Section 3.1: Sequences and Series

Transcription

1 Sectio 3.: Sequeces d Series Sequeces Let s strt out with the defiitio of sequece: sequece: ordered list of umbers, ofte with defiite ptter Recll tht i set, order does t mtter so this is oe wy tht sequece differs from set. Also, repetitio does t mtter i set but does i sequece: if umber is repeted i sequece, it is t cosidered duplicte d cot be removed without chgig the sequece. Sequeces, like sets, c be fiite or ifiite. If sequece is fiite, the either the lst term or the umber of terms must be specified so tht it s cler where the sequece stops. Which of the followig sequeces re ifiite? Which re fiite? ) 7,, 5, 9, b), 4, 9, 6, 5, 36, 00 c) 4,,,,,,, b) d c) re fiite, becuse their lst terms re give. ), however, goes o forever so is ifiite. To begi with, let s exmie some sequeces i detil. We will begi by lookig for ptters i ech sequece. Wht is the ptter for the followig sequeces? Wht is the ext term for ech sequece? ) 7,, 5, 9, b), 4, 9, 6, 5, 36, 00 c) 4,,,,,,,

2 d) 3, 6,, 4, e) 3, 6, 5, 4, ) The ptter is tht you dd 4 to the previous term to get the ext term. The ext term is the 3. b) The ptter is tht if you sy tht is the first term d 4 is the secod term, the will be the th term. So the ext term fter 36 is 49. c) The ptter is to divide ech term by two (or multiply by hlf) to get the ext term. So the term fter /6 will be /3. d) The ptter is to multiply ech term by to get the ext term. The ext term is the 48. e) The ptter is to subtrct 9 from the previous term, so the ext oe is 33. Note tht i this previous exmple, the lst two sequeces looked very similr for three of their first four terms. However, the third term is differet so the ptter for the two sequeces is ot the sme d subsequet terms could look very differet. Nottio We will use the ottio for the th term i sequece, where is the idex. For exmple, the first term would the be, the secod term, d so o. The idex, the, is positive iteger (or turl umber, if you like). Other ottios my strt their coutig with o beig the first term. For the purposes of this course, we ll stick to strtig t =. Defiig Sequece There re three wys to defie sequece: ) List ll of the terms, or eough terms to set up the ptter. If the sequece is fiite, the either the lst term or the umber of terms must be give. ) Give geerl formul for the th term. 3) Give recursive formul for the th term. Let s look t exmples of ech type. For istce, the sequeces 7,, 5, 9, d, 4, 9, 6, 5, 36, 00 re exmples of sequeces defied by listig the terms.

3 Geerl Formul A geerl formul is formul tht gives s fuctio of oly. Let s look t the followig exmples to exmie some sequeces defied i this wy. Give the first four terms of the sequece give by the geerl formul 4 3. = 4 + 3, so = = 7 = = 3 = = 5 4 = = 9 The first four terms re the 7,, 5, d 9. This is the sme sequece tht ws give s prt ) i the first two exmples of this sectio. Give ll terms of the sequece give by the formul for 5. 3 This is fiite sequece, sice restrictios hve bee plced o the vlues of. The terms re the:

4 You c see from the previous exmples tht the geerl formul llows you to clculte for y vlue of. The very useful thig bout the geerl formul is tht you do t eed to kow the previous term to clculte prticulr term. For istce, if you wt to kow the 50 th term of the sequece 7,, 5, 9,, you c determie tht the ptter is to dd 4 to the previous term to get the ext term. However, to get the 50 th term, you d hve to clculte the 49 th first, but the 49 th requires the 48 th, d so o. But if you isted use the expressio 4 3, which gives the sme sequece, the the 50 th term is just d there s o eed to clculte precedig terms. Hdy! Recursive Defiitio A recursive formul gives formul for the ext term i terms of the previous oe. For exmple, i our old fried 7,, 5, 9,, the ext term is foud by ddig 4 to the previous term: 4. However, tht s ot eough iformtio to uiquely defie the series becuse you do t kow where to strt. A complete defiitio must iclude the first term lso. Therefore, the recursive defiitio for our old fried 7,, 5, 9, would be 7 Recursive defiitios, the, must specify the first term or terms d lso the rule which llows you to clculte the ext term from the previous term or terms. 4 Clculte the first four terms of the sequece give by 3 0 The first term is lredy give, 3. The

5 Give recursive formul for the sequece, 6, 8, 54, The ptter is tht the ext term equls the previous term times three. Therefore, 3 Recursive defiitios hve the sme drwbck tht we ve see before: if we wt to kow the 00 th term, we eed to clculte the 99 th first, d so o. Oly the geerl formul llows us to clculte ech term directly without kowig the previous oe. Fibocci sequece The Fibocci sequece is the most fmous exmple of recursive sequece:,,, 3, 5, 8, 3, The ptter c be quite difficult to spot you get the ext term from the sum of the two previous terms. The recursive formul for this sequece is therefore Here, the first two terms must be give to strt off with so tht you re the ble to clculte the third term from the previous two. Series A series is the sum of sequece, which c be fiite or ifiite. Here re two exmples: ) b) Nottio The sum of the first terms of sequece is deoted by S (lso sometimes clled the th prtil sum). If the series is fiite, it could be the sum of ll of the terms. S is how we write the sum of ifiite series, like the secod exmple bove.

6 For the series , clculte S 3 d S 4. S 3 = = 60 S 4 = = 88 However, it s esy to see tht this method becomes very cumbersome for lrge vlues of. We ll develop some more efficiet methods i the ext two sectios. Sigm ottio It s esy to tke sequece i list form d trsform it ito series by chgig ll of the comms to + sigs. However, wht if you re give the geerl formul isted? For exmple, let s tke 7,, 5, 9, which we kow to be 4 3. Sice the geerl form is so useful for fidig whe is lrge, it would be ice if we could reti tht iformtio while writig our sum. To do so, we ll itroduce ew ottio clled sigm ottio. It uses the Greek letter sigm (the uppercse oe): Σ, which is commoly used to me sum of. Let s look t exmple of sigm ottio d discuss wht ll of the prts me. Cosider the followig 5 i (4i 3) The letter i is idex here, d it rus from the vlue give t the bottom of the sigm to the umber t the top of the sigm i steps of. Here, i rus from to 5. We re summig, the, the vlue of 4i + 3 for ech vlue of i s it rus from to 5: 5 i Let s look t more exmples. i i i3 i4 i5 (4i 3)

7 Clculte 3 i (i 5) 3 i i i i 3 (i 5) Clculte 8 j 9 j6 9 j6 j 6 j 7 j 8 j 9 8 j Clculte 6 k 3 6 j k k 3 k 4 k 5 k The tricky thig bout the lst oe is decidig how my terms there re. You my, s is show bove, write out ll of the possible vlues of the idex. Or you my use the followig ifty rule:

8 # terms = lst first + For istce, the lst exmple hd the idex ruig from to 6. The umber of terms, the, for tht series is 6 + = 5. Write the followig series i sigm ottio: Let s pick our idex first. If we wt to be lzy, isted of strtig our idex t, we could strt t d our series would be 0 k Other cceptble swers would ivolve chgig our strtig poit for the idex 9 to give j or i or eve l 55 j fvourite umber. 8 i0 k 65 l57 if 57 hppes to be your Write the followig sequece i sigm ottio: j3 To write ifiite series i sigm ottio, you just replce the fil vlue of the idex with. j

9 Sectio 3.: Sequeces d Series Exercises Predict the ext three terms of the followig sequeces.. 8, 6, 4,., 4, 9, 6, 3., 4, 48, 96, 4. 44, 36, 9, 5.,, 3,, 5, 6, , 0, 0, 7. 3, 5, 37, 49, 8.,,,, Predict the geerl term (or th term ) of the followig sequeces. 9., 4, 9, 6, 0.,, 3,, 5, 6,...., 4, 6, 8,.,,,, Fid the first four terms of the followig recursively defied sequeces

10 6. I ech of the followig, the geerl formul for the th term of sequece is give. Fid the first four terms ! 0. I ech of the followig, the geerl formul for the th term of sequece is give. Clculte the specified terms.. 5 ; ; ; ; 0 Clculte S 3 d S 6 for the followig series Write out ech sum i full d the evlute

11 0 30. j j i0 i k 0 k 0 Write ech series i sigm ottio Evil lert! 37. (sty) Write the sequece, 4, 9, 6, usig recursive defiitio. 38. (tricksy) Write the sequece,, 6, 4, usig geerl formul. 39. (chllegig) Wht s the ext term i the sequece 4, 5, 0, 00, 000? Wht s the recursive formul for this sequece?

12 Sectio 3.: Sequeces d Series Solutios., 0, 8 (ptter is to subtrct ). 5, 36, 49 (th term is equl to 3. 9, 384, 768 (multiply by ) ,, (divide by 4) 5. 7,,3 (th term is ) 6. 40, 80, 60 (multiply by ) 7. 6, 73, 85 (dd ) 8.,, , 7,, , 30, 90, 70 5., 3, 6, ,,, ,, 4, 7 8.,,3,9 3 9.,, 6, 4

13 0.,,, = = = S 3 = 8, S 6 = S 3 = 4, S 6 = 9 7. S 3 = 5, S 6 = S 3 = 9, S 6 = j 30. j i0 5 k 0 0 i j k i k i 3 j k k k

14 37. You could either do or other possibility is 38.! 39. The ext term is 00,

15 Sectio 3.: Arithmetic Sequeces d Series Arithmetic Sequeces Let s strt out with defiitio: rithmetic sequece: sequece i which the ext term is foud by ddig costt (the commo differece d) to the previous term Here re some exmples of rithmetic sequeces: ) 7,, 5, 9, b), 4, 3, 0, 59 c),.3,.6,.9, The first oe hs commo differece of 4, the secod 7, d the third 0.3. Note tht i ech of them, we c fid the commo differece d by tkig y term d subtrctig the previous term from it. For the followig sequeces, stte whether ech of them is rithmetic. ) 3, 0, 7, 4, b) 4, 5, 7, 0, c), 4, 8, 6, d),,,, ) Yes, becuse the commo differece d is 7. b) No, becuse you re ot ddig the sme umber ech time. c) No, becuse you re multiplyig by to get the ext term, ot ddig. d) No, becuse the differece betwee ech pir of terms is differet. Agi, you c defie rithmetic sequece i oe of three wys: by listig the terms, by givig recursive defiitio, or by givig geerl defiitio.

16 Recursive Defiitios for Arithmetic Sequeces Let s look first t exmple. Give recursive defiitio for the sequece, 0, 8, 6, Recll tht recursive defiitio hs two prts: listig the first term d givig the ptter. I this cse, the ptter is ddig d = 8 to the previous term to get the ext term. The recursive defiitio is therefore 8 More geerlly, the recursive formul for y rithmetic sequece is Geerl Formule for Arithmetic Sequeces isert vlue here> d Let s exmie the previous exmple i more detil to see if we c recogize y ptters d come up with geerl formul. Rewritig ech term, we get, 0, 8, 6,...,6,6,63,... So the 3 rd term equls the first plus 6 times, the 4 th term equls the first plus 6 times 3, d the th term will equl the first plus 6 times ( ). More geerlly, the th term will equl the first plus d times ( ). I other words, for y rithmetic sequece. d Write geerl formul for the sequece, 0, 8, 6, This sequece is rithmetic with the first term d commo differece 8.

17 d The geerl formul is the tht = 8 6. Wht is the 50 th term i the sequece i the sequece, 0, 8, 6,? This is the sme sequece from the previous exmple. We my the use the formul we derived, = 8 6, with = The 50 th term is 394. Wht is the commo differece i the rithmetic sequece i which the first term is 8 d the twelfth term is 59? d 59 8 d 77 d d 7 The commo differece is 7. Which term hs vlue of 404 i the sequece 37, 8, 9,?

18 So is 37 d d is +9. The we wt to fid the vlue of for which equls 404. d The fiftieth term is 404. Fid the first four terms of the rithmetic sequece i which the thirteeth term is 97 d the fiftieth term is 393. So 3 = 97 d 50 = 393. The we fid tht d 97 3 d 97 d However, this hs two ukows, d d. Let s look t 50 : d d d We ow hve two ukows, but two equtios, givig us the system 97 d d Solvig this system, we first multiply the top equtio by egtive :

19 97 d d Ad the dd the two equtios together, so tht the terms ccel out d d 8 Now we substitute ito oe of the origil equtios: 97 d Sice = d d = 8, our sequece is the, 9, 7, 5, Arithmetic Series Recll tht S is the sum of the first terms of series. Let s look t couple of exmples of rithmetic series to see if we c idetify y ptters. Suppose we wish to tke some prtil sums of the series Let s first clculte S 6. We could just fid the first six terms d dd them up, but otice the followig: S 6 = The sum of the first d lst umbers is 44. The sum of the secod d secod-to-lst is lso 44. So is the sum of the third d third-lst. So whe you tke the terms i pirs, ech pir hs the sme sum,, d there re / pirs i totl. The S. Wht if, however, there re odd umber of terms? Let s lso clculte S 7 : S 7 =

20 The sum of the first d lst is 5, s is the sum of the ech ier pir. Notice tht the middle, upired vlue, is ½ of 5. So i sese, the middle term is ½ of pir, for totl of 3½ pirs. But tht s just 7/, which is our / i the origil formul! So we re still good. The reltioship still works, for both odd d eve vlues of. S Fid the sum of the first forty terms of the series This is just the sme sequece s before, with = d d = 8. I order to use our previous formul, however, we eed to clculte 40 before we c clculte S d So, S 40 S The sum of the first forty terms is 630. (Much esier th writig out the first forty terms d ddig them up!) I the previous exmple, we used the formul for to clculte the lst term d put its vlue ito the formul for S. We could do tht i more geerl wy: S d d d the lst expressio, which gives S s fuctio of the first term, the umber of terms, d the commo differece, c lso be used to evlute series.

21 Fid the sum of the first oe hudred terms of the sequece 5, 6, 7, 6,. This sum will just be , with = 5 d d =. S d 00 S Clculte j3 The first term will be for j=3 d will equl 5(3)+0=5. Next is j=4 d will equl 5(4)+0=30, j=5 equlig 5(5)=35, d so o. The lst term will be for j=8 d will equl 5(8)+0=00. I other words, our series is Is it rithmetic? Yes, with commo differece d = 5. Wht else do we eed for our clcultio? The umber of terms equls (lst first +), so is (8-3+)=6. The S 6 S Pt the mth istructor sks her studets to do five word problems the first week, six the secod week, seve the third week, d so o, icresig the umber of word problems ech week by oe. ) How my word problems will diliget studets be doig i the lst week of clsses (the th week)?

22 b) How my word problems will diliget studets hve completed durig the course of the term ( weeks)? ) The umber of word problems is sequece: 5, 6, 7,. I fct, it s rithmetic sequece with = 5 d d =. I the eleveth week, the, d 505 Diliget studets will solve 5 word problems i the lst week of clsses. b) The totl umber of word problems solved is S S Diliget studets will hve solved 0 word problems i totl. Summry For rithmetic sequece, the th term is give by d or For rithmetic series, the sum of the first terms (th prtil sum) is S S d

23 Sectio 3.: Arithmetic Sequeces d Series Exercises Stte whether the followig sequeces re rithmetic or ot. If they re, stte the first term d commo differece.. 8, 9,, 3, 6,. 3, 0, 7, 4, 3. 3, 6,, 4, 4.,, 6, 4, 5. 8, 7, 63, 54, ,,,,,, Give both the geerl formul (or th term ) d the recursive formul for the followig rithmetic sequeces. For the geerl formul, be sure to simplify your swer. 7., 3, 5, 7, 8. 5, 6, 7, 8, 9. 40, 37, 34, 3, 0. 4, 8, 3, 36, For the followig rithmetic sequeces, clculte 50 d 6.. 8, 6, 4,,.,.3,.6,.9, Stte whether the followig recursively defied sequeces re rithmetic or ot. (Is there esy wy to tell?)

24 For the followig sequece, clculte the 0 st term: 5, 5, 5, 35, 0. For the followig sequece, which term equls 37?, 9, 7, 5,. Wht is the commo differece for the rithmetic sequece with = 00 d 40?. Clculte the first term for the rithmetic sequece with commo differece 7 whose sixteeth term is Clculte the first four terms of the rithmetic sequece i which the sixth term is 7 d the sixtieth term is Clculte the first four terms of the rithmetic sequece i which the oe hudredth term is 403 d the sixty-fourth term is Give geerl formul for the rithmetic sequece i which the twetieth term is 07 d the thirty-fifth term is Give recursive formul for the rithmetic sequece i which the eleveth term is 44 d the fifty-secod term is Clculte S 0 for the series Evlute the series Evlute the series Clculte S 00 for the series Clculte the sum of the odd umbers betwee 00 d Fid the sum of the turl umbers from 50 to 5, iclusive.

25 Clculte the followig sums k 0 9 j0 40 i30 50 k 3 5k 6 j i 7 7 3k 37. I supermrket disply, there re 37 cs i the bottom lyer, 35 i the ext lyer up, 33 i the ext, d so o. How my lyers re there if there re 7 cs i the top row? 38. I the previous problem, how my cs re there ltogether? 39. I old-fshioed thetre, there re 5 sets i the first row, 6 i the ext, 7 i the oe fter, d so o. If there re 0 rows i totl, how my sets re there ltogether?

26 Sectio 3.: Arithmetic Sequeces d Series Solutios. ot rithmetic. yes, d = 7 3. o 4. o 5. yes, d = 9 6. yes, d = ¼ 7. d 8. 6 d d d , so 50 = 80 d 6 = , so 50 = 6.7 d 6 = first four terms re 5, 9, 3, 7, so rithmetic with d = 4 4. first four terms re, 4, 48, 96, so ot rithmetic 5. first four terms re 75, 55, 35, 5, so rithmetic with d = 0 6. first four terms re 6, 7, 8, 9, so rithmetic with d = 7. first four terms re 7, 5, 7, 5, so ot rithmetic 8. first four terms re 3, 9, 8, 656, so ot rithmetic

27 9. 0 5, so 0 = , so = 8. d = 40. = 3 3. = d d = 3, so the first four terms re, 5, 8, 4. = 7 d d = 4, so the first four terms re 7,, 5, S 0 = S 5 = S = S 00 = S 00 = S 76 = S 53 = S 83 = S = S 500 = 370, = S 6 = S 0 = 690

28 Sectio 3.3: Geometric Sequeces d Series Geometric Sequeces Let s strt out with defiitio: geometric sequece: sequece i which the ext term is foud by multiplyig the previous term by costt (the commo rtio r) Here re some exmples of geometric sequeces: ) 9, 8, 36, 7, b), 8, 7, 8, c) 0, 30, 90, 70, d) 3,, 48, 9, e) 48, 36, 7, The commo rtios of ech of these sequeces, i order from ) to e), is, 3, 3, 4, 3, 4 respectively. Note tht i ech of them, we c fid the commo rtio r by tkig y term d dividig it by the previous term. Like y other sequeces, geometric sequeces c be fiite or ifiite. c) bove is fiite, s the lst term is specified. The others re ifiite sequeces. For ech of the followig sequeces, stte whether it is rithmetic, geometric, or either. ) 45, 5, 5, b) 5, 3,,, c), 8, 7, 64,, 000 d),,,,,, ) Geometric, becuse the commo rtio r is 3.

29 b) Arithmetic, becuse the commo differece d is. c) Neither, becuse there is t either commo differece or rtio betwee terms. 3 (I fct, the ptter is tht.) d) Geometric, becuse the commo rtio r is. Agi, you c defie geometric sequece i oe of three wys: by listig the terms, by givig recursive defiitio, or by givig geerl defiitio. Recursive Defiitios for Geometric Sequeces Let s look t exmple. Give recursive defiitio for the sequece, 0, 50, 50, Recll tht recursive defiitio hs two prts: listig the first term d givig the ptter. I this cse, the ptter is multiplyig the previous term by r = 5 to get the ext term. The recursive defiitio is therefore 5 More geerlly, the recursive defiitio for y geometric sequece is Geerl Formule for Geometric Sequeces <isert vlue here> r Let s exmie the previous exmple i more detil to see if we c recogize y ptters d come up with geerl formul. Rewritig ech term, we get, 0, 50, 50,... 3, 5, 5, 5,... So the 3 rd term equls the first times 5 squred, the 4 th term equls the first times 5 cubed, d the th term will equl the first times 5 rised to the ( ) power. More geerlly, the th term equls the first term times r rised to the ( ) power, mely r

30 for ll geometric sequeces. Write geerl formul for the sequece 3, 6,, This sequece is geometric with the first term 3 d commo rtio. r 3 The geerl formul is the tht = 3 -. Wht is the 0 th term i the sequece i the sequece 3, 6,,? This is the sme sequece from the previous exmple. We my the use the formul we derived bove with = r ,57,864 The 0 th term is,57,864, which provides ice exmple for how fst geometric sequeces c grow, eve for smll vlues of r. Write geerl formul for the sequece 8,, 8, 7,? Wht is the fifteeth term i this sequece? The fiftieth?

31 5 50 r So the geerl formul is d , respectively. d the fifteeth d fiftieth terms re Geometric Series Recll tht S is the sum of the first terms of series. Let s look t how formul for S is derived. S... S r r r r r r Let s tke tht lst expressio for S d multiply it by r to get rs rr r r... r r r The if we dd the rows for S d rs, we get S r r r... r r r 3 4 rs r r r r... r r r S rs r sice ll of the terms i betwee these two ( d r ) will ccel. The S r r d S r r The lst formul bove is the formul for the sum of the first terms for y geometric series.

32 Fid the sum of the first 0 terms of the series This is geometric series with = 3 d r =. We wt to fid S 0. S S 0 r r 3 The sum of the first 0 terms is 3,45, ,45,75 Fid the sum of the first forty terms of the series This is geometric series with = 8 d r = S S 0 r r We wt to fid S The sum of the first forty terms is Sum of Ifiite Geometric Series Let s tke look t the ifiite series Wht hppes whe we try to evlute this sum usig the S formul? We c put = ½, r = ½, d = ito the formul, but we will ru ito rodblock whe we try to evlute (½).

33 Let s tke closer look t the behviour of (½) for lrge vlues of. As gets lrger, the frctio gets ever smller. I fct, s pproches, (½) will pproch zero. This is true for y r provided tht r <. (If you re ot fmilir with the bsolute vlue brs, x, equivlet expressio is tht < r <.) Recllig tht S r r d lettig the r term go to zero, the S r for < r < for y ifiite geometric series, provided tht r meets the restrictio bove. Let s ow revisit the series tht strted this discussio,..., d evlute it i the followig exmple. Evlute This series is geometric with = ½ d r = ½. The The sum of this series is. / / S r / / Evlute This series is geometric with = 4 d r = 3.

34 4 4 3 S 4 7 r 3 3 Evlute This series is ideticl to the previous oe except tht r is ow egtive: = 4 d r = S r Evlute This series is geometric with = d r = 3. You my lredy relize wht s goig o, but i cse you do t, let s ively put the vlues ito the formul d see wht we get: S 4 r 3 Wit! How c the sum of buch of positive umber be egtive? The swer is tht our restrictio for r is tht it must be betwee d, but r =.5. Becuse r does ot stisfy the restrictio, we cot use the bove formul for S. Ideed, if you dd up buch of positive umbers tht re icresig s you go up, you c see tht the sum just keeps gettig bigger s we dd more terms. You could the either sy tht the sum is ifiite (dicey) or does ot exist (sfer). But why is it sfer to sy does ot exist i the lst exmple? Let s look t three sums: )

35 b) 8 7 c) Ech term i ) is gettig more positive, so the sum of tht sequece will be +. Ech term i b) is gettig more d more egtive, so the sum of tht sequece will be. But i the lst term, the sum oscilltes bck d forth: S =, S = 6, S 3 =, S 4 = 9.5, d so o. The sig of S is either positive or egtive depedig o whether the umber of terms you ve dded is eve or odd. Rther th debtig whether ifiity is odd or eve (?!), we will just sy tht the sum does ot exist. Evlute j 7 j0 3. Ick! The best plce to strt is to figure out the first few terms to determie the ptter: whe j = 0, whe j =, whe j =, so our sequece is 7, 9, 3, This is geometric with = 7 d r = 3. The S r 3 3 Evlute k. k 5

36 Oce gi, let s figure out the first few terms to determie the ptter: whe k = 5, k 5.5 whe k = 6, k 6 3 whe k = 7, k so our sequece is.5, 3, 3.5. Wit! This is rithmetic! Not oly tht, but the umbers re icresig. So the sum will be ifiite, or if you prefer, the sum does ot exist. Repetig Decimls Let s exmie 0.7 i some detil to see wht we fid: But this is just the sum of ifiite series with = 0.7 d r = 0.. Rewritig d r i frctio form (you ll see why i miute) gives = 7 0 d r = 0. The S r So 0.7 = 7/9. Iterestig! Fid exct frctio for

37 But this is just the sum of ifiite series with = 6 0 d r = 0. The S r So 0.6 = /3. Fid exct frctio for But this is just the sum of ifiite series with = 8 00 d r = 00. The S r So 0.8 = /. Summry For geometric sequece, the th term is give by r For geometric series, the sum of the first terms (th prtil sum) is S r r For ifiite geometric series, the sum is S, provided tht < r <. r

38 Sectio 3.3: Geometric Sequeces d Series Exercises Stte whether the followig sequeces re geometric or ot. If they re, stte the first term d commo rtio.. 8, 9,, 3, 6,. 3, 0, 7, 4, 3. 3, 6,, 4, 4.,, 6, 4, 5. 8, 7, 63, 54, 6. 7,48,3,... Give both the geerl formul (or th term ) d the recursive formul for the followig geometric sequeces. 7., 3, 9, 7, 8. 64, 6, 4,, 9., 6,, 4, 0. 4,.4, 0.4, For the followig sequeces, clculte 50 d 6.., 8, 7,., 8, 6 3, Stte whether the followig recursively defied sequeces re geometric or ot. (Is there esy wy to tell?)

39 For the followig sequece, clculte the 0 st term: 5, 5, 45, 0. For the followig sequece, clculte the 0 th term: 7, 4, 8,. Clculte S 0 for the series Clculte S 0 for the series Clculte the sum, if it exists, for the followig series Clculte the followig sums, if they exist k 0 j k j i 5 0. j

40 i i0 3. If the umber of vmpires i Trsylvi doubles every moth, the how my vmpires will be i Trsylvi i 3 yers, strtig from oe idividul? Commet o your result if the totl popultio of Trsylvi is millio people. 3. As I ws goig to St. Ives, I met m with seve wives. Ech wife hd seve scks. Ech sck hd seve cts. Ech sck hd seve kits. Kits, cts, scks, wives: does this form geometric sequece? 33. The pper used i the photocopier by Pt s office is sid to be mm thick. If it is folded over repetedly, doublig its thickess ech time, how thick will the pper be if it s folded 7 times? Bous: why, the, were the Mythbusters hvig so my problems tryig to fold the pper this my times?

41 Sectio 3.3: Geometric Sequeces d Series Solutios. o. o 3. yes, r = 4. o 5. o 6. yes, r = 3 7. d d d d , so d , so d o 4. yes, with r = 5. yes, with r = 0 6. o

42 7. yes, with r = 8. o 9. 53, so 0 = 5(3) 00 = , so 0 = 7( ) 9 = 3,670,06. S 0 = 00 (The exct swer is to three decimls, the swer is ) or , but if you roud. S 0 = 04,857, S r / S = S does ot exist (r > ) 6. S = 6 7. S r 888 ( r ) ( ) 8. S =.5 9. S = S does ot exist (r < ) 3. 3 yers is 36 moths, so we hve 36-term sequece strtig with,, 4, 8, The th term will be, so the 36 th term will be ,359,738,368, which is td lrger th the totl popultio of Trsylvi. 3. m 7 wives # scks = #wives # scks/wife = 7 7 = 49

43 # cts = #scks # cts/sck = 49 7 = 343 # kits = #cts # kits/ct = = 40 So kits, cts, scks, d wives is 40, 373, 49, 7, which is geometric sequece with four terms: = 40 d r = / After oe fold, the thickess will be 0.097, fter two folds 0.097, etc. So our strtig term will be d the will double with r = therefter. So 0.097, d the 7 th term will be , d the pper thickess will be.4 mm, or just over cm thick. (The Mythbusters relized tht the problems with pperfoldig lie with the fold itself. If I remember correctly, they resorted to C-clmps d hittig the fold with hmmer to fltte it.)