Arithmetic Series. or The sum of any finite, or limited, number of terms is called a partial sum of the series.

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1 CONDENSED LESSON 11.1 Arithmetic Series In this lesson, yo Learn the terminology and notation associated with series Discover two formlas for the partial sm of an arithmetic series A series is a sm of terms in a seqence. For example, consider the seqence 4 2 where n 2 The sm of the terms in this seqence is the series or The sm of the first n terms in a series is represented by.for example, S The sm of any finite, or limited, nmber of terms is called a partial sm of the series. The notations S 6 and 6 are shorthand ways of writing n To find the sm of the integers from 1 to 100, yo cold add the terms one by one. Yo can se a recrsive formla and yor calclator to do this qickly. First, write a recrsive rle for the seqence of positive integers. 1 1 where n 2 The recrsive formla for the series is then S where n 2 This formla states that the sm of the first n terms is eqal to the sm of the first (n 1) terms, pls the nth term. Becase 1, yo can rewrite this as S where n 2 Enter the two recrsive formlas into yor calclator. The table shows each term in the seqence and the seqence of partial sms. The points on the graph represent the partial sms S 1 throgh S 100.Yo can se either the table or the graph to find that S 100, the sm of the integers from 1 to 100, is [0, 110, 10, 3000, 10000, 1000] (contined) Discovering Advanced Algebra Condensed Lessons CHAPTER

2 Lesson 11.1 Arithmetic Series (contined) In the investigation yo will find a formla for finding a partial sm of an arithmetic series withot finding all the terms and adding. Investigation: Arithmetic Series Formla The investigation in yor book asks yo to select three integers between 2 and 9. Below, yo will se 3, 6, and 7. Step 1 Use 7 as the first term of the seqence and 3 as the common difference. Write the first ten terms of the seqence and the first ten partial sms of the corresponding series. Seqence: {7, 10, 13, 16, 19, 22, 25, 28, 31, 34} Partial sms: {7, 17, 30, 46, 65, 87, 112, 140, 171, 205} Step 2 Use finite differences to find the degree of a polynomial eqation that wold fit the points n,.yo shold find that the second differences are constant, indicating that a polynomial in the form an 2 bn c fits the points. Sbstitting (1, 7), (2, 17), and (3, 30), yo can write the system a b c 7 4a 2b c 17 9a 3b c 30 Solve this system to find a, b, and c. Yo shold get a 1.5, b 5.5, and c 0. Therefore, the polynomial 1.5n 2 5.5n fits the data. Steps 3 and 4 Create a new series by exchanging either the first term or the common difference with another one of the three integers 3, 6, and 7. Find a polynomial that fits the series. This table shows the reslts for all the possible first terms and common differences. Step 5 In the table at right, look for a relationship between the coefficients of each polynomial and the vales of and d. Yo shold find that the coefficient of n 2 is half the common difference and that the coefficient of n is the first term mins half the common difference. So, yo can write the explicit formla d 2 n2 d 2 n Yo can se this formla to find the partial sm of any arithmetic series. Common First term difference Partial sm d n 2 5.5n 7 6 3n 2 4n n 2 3.5n 3 6 3n n 2 0.5n n 2 1.5n n 2 4.5n n 2 2.5n 6 6 3n 2 3n Use the formla from the investigation to verify that the sm of the integers from 1 to 100 is Then read the example in yor book, which shows another way to find this sm. The method of the example can be extended to derive another formla for the nth partial sm of an arithmetic series n 2 where n is the nmber of terms, is the first term, and is the last term. 176 CHAPTER 11 Discovering Advanced Algebra Condensed Lessons

3 CONDENSED LESSON 11.2 Infinite Geometric Series In this lesson, yo Learn that some infinite geometric series converge to a long-rn vale, or sm Discover a formla for finding the sm of a convergent geometric series Find the sm of a geometric series by sing a graph of partial sms In Lesson 11.1, yo fond partial sms of arithmetic series. As the nmber of terms, n, of an arithmetic series increases, the absolte vale of the partial sm,, increases in the long rn. So, the sm of an infinite nmber of terms of an arithmetic series is infinite. In this lesson, yo will discover that this is not always the case with geometric series. A geometric series is the smmation of terms in a geometric seqence. For example, consider the geometric seqence 1 2, 1 4, 1 8, , 6 3, 2 6, ,... This series has a constant ratio of 1, 2 so the terms get smaller and smaller. Yo can add the terms to create a geometric series. Here are some of the partial sms: S S S If yo contine to find partial sms, yo will get 3 3, 6 2 6, , 8 and so on. Althogh the partial sms get larger and larger, they are always less than 1. It appears that if yo add an infinite nmber of terms, the reslt will not be infinite. An infinite geometric series is a geometric series with an infinite nmber of terms. A convergent series is a series for which the seqence of partial sms approaches a long-rn vale as the nmber of terms increases. This long-rn vale is the sm of the series. The series is a convergent series with a long-rn vale, or sm, of 1. Investigation: Infinite Geometric Series Formla The investigation in yor book asks yo to choose three integers between 2 and 9. Below, yo will se 7, 8, and 9. 1 Step 1 Using 7 as the first term in the seqence and 1 0 of 8, or 0.8, as the common ratio gives the recrsive formla where n 2 The seqence of partial sms for the corresponding series is defined by S where n (contined) Discovering Advanced Algebra Condensed Lessons CHAPTER

4 Lesson 11.2 Infinite Geometric Series (contined) Enter these formlas into yor calclator and find S 400 and S 500.Yo shold find S and S , so the long-rn vale, or infinite sm, is 35. Create a new series either by exchanging 7 for 8 or 9 or by exchanging 0.8 for 0.7 or 0.9. Find the infinite sm for the new series. Step 3 The table at right shows the infinite First term Common ratio Sm sm, S, for every possible combination. The r S S 1 last colmn shows the ratio of the first term to the infinite sm Step 4 Look for patterns in the table. Notice that in each case, S 1 1 r. This can be rewritten as S 1. 1 r For the series yo have considered so far, r If r 1, the formla doesn t work becase 1 1 r is ndefined. An example of sch a series is The seqence of partial sms 2, 4, 6, 8, 10,... increases withot bond, so the series does not converge If r 1, then the partial sms get larger and larger as more terms are added. For example, consider ,which has a constant ratio of 2. The seqence of partial sms 3, 9, 21, 45,...increases withot bond, so the series does not converge. For this example, the formla gives S 3, so it clearly does not work. Read Example B in yor book, in which a graph of partial sms is sed to find the sm of a series. Read the example careflly and make sre yo nderstand the method. Then, read the box after that example, which smmarizes the formla for finding the sm of a convergent infinite geometric series. Note that a geometric series converges only if r 1. Finally, read Example C. Here is another example. EXAMPLE Soltion Find the sm of the infinite series 130(0.84) n 1 n 1 In this case, r 0.84 and 130. Using the formla, 130 S CHAPTER 11 Discovering Advanced Algebra Condensed Lessons

5 CONDENSED LESSON 11.3 Partial Sms of Geometric Series In this lesson, yo Discover a formla for partial sms of geometric series In Lesson 11.2, yo fond sms of convergent geometric series. In this lesson, yo will find partial sms of geometric series. Example A in yor book shows yo how to se a calclator graph to find partial sms of a geometric series. Read the example careflly. In Lesson 11.1 yo discovered a formla for partial sms of arithmetic series. In this investigation, yo ll find a formla for partial sms of geometric series. Investigation: Geometric Series Formla Choose two integers between 2 and 9. Let one integer be the starting vale of a geometric seqence, and let one-tenth of the other integer be the common ratio. Work throgh Steps 1 3 of the investigation in yor book. The reslts below se a seqence with starting term 4 and common ratio 0.6. Step 1 The seqence is defined by 4 and where n 2. Here are the first ten terms of the seqence and the first ten partial sms of the corresponding series. Seqence: {4, 2.4, 1.44, 0.864, , , , , , } Partial sms: {4, 6.4, 7.84, 8.704, , , , , , } Step 2 Here is a graph of the data points n, : [0, 10, 1, 0, 12, 1] The eqation that fits these data is in the form L ab n,where L is the long-rn vale. Using the formla from Lesson 11.2, L r To find the vales of a and b, sbstitte the coordinates of the points (1, 4) and (2, 6.4) into 10 ab n to get the system 4 10 ab ab 2 (contined) Discovering Advanced Algebra Condensed Lessons CHAPTER

6 Lesson 11.3 Partial Sms of Geometric Series (contined) Yo can rewrite these eqations as ab 6 and ab Dividing the second eqation by the first gives b 0.6. Sbstitting 0.6 for b in the first eqation gives a 10. So, the eqation is 10 10(0.6) n. The eqation fits the points. Step 3 The eqation in Step 2 is in the form 1 1 r 1 r 1 r n. 1 r n This is eqivalent to. [0, 10, 1, 0, 12, 1] (1 r) Step 4 Will the formla work if the common ratio is greater than 1? Test it by considering the seqence 2 and 2(3) n where n 2, which has a common ratio of 3. Here are the first ten terms of the seqence and the first ten partial sms of the corresponding series: Seqence: {2, 6, 18, 54, 162, 486, 1458, 4374, 13122, 39366} Partial sms: {2, 8, 26, 80, 242, 728, 2186, 6560, 19682, 59048} Step 5 Applying the formla from Step 3 to this seqence gives r 10 S 10 (1 r) This is the correct vale of S 10, so the formla seems to work for r 1. Now yo have an explicit formla for finding a partial sm of any geometric series. Yoeed to know only the first term, the common ratio, and the nmber of terms. To practice sing the formla, solve the problems in Examples B and C in yor book. Then read the example below. EXAMPLE Find each partial sm. a. 11 9(2.75) n 1 n 1 b Soltion a. 9 and r Use the formla for the partial sm S r 11 S 11 (1 r) b. The first term,, is Each term is three-forths the previos term, so r Enter 1024 and into yor calclator and make a table. The last term given, , is 8.So yoeed to find S 8.Using the formla, r 8 S 8 (1 r) CHAPTER 11 Discovering Advanced Algebra Condensed Lessons