 17 factorial portion on bottom could cancel 3 * 2 is 6 and six would go into 18 three times. So now we'd end up with 20 * 19 * 3 or 0 for 1140. Another way to do this, or to think of notation, is to think about 20 factorial being 20 * 19 * 18 * 17 factorial over 3 * 2 * 1 times the 17 factorial. And then you can see that those 17 factorials cancel and we would still reduce the three and the two go into 18, leaving a six. No, leaving a three. And then what I did was 3 * 20 to get 6060 times nineteen 1140. This next one we're going to have 18 * 17 * 16 factorial. I stopped with the largest number in the denominator because those are going to cancel. So the 16 factorial on top cancels with 16 on bottom. The two goes into the 18 nine times, so 9 * 17 153. The next one we're going to write a summation. So if I have 1 ^2 + 2 ^2 + 3 ^2 plus... Plus 15 ^2. I could say I is going to equal 1 to 15 of I ^2. Now we actually could have lots of different possibilities for answers here. We could have had I = 0 to 14 of I -, 1 ^2. We could have had I equal 2 to 16. Oh, the last one would have been I + 1 I -1 here, because 2 - 1 would be 1 ^2. We could have had I equal 5 to 19 of I - 4 ^2. So there are infinitely many summations. Usually one is more obvious than the rest. If we look at the next example, if we let I equal 1 to 14, we'd have I / I + 1, or we could say I equal 2 to 15 and we would have I -, 1 / I. We could do lots of other possibilities. If we wanted I equaling zero, we would have zero to 13 of I + 1 / I + 2 etcetera. Thank you and have a wonderful day.