CCA2_PG_ch10.pdf - SERIES 10.1.1 \u2013 10.1.4 This chapter revisits sequences\u2014arithmetic then geometric\u2014to see how these ideas can be extended and how

CCA2_PG_ch10.pdf - SERIES 10.1.1 – 10.1.4 This chapter...

This preview shows page 1 - 3 out of 10 pages.

116 © 2009, 2014 CPM Educational Program. All rights reserved. Core Connections Algebra 2 SERIES 10.1.1 – 10.1.4 This chapter revisits sequences—arithmetic then geometric—to see how these ideas can be extended, and how they occur in other contexts. A sequence is a list of ordered numbers, whereas a series is the sum of those numbers. Students develop methods for finding those sums and learn a compact way to write the sums, known as summation notation. These ideas are extended with explorations into Pascal’s Triangle, the Binomial theorem, and natural logarithms. For more information, see the Math Notes boxes in Lessons 10.1.2, 10.1.3, and 10.1.4.Example 1 Find the sum of each of the following series. a. 4 + 9 + 14 + 19 + 24 + … + 59. b. Twelve terms with t(1) = 3 and t(12) = 69. c. t(n) = –3n+ 10, for integer values starting at 1 and ending at 15. Each of these problems represents an arithmetic series because there is a constant difference between each consecutive term. They are series because they are the sums of the terms. For part (a), we could simply add each term, filling in the terms represented by the “…,” but that would take some time and there is a good chance for an arithmetic error. Instead, we will use one of the methods developed in the chapter. We will need to know the formula for the nthterm of this sequence, as well as which term the number 59 is. The formula for the nthterm is t(n) = 5(n– 1) + 4 (verify this!) and by setting this equal to 59, we find that 59 is the 12thterm. To find the sum, we write out the sum labeling it S. We repeat this by writing it in reverse order. Adding these two equations gives us a new equation that makes it easier to solve for S. The sum of these twelve terms is 378. S=4 +9+14+19+24+...+59+S=59+54+49+44+39+...+42S=63+63 +63+63+63+...+6312of these2S=12(63)S=378In part (b) we determine the formula for the nthterm by using the given information to write an equation and solve. This gives us the formula t(n) = 6(n– 1) + 3. Knowing the first and last terms is enough information to use the method of the previous example to find the sum. The method generalizes as follows: add the first and last terms, multiply the result by the number of terms in the series, then divide by two. t(1)=3t(12)=69t(n)=d(n1)+t(1)So: t(12)=d(121)+369=11d+366=11dd=6S=12(3+69)2=432
Chapter 10 Parent Guide with Extra Practice © 2009, 2014 CPM Educational Program. All rights reserved. 117 In part (c) we have the formula for the nthterm and we know there will be 15 terms. We will calculate the first and the last term, and use them in the same procedure as above, using S(n)=n t(1)+t(n)()2.

You've reached the end of your free preview.

Want to read all 10 pages?

• Fall '16
• Joseph Begeny
• Summation, Geometric progression, CPM Educational Program