Basic Elements of Arithmetic Sequences and Series

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1 MA40S PRE-CALCULUS UNIT G GEOMETRIC SEQUENCES CLASS NOTES (COMPLETED NO NEED TO COPY NOTES FROM OVERHEAD) Basic Elemets of Arithmetic Sequeces ad Series Objective: To establish basic elemets of arithmetic sequeces ad series Example : Cosider the arithmetic sequece 3, 7,, 5, 9, What does the mea? What is the 7 th term, t 7? What is the 0 st term of the sequece? The key feature of a arithmetic sequece is that there is a commo differece d betwee ay two cosecutive terms. To obtai ay term, add d to the precedig term or subtract d from the followig term. We call it arithmetic series because of this addig ad subtractig. The formula for the th term ( t ) of a arithmetic sequece is give by: a. recursively: which meas that each term is calculated from the immediately precedig term t + = t + d b. explicitly (or directly): Which meas ay term ca be directly calculated t = a + ( )d where a is the first term (t ), ad d is the commo differece betwee terms. I the case of fidig the 0 st term the, we have a = 3, d = 4, ad = 0. t 0 = 3 + (0 )*4 = 403 MA40SP_G_GeoSeries_ClassNotes.doc Revised:20045

2 2 Example 2: How may terms are there i the arithmetic sequece 4, 5, 26,, 2853? We will use the formula t = a + ( )d ad will solve for the variable. 4 + ( ) = 2853 ( ) = = 259 = 260 Example 3: The 3rd term of a arithmetic progressio is 3, ad the 500th term is Fid the first two terms. First, we express 3 ad 2498 i terms of the formulas for t 3 ad t 500 respectively. 3 = a + (3 - )d 2498 = a + (500 - )d This results i a system of two equatios i two ukows: a + 2d = 3 () a + 499d = 2498 (2) Subtractig equatio () from equatio (2) we ca elimiate the variable a ad fid d. Thus, from equatio (), we ca solve for a. 497d = 2485 d = 5 a = 3 2d a = 3 0 = 3 Thus the first two terms of the sequece are 3 ad 8.

3 3 Example 4: Fid the sum of 0 terms of the series The formula for the sum of terms of a arithmetic series. S = 2a + ( )d 2 [ ] S = (0 - )4 2 = [ ] [ ] = The formula is much simpler whe a = ad d =. Try it for that Teacher will prove where this formula comes from: PRACTICE PROBLEMS YOU CAN TRY a. fid the ext arithmetic term ad state the d : (), 3, 5, 7, 9,. d= (2), 5, 9, 3, 7,. d= (3) 22, 33, 44, 55, d= b. fid the 2 th term explicitly: (), 3, 5, 7, 9,. (2), 5, 9, 3, 7,. (3) 22, 33, 44, 55,

4 4 c. Fid the sum of the arithmetic series: (), 3, 5, 7, 9,. For the first 20 terms (2), 5, 9, 3, 7,. For the first 30 terms (3) 22, 33, 44, 55, For the first 0 terms (4) what is the sum of all the umbers from to 00? (5) what is the sum of all the umbers from to,000? (6) what is the sum of all the umbers o a clock? Geometric Sequeces Objective: To lear about the properties of geometric sequeces Ivestigatio: Cosider the sequece 2, 6, 8, 54. The fill i the blaks below with t 5, t 6, ad t 7. 2, 6, 8, 54, 62, 486, 458 t t t t t t t Explai the patter whereby you foud t 5, t 6, ad t 7. Each term is obtaied by multiplyig the precedig term by 3. This sequece of umbers is called a geometric sequece. The distiguishig feature of a geometric sequece is that the ratio formed by dividig ay term (the th term) by the precedig term (the ( ) th term) is a costat. So above, each successive term could be calculated by t + = t *r. This is called the recursive formula for the geometric series. This ratio, r, is called the commo ratio. I this example, the commo ratio is = = = 54 8 = 8 6 = 6 2 = 3 You were able to fid t 5, t 6, ad t 7. Ca you fid t 2? Doig it recursively meas you would eed to calculate all the terms t 8, t 9, all the way up to t 20, so you could fid t 2.

5 5 There are two ways to fid t 2 : a loger tedious way, ad a shorter more efficiet way. Loger, cumbersome way: Multiply cotiuously by 3 util you reach t 2. Shorter, eater way: Look for a patter From this patter, what is t 2? t t 2 t 3 t 4 t 5 t 6 t t 2 = = Geeralizatio of Geometric Sequeces Let the first term of the sequece be a, ad let r equal the commo ratio. The first seve terms ca be writte as follows: t t 2 t 3 t 4 t 5 t 6 t 7... a ar ar 2 ar 3 ar 4 ar 5 ar 6... What is the formula for the 00 th term? The 60 th term? t 00 = ar 00 = ar 99 t 60 = ar 60 = ar 609 What is the explicit formula for computig th term of a geometric sequece directly? t = ar Example : Determie the commo ratio of the sequece, 3, 9, 27. r = 3 = 3

6 6 Example 2: Determie the 4 th term of the geometric sequece 6, 2, 24 We have a = 6 ad r = 2 Sice t = ar we have t 4 = = = 4952 Notice that geometric series are very much just expoetial equatios (from that uit!), the variable is a expoet. I this case though, the variable ca oly take o whole umbers Example 3: Usig a r of less tha (r < ) I the World Domioes touramet 78,25 players are placed i groups of 5 players per table. Oe game is played by these 5 players, ad the wier at each table advaces to the ext roud, ad so o util the fial game of 5 players. How may rouds would the ultimate wier have played (icludig the fial roud)? We have a = 7825; t = 5; r = 5. t = ar 5 = Fially, = 7. The wier would have played 7 rouds. Example 5: Relatig to expoetial equatios The world s populatio grows by 2% per year. The world food productio ca sustai a additioal 200 millio people per year. I 987 the populatio was 5 billio, ad food productio could sustai 6 billio people. a) Calculate the populatio i 998, 2009, ad 209. b) Calculate the populatio that food productio could sustai i 998, 2009, 209. c) Whe will the populatio exceed the food supply?

7 7 a) The world s populatio i each of the years startig with 987 ca be cosidered as a geometric sequece. The first term of the sequece will be 5 (billio) ad the commo ratio r will be.02. Fill i the table below with the world populatio for the first five years, startig with 987. t t 2 t 3 t 4 t billio 5 (.02) = 5. billio 5 (.02) 2 = 5.2 billio 5 (.02) 3 = 5.3 billio 5 (.02) 4 = 5.4 billio So the formula for the world populatio i ay year is P = 5 (.02) where positio of that year i the sequece of years startig with 987. is the What is the value of for each of the followig years? 998: = : = : = 33 Now calculate the world populatio i each of these years P = 5 (.02) = 6.2 billio 22 P = 5 (.02) = 7.7 billio 32 P = 5 (.02) = 9.4 billio b) The world s populatio that food productio ca sustai i each of the years startig with 987 ca be cosidered as a arithmetic sequece. The first term t = a of the sequece will be 6 (billio) ad the commo differece d will be 0.2 (billio). Fill i the table below with the world sustaiable populatio for the first five years, startig with 987. t t 2 t 3 t 4 t billio 6 + (2 )0.2 = 6.2 billio 6 + (3 )0.2 = 6.4 billio 6 + (4 )0.2 = 6.6 billio 6 + (5 )0.2 = 6.8 billio

8 8 So the formula for the world populatio i ay year is P = 6 + ( )0.2 where is the positio of that year i the sequece of years startig with 987. Like before, fill i the value of for each of the followig years. 998: = : = : = 33 Now calculate the world sustaiable populatio i each of these years. 998 P = 6 + (2 )0.2 = 8.2 billio 2009 P = 6 + (23 )0.2 = 0.4 billio 209 P = 6 + (33 )0.2 = 2.4 billio c) To fid whe the world populatio will exceed the sustaiable populatio, we equate the two formulas ad solve for. 5 (.02) = 6 + ( )(0.2) We ll use the graphig calculator to solve this rather complex equatio. Sketch the itersectig curves i the blak grid below. Usig the itersect tool, we fid that equals approximately 7 years. Therefore, the world populatio will exceed sustaiable populatio i about the year (7 ) = 2057 whe the world populatio is about

9 9 20 billio. YOU TRY: a. Fid the 8 th term of the sequece 3,6,2, 24, b. A sum of moey is ivested ad compouds aually. Fid the value of $,000 ivested at 6% after 0 years. Geometric Series A series is just the sum of the terms of a sequece Objective: To derive ad apply expressios represetig sums for geometric growth ad to solve problems ivolvig geometric series Defiitio: A geometric series is the sum of the terms of a geometric sequece. Cosider the geometric sequece, 2, 4, 8, 6, 32, 64, 28, 256, 52, 024. If we add the terms of the sequece, we ca write the geometric series as Sum = S = Istead of addig the terms directly, let s evaluate S usig a trick as follows. We will multiply each term of the series by 2 (which is the value of the commo ratio r), realig the result uder the origial series, ad subtract equatio (2) from equatio (). S = () 2S = (2) S 2S = 2048 S = 2047 S = 2047 The above is ot ulike the elimiatio method we used previously i Systems of Liear Equatios.

10 0 We ca use this process to develop a geeral formula for S, the sum of the first terms of a geometric series. I geeral, a geometric series ca be writte as follows, with rs writte udereath. S = a + ar + ar 2 + ar ar -2 + ar -...() rs = ar + ar 2 + ar ar -2 + ar - + ar...(2) S rs = a ar Factorig S, we have S ( r) = a ar Fially, S = a ar = a( r ) r r where r Alteratively, sice t = ar - we ca write S as follows: S = a( r ) r ( ) = a ar = a ar - r = a rt = a rl r r r r where l is the last term of the geometric series (i.e., t or ar ) Summary: The sum of the first terms of a geometric series is give by: S = a( r ) r or where a is the first term, l is the last term, ad r. S = a lr r Example : Determie the sum of 4 terms of the geometric series: S = S = a( r ) r = 6( 34 ) =

11 Example 2: Determie the th term, ad the sum of the first terms of the geometric sequece which has 2, 6, ad 8 as its first three umbers. t = ar - t = 2(3) - a( - r ) S = ( - r) 2( - 3 ) S = = 3 - ( - 3) Example 3: For the geometric series , fid the sum of the first 20 terms. ( ) S = a r r ( ) S 20 = 20 3 = = 3 4 Example 4: While traiig for a race, a ruer icreases her distace by 0% each day. If she rus 2 km o the first day, what will be her total distace for 26 days of traiig? (accurate to 2 decimal places) a = 2 r =. = 26 S 26 = 2 + 2(.) + 2(.) 2 + 2(.) (.) (.) 25 S 26 = a( r ) r = ( ) Ğ = km

12 2 Sigma Notatio Objective: To use sigma otatio to write ad evaluate series A series ca be writte usig the Greek capital letter ( sigma ). This otatio provides us with a mathematical shorthad for writig ad workig with various kids of series. A series writte with sigma otatio could have the followig form. 7 2 = This otatio meas simply that we replace the variable with the umbers through 7 respectively, ad the add the resultig terms. 7 So, 2 = = 40 = Part I: Covertig sigma otatio to expaded form Example : 5 Write the series 2 k i expaded form, ad compute the sum. k= We replace the variable k with the umbers through 5 respectively, ad the add the resultig terms. Example 2: Write the series 5 2 k = = 62 k= 8 (4 5) i expaded form, ad compute the sum. = 3 We replace the variable with the umbers 3 through 8 respectively, ad the add the resultig terms. Note that the bottom idex does ot ecessarily have to equal.

13 3 8 = 3 (4 5) = (4 3 5) + (4 4 5) + (4 5 5) + (4 6 5) + (4 7 5) + (4 8 5) = = 6 2 (7 + 27) = 02 Note that this is a arithmetic series, just a arithmetic sequece with the terms added together. Example 3: 4 Write the series ( ) i 4i i expaded form, ad compute the sum. i = As before, we replace the variable i with the umbers through 4 iclusive, ad the add the resultig terms. 4 i = ( ) i 4i = ( ) 4( )+ ( ) 2 4( 2)+ ( ) 3 4( 3) + ( ) 4 4( 4) = ( ) 0 4( ) + ( ) 4( 2)+ ( ) 2 4( 3) + ( ) 3 4( 4) = 4 + ( 8) ( 6) = 8 Part II: Covertig expaded form to sigma otatio Example 4: Covert the series to sigma otatio. There are two steps i the coversio process. Step Fid a formula for the geeral term t. Sice this is a arithmetic series, we ca compute t as follows. t = a + ( )d = 5 + ( )4 = 4 +

14 4 Step 2 Sice there are 7 terms, we place = uder the letter sigma, ad the idex 7 above the letter sigma. The we write the expressio ( 4 ) to the right of the symbol. Example 5: 7 = (4 ) Covert the series to sigma otatio. Step Fid a formula for the geeral term t. Sice this is a geometric series, we ca compute t as follows. t = ar t = 3 4 Step 2 Sice there are 6 terms, we place = uder the letter sigma, ad the idex 6 above the letter sigma. The we write the expressio 3 4 to the right of the symbol 6 = 3 4 Example 6: How may terms are there i the series 3 k +2? 50 k = 50 terms Example 7: How may terms are there i the series 3 k +2? 9 k=4 You might give the aswer 9 4 = 5 terms, but you ca see by expadig the series that the aswer is actually 6 terms. That is because we start coutig (zero), at the 4 th term which is icluded)

15 5 Example 8: How may terms are there i the series 3 k +2? You Try some problems 800 k=4 Rule: Subtract the bottom idex from the top idex, ad add. The umber of terms is (800 4) + = 797. Write each of these i expaded otatio ad the calculate their sums 5 a. 2k k= 4 b. k k= 6 c. (2k 4) k= 4 d. 2 k= 2 k 2. Write the followig series usig Sigma otatio a b c Cosider the sequece 3, 6, 2, 24,. a. fid the 8 th term b. fid the sum of the first 8 terms 4. Cosider the geometric sequece with t = 000 ad r =.05 a. write the first three terms b. fid the sum of the first 20 terms (to two decimal places)

16 6 5. Evaluate: 7 a. 8(.2) k= 3 k 5 5 b. k k= 2 Ifiite Geometric Series Objective: To ivestigate the cocept of ifiite geometric series Ivestigatio: Cosider the geometric series S = Now compute the followig partial sums for the series. S 2 = = 3 4 S 3 = = 7 8 S 4 = = 5 6 S 5 = = 3 32 S 6 = = S 7 = = S 8 = = S 9 = = 5 52 S 0 = = What do you otice about the sums for S as gets larger ad larger?

17 7 The values get closer ad closer to. Whe we icrease the value of eve more, what happes to the values for S? The sums would get eve closer to. To further ivestigate this pheomeo, let s look at the algebraic expressio for S for this particular geometric series. We have a = 2 ad r = 2. S = a( r ) r 2 2 = 2 = 2 As gets larger ad larger, what happes to 2? It gets closer ad closer to 0. As grows larger ad larger, how close to zero will 2 get? As close as you wat Ca 2 ever equal 0? No I other words, by makig sufficietly large, you ca make 2 as close to 0 as you like. This is the foudatio of the mathematical cocept called the limit. (You will lear i depth about limits whe you study calculus.)

18 8 Formally, we say that as icreases without boud, the umber 2 approaches 0, ad therefore, the value of S = 2 approaches. If you were graphig it you see a asymptote. How close ca S = 2 get to? By makig sufficietly large, S ca get as close as you wat to. If the series S = does ot ed at the th term, but rather cotiues o idefiitely, we call it a ifiite series. We also say that the sum of the ifiite series S = is. Geeralizatio: Let s look at the cocept of the sum of a ifiite geometric series usig the formula S = a( r ) r ( r ). If r <, what happes to r as icreases without boud? r gets closer ad closer to 0 How close to 0 ca r get? By makig sufficietly large, as close as you wat Thus if r <, we say that the sum of a ifiite series is S = a( 0) r = a r If r >, what happes to the value of S as gets larger ad larger? S also gets larger ad larger. I other words, it does ot have a fiite sum. e.g., S = We say that the sequece associated with this series diverges, it is a diverget sequece

19 9 Example : Fid the sum of the ifiite series a = 2 ad r = 5 So S = a r = 2 = Example 2: Evaluate: k = 3 k S = We have a = 3 ad r = 3 so S = a 3 r = 3 = 2 Example 3: Fid the sum of the ifiite geometric series S = We have a = 3 0 ad r = 0. So, S = a r = = Note that the series S = is the expaded form of the repeatig decimal = 3

20 20 Example 4: A oil well produces barrels of oil durig its first moth of productio. If its productio drops by 5% each moth, estimate the total productio before the well rus dry. S = (0.95) (0.95) (0.95) (0.95) We have a = ad r = So, S = = barrels Example 5: A ball is dropped from a height of 6 metres. The ball rebouds a half of the height after each bouce. Calculate the total vertical distace the ball travels before comig to rest. We igore the first 6 m i computig the sum of the ifiite series. S = = We have a = 6 ad a = 2. So, S = a r = 6 2 = 32 metres. The total vertical distace travelled by the ball is = 48 metres.

21 2 SOME FOR YOU TO TRY. Cosider the series a. fid the 6 th term b. fid the sum of the six terms c. fid, to 4 decimal places, the sum of the first six terms d. fid the sum to ifiity 2. Fid the sum of each of these ifiite geometric series: 8 8 a b c Here is a good oe relatig to fractals! A 4 x 4 uit square is divided ito four equal squares. The bottom left square is shaded. The top right square is divided ito 4 equal squares ad its bottom left is shaded, the etc. This is doe forever. a. what is the total area shaded? b. what is the perimeter of all the shaded squares?

22 22 4. Ivestigate this series (it is of course ot arithmetic or geometric): x x x x f ( x) = + + 2! 4! 6! 8! Try pluggig i 2 ad 5 (ie: fid f(2) ad f(5)). Compare that with cosie of 2 ad cosie of 5 (i radias of course). Try graphig f(x) ad cos(x). Compare!

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