Our first case consists of those sequences, which are obtained by adding a constant number d to obtain subsequent elements:

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1 Week 13 Sequences and Series Many images below are excerpts from the multimedia textbook. You can find them there and in your textbook in sections 7.2 and 7.3. We have encountered the first sequences and series last week. This week we will study two different cases in detail. We will be able to compute the values of the individual terms and the value of the partial sums in both cases. For the geometric series we will even be able to compute the values for certain infinite sums. Arithmetic Sequences Our first case consists of those sequences, which are obtained by adding a constant number d to obtain subsequent elements: Work through Example 1 in section 7.2 starting on page 479.

2 Work through Examples 2 and 3 in section 7.2 on page 480. Discussion Question 1: Consider the arithmetic sequence 7, 11, 15,.. a. Find the formula for the n th term. b. Find the 12 th term of the sequence. c. Which term equals 131? Notice that the notation is hiding a bit of what we already know. An arithmetic sequence is simply a linear function. So far we have denoted a linear function by y = mx + b. Comparing this to the formula for the sequence here, we see that for the sequence we require n to be greater than 1 whereas x can be any real number in particular x = 0. Identifying x with (n-1) we notice that the formula a n = a 1 + d(n - 1) looks like y = b + mx. Now d is the slope and a 1 corresponds to the y-intercept. The biggest problem in chapter 7 for us will be to get used to the new notation and to see the old concepts in the disguises of the different way of writing. Do the Interactive Discovery in section 7.2 on page 481. How does the graph of an arithmetic sequence look? Arithmetic Series We now will compute the sum of the elements of an arithmetic sequence. That's what we call an arithmetic series. There is actually a formula to compute that sum easily. To understand how that formula works let's consider the sum of the following 5 elements: 1, 3, 5, 7, 9. Since this is just a sum of 5 elements we can easily compute it directly. But let's use it as an example. The trick is that we can write the sequence down twice, at first in the usual order

3 and then again in reverse order. Now if we add up matching elements we obtain the same sum in each case: First sequence Backwards Sum of both Now it is easy to add the total sum of both: 5 (10) = 50. But since the original sequence only contributes half of the elements we need to divide this by 2 and be obtain a sum of 25. The formula for the first n terms of the sequence is obtained in a similar way. You find the explanation in the text on page 482. Work through examples 5 and 6 on page 583. Discussion Question 2: Find the sum of the first 50 terms of the arithmetic sequence 1, 3, 5,. Last week we introduced sigma notation to denote sums of sequences. The sum we are asked to find in discussion question 2 can be written as: (k 1). Notice we could also write it as 2k 1. One of our challenges with computing sums of sequences will be to realize that a given formula for a general element represents an arithmetic sequence. In both notations we see that k appears as a linear term. That is the criterion to alert us that we are dealing with an arithmetic sequence. 50 Work through example 7 on page 584. Discussion Question 3: Find the sum: k Watch the video for section 7.2.

4 Geometric Sequences We are now turning to the second case of special sequences we are studying. In contrast to the arithmetic sequences, which are obtained by adding a constant term, a geometric sequence is obtained by multiplying a starting element consecutively with the same number r. This leads to the elements having a common ratio. We find now that the formula for the general term of such a sequence is an exponential expression in n. Work through Examples 1, 2 and 3 in section 7.3 starting on page 489. Discussion Question 4: Consider the geometric sequence 3, 9, 27.. a. Find the formula for the n th term. b. Find the 12 th term of the sequence. c. Which term equals 59049? Do the Interactive Discovery in section 7.3 on page 490. How does the graph of a geometric sequence look? We saw that we can identify the arithmetic sequences with linear functions and now we experience that the geometric sequences correspond to exponential functions.

5 Geometric Series Again we can find a formula to compute the sum of the first n elements of a geometric sequence. The formula is developed on page 491. Notice the r cannot equal 1. That is not a big restriction, because that would really be an arithmetic sequence, since all elements would be identical, and in that case we can find the sum easily through direct computation. Work through Example 4 in section 7.3 starting on page 491. Discussion Question 5: Consider the geometric sequence 3, 9, 27.. Find the sum of the first 10 terms. Again we have to adjust when we express the same ideas in sigma notation: Work through example 5 on page 592. Discussion Question 6: Find the sum: 9 (0.9) k You can already anticipate a problem here. The last question looks just like the expression in discussion question 3, but now we are to use a totally different approach. The key difference is to realize that the term (0.9) k is an exponential expression and that alerts us that we are dealing with a geometric sequence. We also notice in the last problem that the expression (0.9) k approaches 0 as k gets large. This means that if we add up many terms in the series the later terms add less and less to the whole sum. In fact if the terms get small "fast enough" we can even "add" infinitely many terms. Let's consider the following example. Imagine a square sheet of paper with side of length 1 meter. Now cut the paper in half, put the one half to the side and cut the remaining half again in half. As before put one of the

6 halfs to the side and cut the other half in two equal pieces. Assuming that we can keep on doing this we get the following: After k steps the size of the k th piece of paper we are putting to the side k 1 is 2. Adding all the pieces that pile up on the side will obviously give us never more than the original sheet of 1 square meter and the longer we are adding the closer to the total of 1 we will get. So, imagining that we can cut the paper for infinitely many steps, we can say that: k 1 2 =1. To be exact we are talking of a limit, because we can never actually add these infinitely many pieces of paper. In fact after a few steps we will have pieces, which are smaller than a molecule and we can't even talk about them being "paper" any more. But in the setting of mathematics this thought experiment makes sense and illustrates an important point. It makes sense to talk about the sum of infinitely many terms, as long as the end result does not change very much when "later" elements are added. If we had doubled the pieces of paper, there would have been no chance of getting a sum of infinitely elements, because the size of each subsequent piece would grow very large. The amazing thing is that we can actually compute the value of such infinite geometric series by a simple formula. If we consider the formula for the finite sum S n = a (1 1 rn ) we notice that the term r 1 r n goes to 0 when n grows large, provided that the absolute value of r is less than 1. We get the formula for the infinite sum by just replacing the r n by 0. We can apply this formula to our paper problem: 1 1 k 1 2 = = =1

7 It is very important to remember that the formula only applies when the absolute value of r is less than 1.