Series. 4 Sequences and. Introduction. 4.1 Sequences

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1 4 Sequeces ad Series Assessmet statemets. Arithmetic sequeces ad series; sum of fiite arithmetic sequeces; geometric sequeces ad series; sum of fiite ad ifiite geometric series. Sigma otatio..3 Coutig priciples, icludig permutatios ad combiatios. The biomial theorem: expasio of (a b), N..4 Proof by mathematical iductio. Itroductio The heights of cosecutive bouds of a ball, compoud iterest, ad Fiboacci umbers are oly a few of the applicatios of sequeces ad series that you have see i previous courses. I this chapter you will review these cocepts, cosolidate your uderstadig ad take them oe step further. 4. Sequeces Take the followig patter as a example: The first figure represets dot, the secod represets 3 dots, etc. This patter ca also be described differetly. For example, i fuctio otatio: f () 5, f () 5 3, f (3) 5 6, etc., where the domai is Z Here are some more examples of sequeces: 6,, 8, 4, 30 3, 9, 7,, 3 k, 3 { i ; i 5,, 3,, 0 } 4 {b, b,, b, }, sometimes used with a abbreviatio {b } 5

2 4 Sequeces ad Series The first ad third sequeces are fiite ad the secod ad fourth are ifiite. Notice that, i the secod ad third sequeces, we were able to defie a rule that yields the th umber i the sequece (called the th term) as a fuctio of, the term s umber. I this sese, a sequece is a fuctio that assigs a uique umber (a ) to each positive iteger. Example Fid the first five terms ad the 50th term of the sequece {b } such that b 5. Solutio Sice we kow a explicit expressio for the th term as a fuctio of its umber, we oly eed to fid the value of that fuctio for the required terms: b 5 5 ; b ; b ; b ; b ; ad b So, iformally, a sequece is a ordered set of real umbers. That is, there is a first umber, a secod, ad so forth. The otatio used for such sets is show above. The way we defied the fuctio i Example is called the explicit defiitio of a sequece. There are other ways to defie sequeces, oe of which is the recursive defiitio. The followig example will show you how this is used. Example Fid the first five terms ad the 0th term of the sequece {b } such that b 5 5 ad b 5 (b 3). Plot Plot Plot3 Mi U() (u( ) 3 ) U(Mi) 5 V() V(Mi) W() U(5) 70 U(0) Solutio The defiig formula for this sequece is recursive. It allows us to fid the th term b if we kow the precedig term b. Thus, we ca fid the secod term from the first, the third from the secod, ad so o. Sice we kow the first term, b 5 5, we ca calculate the rest: b 5 (b 3) 5 (5 3) 5 6 b 3 5 (b 3) 5 (6 3) 5 38 b 4 5 (b 3 3) 5 (38 3) 5 8 b 5 5 (b 4 3) 5 (8 3) 5 70 Thus, the first five terms of this sequece are 5, 6, 38, 8, 70. However, to fid the 0th term, we must first fid all 9 precedig terms. This is oe of the drawbacks of the recursive defiitio, uless we ca chage the defiitio ito explicit form. This ca easily be doe usig a GDC. 5

3 Example 3 A Fiboacci sequece is defied recursively as 5 F 5 5 F F. a) Fid the first 0 terms of the sequece. b) Evaluate S 5 i 5 F i for 5,, 3,, 0. c) By observig that F 5 F 3 F, F 5 F 4 F 3, ad so o, derive a formula for the sum of the first Fiboacci umbers. Fiboacci umbers are a sequece of umbers amed after Leoardo of Pisa, kow as Fiboacci (a short form of filius Boaccio, so of Boaccio ). Solutio a),,, 3, 5, 8, 3,, 34, 55 b) S 5, S 5, S 3 5 4, S 4 5 7, S 5 5, S 6 5 0, S 7 533, S , S , S c) Sice F 3 5 F F, the F 5 F 3 F F 5 F 4 F 3 F 3 5 F 5 F 4 F 4 5 F 6 F 5 F 5 F F S 5 F F Notice that S F 7 F 5 3 ad S F 0 F Note: parts a) ad b) ca be made easy by usig a spreadsheet. Here is a example: A F() B S() C D Let this cell be A A3 The copy it dow Let this cell be B0 A The copy it dow 53

4 4 Sequeces ad Series Notice that ot all sequeces have formulae, either recursive or explicit. Some sequeces are give oly by listig their terms. Amog the may kids of sequeces that there are, two types are of iterest to us: arithmetic ad geometric sequeces, which we will discuss i the ext two sectios. Exercise 4. Fid the first five terms of each ifiite sequece defied i questios 6. s() 5 3 g(k) 5 k 3 3 f () a 5 5 a 5 a 3; for. 5 a 5 () ( ) 3 6 b 5 3 b 5 b ; for > Fid the first five terms ad the 50th term of each ifiite sequece defied i questios a b u 5 () 0 a 5 a 5 a 5 ad a 5 3 u 5 3 u ad u b 5 3. b ad b 5 4 a 5 a ad a 5 Suggest a recursive defiitio for each sequece i questios _ 3,, 48, 9, 6 _ a, _ 3 a3, 8_ 9 a5, 3 7 a7, 7 a 5k, a 4k, 3a 3k, 4a k, 5a k, I questios 8, write dow a possible formula that gives the th term of each sequece. 8 4, 7,, 9, 9, 5, 8,, 0, 3_ 4, 5_ 9, 7 6, 9 5, _ 4, 3_ 5, 5_ 6,, 9_ 8, Defie a 5 F,., where F is a member of a Fiboacci sequece. F a) Write the first 0 terms of a. b) Show that a 5 a 3 Defie the sequece F 5 ( 5 ( 5 ) ( 5 ) ) a) Fid the first 0 terms of this sequece ad compare them to Fiboacci umbers. b) Show that ( 6 5 ). c) Use the result i b) to verify that F satisfies the recursive defiitio of Fiboacci sequeces. 54

5 4. Arithmetic sequeces Examie the followig sequeces ad the most likely recursive formula for each of them. 7, 4,, 8, 35, 4, a 5 7 ad a 5 a 7, for.,, 0, 9, 38, 47, a 5 ad a 5 a 9, for. 48, 39, 30,,, 3, 6, a 5 48 ad a 5 a 9, for. Note that i each case above, every term is formed by addig a costat umber to the precedig term. Sequeces formed i this maer are called arithmetic sequeces. Defiitio of a arithmetic sequece A sequece a, a, a 3, is a arithmetic sequece if there is a costat d for which a 5 a d for all itegers.. d is called the commo differece of the sequece, ad d 5 a a for all itegers.. So, for the sequeces above, 7 is the commo differece for the first, 9 is the commo differece for the secod ad 9 is the commo differece for the third. This descriptio gives us the recursive defiitio of the arithmetic sequece. It is possible, however, to fid the explicit defiitio of the sequece. Applyig the recursive defiitio repeatedly will eable you to see the expressio we are seekig: a 5 a d; a 3 5 a d 5 a d d 5 a d; a 4 5 a 3 d 5 a d d 5 a 3d; So, as you see, you ca get to the th term by addig d to a, ( ) times, ad therefore: th term of a arithmetic sequece The geeral (th) term of a arithmetic sequece, a, with first term a ad commo differece d, may be expressed explicitly as a 5 a ( )d This result is useful i fidig ay term of the sequece without kowig all the previous terms. Note: The arithmetic sequece ca be looked at as a liear fuctio as explaied i the itroductio to this chapter, i.e. for every icrease of oe uit i, the value of the term will icrease by d uits. As the first term is a, the poit (, a ) belogs to this fuctio. The costat icrease d ca be cosidered to be the gradiet (slope) of this liear model; hece, the th term, the depedet variable i this case, ca be foud by usig the poitslope form of the equatio of a lie: y y 5 m(x x ) a a 5 d( ) a 5 a ( )d This agrees with our defiitio of a arithmetic sequece. 55

6 4 Sequeces ad Series Example 4 Fid the th ad the 50th terms of the sequece,, 0, 9, 38, 47, Solutio This is a arithmetic sequece whose first term is ad commo differece is 9. Therefore, a 5 a ( )d 5 ( ) a Example 5 Fid the recursive ad the explicit forms of the defiitio of the followig sequece, the calculate the value of the 5th term. 3, 8, 3,, Solutio This is clearly a arithmetic sequece, sice we observe that 5 is the commo differece. Recursive defiitio: a 5 3 a 5 a 5 Explicit defiitio: a 5 3 5( ) 5 8 5, ad a Example 6 Fid a defiitio for the arithmetic sequece whose first term is 5 ad fifth term is. Solutio Sice the fifth term is give, usig the explicit form, we have a 5 5 a (5 )d 5 5 4d d 5 _ 3 This leads to the geeral term, a 5 5 _ 3 ( ), or, equivaletly, the recursive form a 5 5 a 5 a 3 _,. Hit: Defiitio: I a fiite arithmetic sequece a, a, a 3,..., a k, the terms a, a 3..., a k are called arithmetic meas betwee a ad a k. Example 7 Isert four arithmetic meas betwee 3 ad 7. Solutio Sice there are four meas betwee 3 ad 7, the problem ca be reduced to a situatio similar to Example 6 by cosiderig the first term to be 3 ad the sixth term to be 7. The rest is left as a exercise for you! 56

7 Exercise 4. Isert four arithmetic meas betwee 3 ad 7. Say whether each give sequece is a arithmetic sequece. If yes, fid the commo differece ad the 50th term; if ot, say why ot. a) a 5 3 b) b 5 c) c 5 c, ad c 5 d) u 5 3u e), 5, 7,, 9, f ), 5,, 9, For each arithmetic sequece i questios 3 8, fid: a) the 8th term b) a explicit formula for the th term c) a recursive formula for the th term. 3,, 6, 0, 4 9, 5,, 7, 5 6, 3,,, , 9.95, 9.83, 9.7, 7 00, 97, 94, 9, 8, 3_ 4, _, 7_ 4, 9 Fid five arithmetic meas betwee 3 ad 3. 0 Fid three arithmetic meas betwee 99 ad 300. I a arithmetic sequece, a ad a Fid a explicit formula for the th term of this sequece. I a arithmetic sequece, a ad a Fid a explicit formula for the th term of this sequece. I each of questios 3 7, the first 3 terms ad the last term of a arithmetic sequece are give. Fid the umber of terms. 3 3, 9, 5,, , 3, 3,, _ 8, 4 _ 4, 5 3_ 8,, 4 3_ 8 6 _ 3, _, _ 3,, 5_ 6 7 k, k, 3k,, 9k 8 Fid five arithmetic meas betwee 5 ad. 9 Fid three arithmetic meas betwee 99 ad I a arithmetic sequece, a 3 5 ad a Fid a explicit formula for the th term of this sequece. I a arithmetic sequece, a ad a Fid a explicit formula for the th term of this sequece. The 30th term of a arithmetic sequece is 47 ad the commo differece is 4. Fid a formula for the th term. 3 The first term of a arithmetic sequece is 7 ad the commo differece is 3. Is 9803 a term of this sequece? If so, which oe? 4 The first term of a arithmetic sequece is 9689 ad the 00th term is Show that the 0th term is 896. Is a term of this sequece? If so, which oe? 5 The first term of a arithmetic sequece is ad the 30th term is 47. Is 995 a term of this sequece? If so, which oe? 57

8 4 Sequeces ad Series 4.3 Geometric sequeces Examie the followig sequeces ad the most likely recursive formula for each of them. 7, 4, 8, 56,, 4, a 5 7 ad a 5 a 3, for., 8, 6, 458, 3, a 5 ad a 5 a 3 9, for. 48, 4,, 6, 3,.5, a 5 48 ad a 5 a 3 0.5, for. Note that i each case above, every term is formed by multiplyig a costat umber with the precedig term. Sequeces formed i this maer are called geometric sequeces. Defiitio of a geometric sequece A sequece a, a, a 3, is a geometric sequece if there is a costat r for which a 5 a 3 r for all itegers.. r is called the commo ratio of the sequece, ad r 5 a 4 a for all itegers.. So, for the sequeces above, is the commo ratio for the first, 9 is the commo ratio for the secod ad 0.5 is the commo ratio for the third. This descriptio gives us the recursive defiitio of the geometric sequece. It is possible, however, to fid the explicit defiitio of the sequece. Applyig the recursive defiitio repeatedly will eable you to see the expressio we are seekig: a 5 a 3 r ; a 3 5 a 3 r 5 a 3 r 3 r 5 a 3 r ; a 4 5 a 3 3 r 5 a 3 r 3 r 5 a 3 r 3 ; So, as you see, you ca get to the th term by multiplyig a with r, ( ) times, ad therefore: th term of geometric sequece The geeral (th) term of a geometric sequece, a, with commo ratio r ad first term a, may be expressed explicitly as ( ) a 5 a 3 r This result is useful i fidig ay term of the sequece without kowig all the previous terms. Example 8 a) Fid the geometric sequece with a 5 ad r 5 3. b) Describe the sequece 3,, 48, 9, 768, c) Describe the sequece, _, _ 4, _ 8, d) Graph the sequece a 5 _

9 Solutio a) The geometric sequece is, 6, 8, 54,, 3 3. Notice that the ratio of a term to the precedig term is 3. b) This is a geometric sequece with a 5 3 ad r 5 4. The th term is a (4). Notice that, whe the commo ratio is egative, the terms of the sequece alterate i sig. c) The th term of this sequece is a 5 ( _ ). Notice that the ratio of ay two cosecutive terms is _. Also, otice that the terms decrease i value. d) The graph of the geometric sequece is show o the left. Notice that the poits lie o the graph of the fuctio y 5 _ 4 3x. Example 9 At 8:00 a.m., 000 mg of medicie is admiistered to a patiet. At the ed of each hour, the cocetratio of medicie is 60% of the amout preset at the begiig of the hour. a) What portio of the medicie remais i the patiet s body at oo if o additioal medicatio has bee give? b) If a secod dosage of 000 mg is admiistered at 0:00 a.m., what is the total cocetratio of the medicatio i the patiet s body at oo? Solutio a) We use the geometric model, as there is a costat multiple by the ed of each hour. Hece, the cocetratio at the ed of ay hour after admiisterig the medicie is give by: a 5 a 3 r ( ), where is the umber of hours Thus, at oo 5 5, ad a (5 ) b) For the secod dosage, the amout of medicie at oo correspods to 5 3, ad a (3 ) So, the cocetratio of medicie is mg. Compoud iterest Iterest compouded aually Whe we borrow moey we pay iterest, ad whe we ivest moey we receive iterest. Suppose a amout of e000 is put ito a savigs accout that bears a aual iterest of 6%. How much moey will we have i the bak at the ed of four years? 59

10 4 Sequeces ad Series It is importat to ote that the 6% iterest is give aually ad is added to the savigs accout, so that i the followig year it will also ear iterest, ad so o. Time i years Amout i the accout ( 0.06) 000( 0.06) (000( 0.06)) ( 0.06) ( 0.06) 5 000( 0.06) 3 000( 0.06) (000( 0.06) ) ( 0.06) ( 0.06) 5 000( 0.06) ( 0.06) 3 (000( 0.06) 3 ) ( 0.06) 3 ( 0.06) 5 000( 0.06) 4 Table 4. Compoud iterest. Table 4. Compoud iterest formula. This appears to be a geometric sequece with five terms. You will otice that the umber of terms is five, as both the begiig ad the ed of the first year are couted. (Iitial value, whe time 5 0, is the first term.) I geeral, if a pricipal of P euros is ivested i a accout that yields a iterest rate r (expressed as a decimal) aually, ad this iterest is added at the ed of the year, every year, to the pricipal, the we ca use the geometric sequece formula to calculate the future value A, which is accumulated after t years. If we repeat the steps above, with A 0 5 P 5 iitial amout r 5 aual iterest rate t 5 umber of years it becomes easier to develop the formula: Time i years 0 A 0 5 P Amout i the accout A 5 P Pr 5 P( r) A 5 A ( r) 5 P( r) t A t 5 P( r) t Notice that sice we are coutig from 0 to t, we have t terms, ad hece usig the geometric sequece formula, a 5 a 3 r ( ) A t 5 A 0 3 ( r) t Iterest compouded times per year Suppose that the pricipal P is ivested as before but the iterest is paid times per year. The r is the iterest paid every compoudig period. Sice every year we have periods, for t years, we have t periods. The amout A i the accout after t years is A 5 P ( r ) t 60

11 Example 0 E000 is ivested i a accout payig compoud iterest at a rate of 6%. Calculate the amout of moey i the accout after 0 years if a) the compoudig is aual b) the compoudig is quarterly c) the compoudig is mothly. Solutio a) The amout after 0 years is A 5 000( 0.06) 0 5 E b) The amout after 0 years quarterly compoudig is A ( ) 40 5 E84.0. c) The amout after 0 years mothly compoudig is A ( 0.06 ) 0 5 E Example You ivested E000 at 6% compouded quarterly. How log will it take this ivestmet to icrease to E000? Solutio Let P 5 000, r , 5 4 ad A i the compoud iterest formula: A 5 P ( r ) t The solve for t: ( ) 4t t Usig a GDC, we ca graph the fuctios y 5 ad y t ad the fid the itersectio betwee their graphs. As you ca see, it will take the E000 ivestmet.64 years to double to E000. This traslates ito approximately 47 quarters. You ca check your work to see that this is accurate by usig the compoud iterest formula: A ( ) 47 5 E03.8 Later i the book, you will lear how to solve the problem algebraically. Y.05 (4x) y y Itersectio X Y x x Example You wat to ivest 000. What iterest rate is required to make this ivestmet grow to 000 i 0 years if iterest is compouded quarterly? Solutio Let P 5 000, 5 4, t 5 0 ad A i the compoud iterest formula: A 5 P ( r ) t 6

12 4 Sequeces ad Series Now solve for r: ( r 4 ) 40 5 ( r ) 40 r r 54( ) 4 So, at a rate of 7% compouded quarterly, the 000 ivestmet will grow to at least 000 i 0 years. You ca check to see whether your work is accurate by usig the compoud iterest formula: A ( ) Populatio growth The same formulae ca be applied whe dealig with populatio growth. Example 3 The city of Bade i Lower Austria grows at a aual rate of 0.35%. The populatio of Bade i 98 was What is the estimate of the populatio of this city for 03? Solutio This situatio ca be modelled by a geometric sequece whose first term is 3 40 ad whose commo ratio is Sice we cout the populatio of 98 amog the terms, the umber of terms is is equivalet to the 33rd term i this sequece. The estimated populatio for Bade is, therefore, Populatio (03) 5 a (.0035) Note: Later i the book, more realistic populatio growth models will be explored ad more efficiet methods will be developed, as well as the ability to calculate iterest that is cotiuously compouded. Exercise 4.3 I each of questios 5 determie whether the sequece i each questio is arithmetic, geometric, or either. Fid the commo differece for the arithmetic oes ad the commo ratio for the geometric oes. Fid the commo differece or ratio ad the 0th term for each arithmetic or geometric oe as appropriate. 3, 3 a, 3 a, 3 3a, a b 5 4 c 5 c, ad c 5 5 u 5 3u, u 5 4 6, 5,.5, 3.5, 78.5, 7, 5,.5, 3.5, 78.5, 8,.75, 3.5, 4.5, 5, 9 8,, 8, 6, 3 9, 0 5, 55, 58, 6, 3, 3, 9, 7, 8, 0., 0., 0.4, 0.8,.6, 3., 3 3, 6,, 8,, 7, 4 6, 4, 0, 8, 34, 5.4, 3.7, 5, 6.3, 7.6, 6

13 For each arithmetic or geometric sequece i questios 6 3 fid a) the 8th term b) a explicit formula for the th term c) a recursive formula for the th term. 6 3,, 7,, 7 9, 5,, 7, 8 8, 3, 4, 5, , 9.95, 9.85, 9.75, 0 00, 99, 98, 97,, _,, 5_, 3, 6,, 4, 3 4,, 36, 08, 4 5, 5, 5, 5, 5 3, 6,, 4, 6 97, 34, 08, 36, 7, 3, 9_, , 5, 5 7, 65 49, 9 6, 3, 3_, 3_ 4, , 9, 38, 76, 3 00, 95, 90.5, 3, 3_ 4, 9 3, 7 56, 33 Isert 4 geometric meas betwee 3 ad 96. Hit: Defiitio: I a fiite geometric sequece a, a, a 3,, a k, the terms a, a 3,, a k are called geometric meas betwee a ad a k. 34 Fid 3 geometric meas betwee 7 ad Fid a geometric mea betwee 6 ad 8. Hit: This is also called the mea proportioal. 36 Fid 4 geometric meas betwee 7 ad Fid a geometric mea betwee 9 ad The first term of a geometric sequece is 4 ad the fourth term is 3, fid the fifth term ad a expressio for the th term. 39 The first term of a geometric sequece is 4 ad the third term is 6, fid the fourth term ad a expressio for the th term. 40 The commo ratio i a geometric sequece is _ 7 ad the fourth term is 4 3. Fid the third term. 4 Which term of the geometric sequece 6, 8, 54, is 8 098? 4 The 4th term ad the 7th term of a geometric sequece are 8 ad Is a term of this sequece? If so, which term is it? 43 The 3rd term ad the 6th term of a geometric sequece are 8 ad Is a term of this sequece? If so, which term is it? 44 Jim put 500 ito a savigs accout that pays 4% iterest compouded semiaually. How much will his accout hold 0 years later if he does ot make ay additioal ivestmets i this accout? 45 At her daughter Jae s birth, Charlotte set aside 500 ito a savigs accout. The iterest she eared was 4% compouded quarterly. How much moey will Jae have o her 6th birthday? 46 How much moey should you ivest ow if you wish to have a amout of 4000 i your accout after 6 years if iterest is compouded quarterly at a aual rate of 5%? 47 I 007, the populatio of Switzerlad was estimated to be 7554 (i thousads). How large would the Swiss populatio be i 0 if it grows at a rate of 0.5% aually? 63

14 4 Sequeces ad Series 48 The commo ratio i a geometric sequece is 3_ 7 ad the fourth term is 4 3. Fid the third term. 49 Which term of the geometric sequece 7,, 63, is 37 78? 50 Tim put 500 ito a savigs accout that pays 4% iterest compouded semiaually. How much will his accout hold 0 years later if he does ot make ay additioal ivestmets i this accout? 5 At her so William s birth, Jae set aside 000 ito a savigs accout. The iterest she eared was 6% compouded quarterly. How much moey will William have o his 8th birthday? 4.4 Series The word series i commo laguage implies much the same thig as sequece. But i mathematics whe we talk of a series, we are referrig i particular to sums of terms i a sequece, e.g. for a sequece of values a, the correspodig series is the sequece of S with S 5 a a a a If the terms are i a arithmetic sequece, we call the sum a arithmetic series. Sigma otatio Most of the series we cosider i mathematics are ifiite series. This ame is used to emphasize the fact that the series cotai ifiitely may terms. Ay sum i the series S k will be called a partial sum ad is give by S k 5 a a a k a k For coveiece, this partial sum is writte usig the sigma otatio: i = k S k 5 a i 5 a a a k a k i 5 Sigma otatio is a cocise ad coveiet way to represet log sums. Here, the symbol S is the Greek capital letter sigma that refers to the iitial i 5 k letter of the word sum. So, the expressio a i meas the sum of all the terms a i, where i takes the values from to k. We ca also write a i to i 5 i 5 m mea the sum of the terms a i, where i takes the values from m to. I such a sum, m is called the lower limit ad the upper limit. Example 4 Write out what is meat by: 5 a) i 4 i 5 7 b) 3 r r 5 3 c) x j p(x j ) j 5 64

15 Solutio 5 a) i i 5 7 b) 3 r r 5 3 c) x j p(x j ) 5 x p(x ) x p(x ) x p(x ) j 5 Example 5 Evaluate Solutio Example 6 Write the sum _ _ 3 _ 3 4 4_ 5 99 i sigma otatio. Solutio We otice that each term s umerator ad deomiator are cosecutive itegers, so they take o the absolute value of k or ay equivalet form. k We also otice that the sigs of the terms alterate ad that we have 99 terms. To take care of the sig, we use some power of () that will start with a positive value. If we use () k, the first term will be egative, so we ca use () k istead. We ca, therefore, write the sum as () _ () _ 3 ()3 _ 3 4 () k 5 () k k k Properties of the sigma otatio There are a umber of useful results that we ca obtai whe we use sigma otatio. For example, suppose we had a sum of costat terms 5 i 5 What does this mea? If we write this out i full, we get 5 i

16 4 Sequeces ad Series I geeral, if we sum a costat times the we ca write i 5 k 5 k k k 5 3 k 5 k. Suppose we have the sum of a costat times i. What does this give us? For example, 5 i 5 5i ( 3 4 5) However, this ca also be iterpreted as follows 5 i 5 5i ( 3 4 5) 5 5 i which implies that 5 i 5 5 5i 5 5 i i 5 I geeral, we ca say i 5 ki 5 k 3 k 3 k 3 5 k 3 ( ) 5 k i i 5 3 Suppose that we eed to cosider the summatio of two differet fuctios, such as k 5 I geeral, k 5 (k k 3 ) 5 ( 3 ) ( 3 ) 3 5 ( ) ( ) 5 (k ) k 5 k 5 (k 3 ) (f (k) g (k)) 5 f (k) g (k) Arithmetic series k 5 I arithmetic series, we are cocered with addig the terms of arithmetic sequeces. It is very helpful to be able to fid a easy expressio for the partial sums of this series. Let us start with a example: Fid the partial sum for the first 50 terms of the series We express S 50 i two differet ways: S , ad S S k 5 5 i 5 66

17 There are 50 terms i this sum, ad hece S S (5) This reasoig ca be exteded to ay arithmetic series i order to develop a formula for the th partial sum S. Let {a } be a arithmetic sequece with first term a ad a commo differece d. We ca costruct the series i two ways: Forward, by addig d to a repeatedly, ad backwards by subtractig d from a repeatedly. We get the followig two expressios for the sum: S 5 a a a 3 a 5 a (a d) (a d) (a ( )d ) ad S 5 a a a a 5 a (a d) (a d) (a ( )d ) By addig, term by term vertically, we get S 5 a (a d ) (a d ) (a ( )d ) S 5 a (a d ) (a d ) (a ( )d ) S 5 (a a ) (a a ) (a a ) (a a ) Sice we have terms, we ca reduce the expressio above to S 5 (a a ), which ca be reduced to S 5 (a a ), which i tur ca be chaged to give a iterestig perspective of the sum, i.e. S 5 ( a a ) is times the average of the first ad last terms! If we substitute a ( )d for a the we arrive at a alterative formula for the sum: S 5 (a a ( )d ) 5 (a ( )d ) Sum of a arithmetic series The sum, S, of terms of a arithmetic series with commo differece d, first term a, ad th term a is: S = _ (a + a ) or S = _ (a + ( )d ) Example 7 Fid the partial sum for the first 50 terms of the series Solutio Usig the secod formula for the sum, we get S ( 3 3 (50 )5) Usig the first formula requires that we kow the th term. So, a , which ow ca be used: S (3 48)

18 4 Sequeces ad Series Geometric series As is the case with arithmetic series, it is ofte desirable to fid a geeral expressio for the th partial sum of a geometric series. Let us start with a example: Fid the partial sum for the first 0 terms of the series We express S 0 i two differet ways ad subtract them: S S S S This reasoig ca be exteded to ay geometric series i order to develop a formula for the th partial sum S. Let {a } be a geometric sequece with first term a ad a commo ratio r. We ca costruct the series i two ways as before ad usig the defiitio of the geometric sequece, i.e. a 5 a 3 r, the S 5 a a a 3 a a, ad rs 5 ra ra ra 3 ra ra 5 a a 3 a a ra Now, we subtract the first ad last expressios to get S rs 5 a ra S ( r) 5 a ra S 5 a ra r ; r. This expressio, however, requires that r, a, as well as a be kow i order to fid the sum. However, usig the th term expressio developed earlier, we ca simplify this sum formula to S 5 a ra r 5 a ra r 5 a ( r ) ; r. r r Sum of a geometric series The sum, S, of terms of a geometric series with commo ratio r (r ) ad first term a, is: S = a ( r ) r [ equivalet to S = a (r ) r ] Example 8 Fid the partial sum for the first 0 terms of the series i the opeig example for this sectio. Solutio S 0 5 3( 0 ) 5 3( ) Ifiite geometric series Cosider the series k 5 ( _ ) k 5 _ _ 4 _ 8

19 Cosider also fidig the partial sums for 0, 0 ad 00 terms. The sums we are lookig for are the partial sums of a geometric series. So, 0 k 5 0 k 5 ( _ ) k 5 3 ( _ ) 0 ( _ ) k 5 3 ( _ ) 0 _ _ ( _ ) k 5 3 ( _ ) 00 k 5 _ 4 As the umber of terms icreases, the partial sum appears to be approachig the umber 4. This is o coicidece. I the laguage of limits, k 5 lim ( _ ) k 5 lim 3 ( _ ) k _ _ 5 4, sice lim ( _ ) 5 0. This type of problem allows us to exted the usual cocept of a sum of a fiite umber of terms to make sese of sums i which a ifiite umber of terms is ivolved. Such series are called ifiite series. Oe thig to be made clear about ifiite series is that they are ot true sums! The associative property of additio of real umbers allows us to exted the defiitio of the sum of two umbers, such as a b, to three or four or umbers, but ot to a ifiite umber of umbers. For example, you ca add ay specific umber of 5s together ad get a real umber, but if you add a ifiite umber of 5s together, you caot get a real umber! The remarkable thig about ifiite series is that, i some cases, such as the example above, the sequece of partial sums (which are true sums) approach a fiite limit L. The limit i our example is 4. This we write as lim a k 5 lim (a a a ) 5 L. k 5 We say that the series coverges to L, ad it is coveiet to defie L as the sum of the ifiite series. We use the otatio k 5 a k 5 lim k 5 a k 5 L. We ca, therefore, write the limit above as k 5 ( _ ) k 5 lim k 5 ( _ ) k 5 4. If the series does ot have a limit, it diverges ad does ot have a sum. We are ow ready to develop a geeral rule for ifiite geometric series. As you kow, the sum of the geometric series is give by S 5 a ra r 5 a ra r 5 a ( r ) ; r. r r If r,, the lim r 5 0 ad a S 5 S 5 lim a ( r ) 5 r r. 69

20 4 Sequeces ad Series We will call this the sum of the ifiite geometric series. I all other cases the series diverges. The proof is left as a exercise. ( _ ) k 5 k 5 _ 5 4, as already show. Sum of a ifiite geometric series The sum, S, of a ifiite geometric series with first term a, such that the commo ratio r satisfies the coditio r < is give by: S = a r Example 9 A ratioal umber is a umber that ca be expressed as a quotiet of two itegers. Show that 0. _ is a ratioal umber. Solutio 0. _ ( 0 ) 6 0 ( 0 ) 3 This is a ifiite geometric series with a ad r 5 0 ; therefore, 0. _ _ 3 Example 0 If a ball has elasticity such that it bouces up 80% of its previous height, fid the total vertical distaces travelled dow ad up by this ball whe it is dropped from a altitude of 3 metres. Igore frictio ad air resistace. Solutio 3 m.4 m.9 m 70 After the ball is dropped the iitial 3 m, it bouces up ad dow a distace of.4 m. Each bouce after the first bouce, the ball travels 0.8 times the previous height twice oce upwards ad oce dowwards. So, the total vertical distace is give by h 5 3 (.4 ( ) ( ) ) l The amout i parethesis is a ifiite geometric series with a 5.4 ad r The value of that quatity is l

21 Hece, the total distace required is h 5 3 () 5 7 m. Applicatios of series to compoud iterest calculatios Auities A auity is a sequece of equal periodic paymets. If you are savig moey by depositig the same amout at the ed of each compoudig period, the auity is called ordiary auity. Usig geometric series you ca calculate the future value (FV) of this auity, which is the amout of moey you have after makig the last paymet. You ivest e000 at the ed of each year for 0 years at a fixed aual iterest rate of 6%. See table below. Year Amout ivested Future value ( 0.06) ( 0.06) ( 0.06) 9 The future value of this ivestmet is the sum of all the etries i the last colum, so it is FV ( 0.06) 000( 0.06) 000( 0.06) 9 This sum is a partial sum of a geometric series with 5 0 ad r Hece, FV 5 000( ( 0.06)0 ) 5 000(( 0.06)0 ) ( 0.06) 0.06 This result ca also be produced with a GDC, as show. We ca geeralize the previous formula i the same maer. Let the periodic paymet be R ad the periodic iterest rate be i, i.e. i 5 r. Let the umber of periodic paymets be m. Period Amout ivested Future value m R R m R R( i ) m R R( i ) R R( i ) m Table 4.3 Calculatig the future value. Plot Plot Plot3 Mi U() U( )* 0.06) ( U(Mi) 000 V() V(Mi) W() sum(seq(u(),,, 0) Table 4.4 Calculatig the future value formula. The future value of this ivestmet is the sum of all the etries i the last colum, so it is FV 5 R R( i) R( i) R( i) m 7

22 4 Sequeces ad Series This sum is a partial sum of a geometric series with m terms ad r 5 i. Hece, FV 5 R( ( i)m ) 5 R ( ( i)m ) 5 R ( i) i ( ( i)m i ) Note: If the paymet is made at the begiig of the period rather tha at the ed, the auity is called auity due ad the future value after m periods will be slightly differet. The table for this situatio is give below. Table 4.5 Calculatig the future value (auity due). Period Amout ivested Future value m R R ( i ) m R R ( i ) m R R ( i ) 3 R R ( i ) m The future value of this ivestmet is the sum of all the etries i the last colum, so it is FV 5 R( i) R( i) R( i) m R( i) m This sum is a partial sum of a geometric series with m terms ad r 5 i. Hece, FV 5 R ( i( ( i)m )) 5 R ( i ( i)m ) ( i) i 5 R ( ( i)m i ) If the previous ivestmet is made at the begiig of the year rather tha at the ed, the i 0 years we have FV 5 R ( ( i)m i ) ( ( 0.06) ) Exercise 4.4 Fid the sum of the arithmetic series Fid the sum: Evaluate ( 0.3k). k Evaluate Evaluate Express each repeatig decimal as a fractio: a) 0. 5 b) c) At the begiig of every moth, Maggie ivests 50 i a accout that pays 6% aual rate. How much moey will there be i the accout after six years? 7

23 I questios 8 0, fid the sum The kth term of a arithmetic sequece is 3k. Fid, i terms of, the sum of the first terms of this sequece. How may terms should we add to exceed 678 whe we add 7 0 3? 3 How may terms should we add to exceed 335 whe we add 8 4? 4 A arithmetic sequece has a as first term ad d as commo differece, i.e., a, a d, a 4d,. The sum of the first 50 terms is T. Aother sequece, with first term a d, ad commo differece d, is combied with the first oe to produce a ew arithmetic sequece. Let the sum of the first 00 terms of the ew combied sequece be S. If T 00 5 S, fid d. 5 Cosider the arithmetic sequece 3, 7,,, 999. a) Fid the umber of terms ad the sum of this sequece. b) Create a ew sequece by removig every third term, i.e.,, 3,. Fid the sum of the terms of the remaiig sequece. 6 The sum of the first 0 terms of a arithmetic sequece is 35 ad the sum of the secod 0 terms is 735. Fid the first term ad the commo differece. I questios 7 9, use your GDC or a spreadsheet to evaluate each sum. 0 7 (k ) k i 3 i () Fid the sum of the arithmetic series Fid the sum Evaluate (3 0.k). k Evaluate Evaluate I questios 5 7, fid the first four partial sums ad the the th partial sum of each sequece. 5 u v u 5 Hit: Show that v 5 73

24 4 Sequeces ad Series 8 A ball is dropped from a height of 6 m. Every time it hits the groud it bouces 8% of its previous height. a) Fid the maximum height it reaches after the 0th bouce. b) Fid the total distace travelled by the ball till it rests. (Assume o frictio ad o loss of elasticity) The sides of a square are 6 cm i legth. A ew square is formed by joiig the midpoits of the adjacet sides ad two of the resultig triagles are coloured as show. a) If the process is repeated 6 more times, determie the total area of the shaded regio. b) If the process is repeated idefiitely, fid the total area of the shaded regio. 4 cm cm cm cm The largest rectagle has dimesios 4 by, as show; aother rectagle is costructed iside it with dimesios by. The process is repeated. The regio surroudig every other ier rectagle is shaded, as show. a) Fid the total area for the three regios shaded already. b) If the process is repeated idefiitely, fid the total area of the shaded regios. I questios 3 34, fid each sum Coutig priciples Simple coutig problems This sectio will itroduce you to some of the basic priciples of coutig. I Sectio 4.6 you will apply some of this i justifyig the biomial theorem ad i Chapter you will use these priciples to tackle may probability problems. We will start with two examples. 74

25 Example Nie paper chips each carryig the umerals 9 are placed i a box. Two chips are chose such that the first chip is chose, the umber is recorded ad the chip is put back i the box, the the secod chip is draw. The umbers o the chips are added. I how may ways ca you get a sum of 8? Solutio To solve this problem, cout the differet umber of ways that a total of 8 ca be obtaied: st chip d chip From this list, it is clear that you ca have 7 differet ways of receivig a sum of 8. Example Suppose ow that the first chip is chose, the umber is recorded ad the chip is ot put back i the box, the the secod chip is draw. I how may ways ca you get a sum of 8? Solutio To solve this problem too, cout the differet umber of ways that a total of 8 ca be obtaied: st chip d chip From this list, it is clear that you ca have 6 differet ways of receivig a sum of 8. The differece betwee the two situatios is described by sayig that the first radom selectio is doe with replacemet, while the secod is without replacemet, which ruled out the use of two 4s. Fudametal priciple of coutig The above examples show you simple coutig priciples i which you ca list each possible way that a evet ca happe. I may other cases, listig the ways a evet ca happe may ot be feasible. I such cases we eed to rely o coutig priciples. The most importat of which is the fudametal priciple of coutig, also kow as the multiplicatio priciple. Cosider the followig situatios: Example 3 You ca make a sadwich from oe of three types of bread ad oe of four kids of cheese, with or without pickles. How may differet kids of sadwiches ca be made? 75

26 4 Sequeces ad Series Solutio With each type of bread you ca have 4 sadwiches. There are possible sadwiches altogether. These are without pickles; if you wat sadwiches with pickles, the you have 4 possible oes. That is, there are possible sadwiches. Example 4 How may 3-digit eve umbers are there? Solutio The first digit caot be zero, sice the umber has to be a 3-digit umber, so there are 9 ways the hudred s digit ca be. There is o coditio o what the te s digit should be, so we have 0 possibilities, ad to be eve, the umber must ed with 0,, 4, 6, or 8. Therefore, we have digit eve umbers. Examples 3 ad 4 are examples of the followig priciple: Fudametal priciple of coutig If there are m ways a evet ca occur followed by ways a secod evet ca occur, the there are a total of (m)() ways that the two ca occur. This priciple ca be exteded to more tha two evets or processes: If there are k evets tha ca happe i,,, k ways, the the whole sequece ca happe i k ways. Example 5 A large school issues special coded idetificatio cards that cosist of two letters of the alphabet followed by three umerals. For example, AB 737 is such a code. How may differet ID cards ca be issued if the letters or umbers ca be used more tha oce? Solutio As the letters ca be used more tha oce, the each letter positio ca be filled i 6 differet ways, i.e. the letters ca be filled i ways. Each umber positio ca be filled i 0 differet ways; hece, the umerals ca be filled i differet ways. So, the code ca be formed i differet ways. Permutatios Oe major applicatio of the fudametal priciple is i determiig the umber of ways the objects ca be arraged. Cosider the followig situatio for example. You have 5 books you wat to put o a shelf: maths (M), physics (P), Eglish (E), biology (B), ad history (H). I how may ways ca you do this? 76

27 To fid this out, umber the positios you wat to place the books i as show If we decide to put the maths book i positio, the there are four differet ways of puttig a book i positio. M P M E M B M H Sice we ca put ay of the 5 books i the first positio, the there will be ways of shelvig the first two books. Oce you place the books i positios ad, the third book ca be ay oe of three books left. M P E M P B M P H Oce you use three books, there are two books for the fourth positio ad oly oe way of placig the fifth book. So, the umber of ways of arragig all 5 books is ! Factorial otatio The product of the first positive itegers is deoted by! ad is called factorial:! ( ) 3 ( ) 3 We also defie 0! 5. Permutatios A arragemet is called a permutatio. It is the reorgaizatio of objects or symbols ito distiguishable sequeces. Whe we place thigs i order, we say we have made a arragemet. Whe we chage the order, we say we have chaged the arragemet. So each of the arragemets that ca be made by takig some or all of a umber of thigs is kow as a permutatio. Number of permutatios of objects The previous set up ca be applied to objects rather tha oly 5. The umber of ways of fillig i the first positio ca be doe i ways. Hit: A permutatio of differet objects ca be uderstood as a orderig (arragemet) of the objects such that oe object is first, oe is secod, oe is third, ad so o Oce the first positio is filled, the secod positio ca be filled by ay of the objects left, ad hece usig the fudametal priciple there will be? ( ) differet ways for fillig the first two positios. Repeatig the same procedure till the th positio is filled is therefore? ( )? ( )? 5! Frequetly, we are egaged i arragig a subset of the whole collectio 77

28 4 Sequeces ad Series rather tha the etire collectio. For example, suppose we wat to shelve 3 of the books rather tha all 5 of them. The discussio will be aalogous to the previous situatio. However, we have to limit our search to the first three positios oly, i.e. the umber of ways we ca shelve three out of the 5 books is To chage this product ito factorial otatio, we do the followig: ! !!!! 5 5! (5 3)! This leads us to the followig geeral result. Number of permutatios of objects take r at a time The umber of permutatios of objects take r at a time is P r 5 P r 5 P r 5 P(, r) 5! ( r)! ; > r To verify the formula above, you ca proceed i the same maer as with the permutatio of objects. 3 (r ) 3 4 r Whe you arrive to the rth positio, you would have used r objects already, ad hece you are left with (r ) 5 r objects to fill this positio. So, the umber of ways of arragig objects take r at a time is P r 5? ( )? ( ) ( r ) Here agai, to make the expressio more maageable, we ca write it i factorial otatio: P r 5? ( )? ( ) ( r ) 5? ( )? ( ) ( r ) ( r)! ( r)!? ( )? ( ) ( r )? ( r)! 5 5! ( r)! ( r)! Example 6 5 drivers are takig part i a Formula car race. I how may differet ways ca the top 6 positios be filled? 78

29 Solutio Sice the drivers are all differet, this is a permutatio of 5 objects take 6 at a time. 5 P 6 5 5! (5 6)! This ca also be easily calculated usig a GDC. 5 Pr !/9! Combiatios A combiatio is a selectio of some or all of a umber of differet objects. It is a uordered collectio of uique sizes. I a permutatio, the order of occurrece of the objects or the arragemet is importat, but i combiatio the order of occurrece of the objects is ot importat. I that sese, a combiatio of r objects out of objects is a subset of the set of objects. For example, there are 4 permutatios of three letters out of ABCD, while there are oly 4 combiatios! Here is why: ABC ACB BAC BCA CAB CBA ABD ADB BAD BDA DAB DBA ACD ADC CAD CDA DAC DCA BCD BDC CBD CDB DBC DCB For oe combiatio, ABC for example, there are 3! 5 6 permutatios. This is true for all combiatios. So, the umber of permutatios is 6 times the umber of combiatios, i.e. 4 P 3 5 3! 4 C 3 where 4 C 3 is the umber of combiatios of the 4 letters take 3 at a time. Accordig to the previous result, we ca write 4 4! 4 P C ! 5 (4 3)! 5 4! 3! 3!(4 3)! The last result ca also be geeralized to elemets combied r at a time. (The ISO otatio for this quatity, which is also used by the IB is ( r ). I this book, we will follow the ISO otatio.) Every subset of r objects (combiatio), gives rise to r! permutatios. So, if you have ( r ) combiatios, these will result i r! ( r ) permutatios. Therefore, P r 5 r! ( r ) ( r ) 5 P r! r! 5 ( r)! 5! r! ( r)!r! ( r ) =! = ( r!( r)! r ). This symmetry is obvious as whe we pick r objects, we leave r objects behid, ad hece the umber of ways of choosig r objects is the same as the umber of ways of r objects ot chose. 79

30 4 Sequeces ad Series 45 Cr Example 7 A lottery has 45 umbers. If you buy a ticket, the you choose 6 of these umbers. How may differet choices does this lottery have? Solutio Sice 6 umbers will have to be chose ad order is ot a issue here, this is a combiatio case. The umber of possible choices is ( 45 6 ) This ca also be calculated usig a GDC. 5 Cr 5 3 Cr 3 3 Cr Example 8 I poker, a deck of 5 cards is used, ad a had is made up of 5 cards. a) How may hads are there? b) How may hads are there with 3 diamods ad hearts? Solutio a) Sice the order is ot importat, as a player ca reorder the cards after receivig them, this is a combiatio of 5 cards take 5 at a time: ( 5 5 ) b) Sice there are 3 diamods ad we wat 3 of them, there are ( 3 3 ) 5 86 ways to get the 3 diamods. Sice there are 3 hearts ad we wat of them, there are ( 3 ) 5 78 ways to get the hearts. Sice we wat them both to occur at the same time, we use the fudametal coutig priciple ad multiply 86 ad 78 together to get 308 possible hads. Example 9 A code is made up of 6 differet digits. How may possible codes are there? Solutio Sice there are 0 digits ad we are choosig 6 of them, ad sice the order we use these digits makes a differece i the code, the this is a permutatio case. The umber of possible codes is 0 P Exercise 4.5 Evaluate each of the followig expressios. a) 5 P 5 b) 5! c) 0 P d) 8 P 3 80

31 Evaluate each of the followig expressios. a) ( 5 5 ) b) ( 5 0 ) c) ( 0 3 ) d) ( 0 7 ) 3 Evaluate each of the followig expressios. a) ( 7 3 ) ( 7 4 ) b) ( 8 4 ) c) ( 0 6 ) ( 0 7 ) d) ( 7 ) 4 Evaluate each of the followig expressios. a) ( 8 5 ) ( 8 3 ) b)? 0! c) ( 0 3 ) ( 0 7 ) d) ( 0 ) 5 Tell whether each of the followig expressios is true. a) 0! 5! 5! b) (5!) 5 5! c) ( 0 8 ) 5 ( 0 93 ) 6 You are buyig a computer ad have the followig choices: three types of HD, two types of DVD players, four types of graphic cards. How may differet systems ca you choose from? 7 You are goig to a restaurat with a set meu. They have three starters, four mai meals, two driks, ad three deserts. How may differet choices are available for you to choose your meal from? 8 A school is i eed of three teachers: PE, maths, ad Eglish. They have 8 applicats for the PE positio, 3 applicats for the maths positio ad 3 applicats for Eglish. How may differet combiatios of choices do they have? 9 You are give a multiple choice test where each questio has four possible aswers. The test is made up of questios ad you are guessig at radom. I how may ways ca you aswer all the questios o the test? 0 The test i questio 9 is divided ito two parts, the first six are true/false questios ad the last six are multiple choice as described. I how may differet ways ca you aswer all questios o that test? Passwords o a etwork are made up of two parts. Oe part cosists of three letters of the alphabet, ot ecessarily differet, ad five digits, also ot ecessarily differet. How may passwords are possible o this etwork? How may 5-digit umbers ca be made if the uits digit caot be 0? 3 Four couples are to be seated i a theatre row. I how may differet ways ca they be seated if a) o restrictios are made b) every two members of each couple like to sit together? 4 Five girls ad three boys should go through a doorway i sigle file. I how may orders ca they do that if a) there are o costraits b) the girls must go first? 5 Write all the permutatios of the letters i JANE. 6 Write all the permutatios of the letters i MAGIC take three at a time. 8

32 4 Sequeces ad Series 7 A computer code is made up of three letters followed by four digits. a) I how may ways is the code possible? b) If 97 of the three-letter combiatios caot be used because they are offesive, how may codes are still possible? 8 A local bridge club has 7 members, 0 females ad 7 males. They have to elect three officers: presidet, deputy, ad treasurer. I how may ways is this possible if a) there are o restrictios b) the presidet is a male c) the deputy must be a male, the presidet ca be ay geder, but the treasurer must be a female d) the presidet ad deputy are of the same geder e) all three officers are ot the same geder. 9 The research ad developmet departmet for a computer maufacturer has 6 employees: 8 mathematicias, computer scietists, ad 6 electrical egieers. They eed to select three employees to be leaders of the group. I how may ways ca they do this if a) the three officers are of the same specializatio b) at least oe of them must be a egieer c) two of them must be mathematicias? 0 A combiatio lock has three umbers, each i the rage to 50. a) How may differet combiatios are possible? b) How may combiatios do ot have duplicates? c) How may have the first ad secod umbers matchig? d) How may have exactly two of the umbers matchig? I how may ways ca five married couples be seated aroud a circle so that spouses sit together? a) How may subsets of {,, 3,, 9} have two elemets? b) How may subsets of {,, 3,, 9} have a odd umber of elemets? 3 Nie seiors ad juiors make up the maths club at a school. They eed four members for a upcomig competitio. a) How may 4-member teams ca they form? b) How may of these 4-member teams have the same umber of juiors ad seiors? c) How may of these 4-member teams have more juiors tha seiors? 4 This problem uses the same data as questio 3 above. Tim, a juior, is the strogest mathlete amog his group while seior Gwe is the strogest amog her group. Either Tim or Gwe must be o the team, but they caot both be o the team. Aswer the same questios as above. 5 A shipmet of 00 hard disks cotais 4 defective disks. We choose a sample of 6 disks for ispectio. a) How may differet possible samples are there? b) How may samples could cotai all 4 defective disks? What percetage of the total is that? c) How may samples could cotai at least defective disk? What percetage of the total is that? 8

33 6 There are three political parties represeted i a parliamet: 0 coservatives, 8 liberals, ad 4 idepedets. A committee of 6 members is eeded to be set up. a) How may differet committees are possible? b) How may committees with equal represetatio are possible? 7 How may ways are there for 9 boys ad 6 girls to stad i a lie so that o two girls stad ext to each other? 4.6 The biomial theorem A biomial is a polyomial with two terms. For example, x y is a biomial. I priciple, it is easy to raise x y to ay power; but raisig it to high powers would be tedious. We will fid a formula that gives the expasio of (x y) for ay positive iteger. The proof of the biomial theorem is give i Sectio 4.7. Let us look at some special cases of the expasio of (x y) : (x y) 0 5 (x y) 5 x y (x y) 5 x xy y (x y) 3 5 x 3 3x y 3xy y 3 (x y) 4 5 x 4 4x 3 y 6x y 4xy 3 y 4 (x y) 5 5 x 5 5x 4 y 0x 3 y 0x y 3 5xy 4 y 5 (x y) 6 5 x 6 6x 5 y 5x 4 y 0x 3 y 3 5x y 4 6xy 5 y 6 There are several thigs that you will have oticed after lookig at the expasio: There are terms i the expasio of (x y). The degree of each term is. The powers o x begi with ad decrease to 0. The powers o y begi with 0 ad icrease to. The coefficiets are symmetric. For istace, otice how the expoets of x ad y behave i the expasio of (x y) 5. The expoets of x decrease: (x y) 5 5 x 5u 5x 4u y 0x 3u y 0x u y 3 5x u y 4 x 0u y 5 The expoets of y icrease: (x y) 5 5 x 5 y 0u 5x 4 y u 0x 3 y u 0x y 3u 5xy 4u y 5u Usig this patter, we ca ow proceed to expad ay biomial raised to power : (x y). For example, leavig a blak for the missig coefficiets, the expasio for (x y) 7 ca be writte as (x y) 7 5 ux 7 ux 6 y ux 5 y ux 4 y 3 ux 3 y 4 ux y 5 uxy 6 uy 7 83

34 4 Sequeces ad Series To fiish the expasio we eed to determie these coefficiets. I order to see the patter, let us look at the coefficiets of the expasio we started the sectio with. (x y) 0 row 0 (x y) row (x y) row (x y) row 3 (x y) row 4 (x y) row 5 (x y) row colum colum colum colum colum colum colum A triagle like the oe above is kow as Pascal s triagle. Notice how the first ad secod terms i row 3 give you the secod term i row 4; the third ad fourth terms i row 3 give you the fourth term of row 4; the secod ad third terms i row 5 give you the third term i row 6; ad the fifth ad sixth terms i row 5 give you the sixth term i row 6, ad so o. So ow we ca state the key property of Pascal s triagle. Pascal s triagle was kow to Persia ad Chiese mathematicas i the 3th cetury. Pascal s triagle Every etry i a row is the sum of the term directly above it ad the etry diagoally above ad to the left of it. Whe there is o etry, the value is cosidered zero. Take the last etry i row 5, for example; there is o etry directly above it, so its value is 0 5. From this property it is easy to fid all the terms i ay row of Pascal s triagle from the row above it. So, for the expasio of (x y) 7, the terms are foud from row 6 as follows: So, (x y) 7 5 x 7 u7 x 6 y u x 5 y u35 x 4 y 3 u35 x 3 y 4 u x y 5 u7 xy 6 y 7. Note: Several sources use a slightly differet arragemet for Pascal s triagle. The commo usage cosiders the triagle as isosceles ad uses the priciple that every two etries add up to give the etry diagoally below them, as show i the followig diagram

35 Example 30 Use Pascal s triagle to expad (k 3) 5. Solutio We ca fid the expasio above by replacig x by k ad y by 3 i the biomial expasio of (x y) 5. Usig the fifth row of Pascal s triagle for the coefficiets will give us the followig: (k) 5 5(k) 4 (3) 0(k) 3 (3) 0(k) (3) 3 5(k)(3) 4 (3) 5 5 3k 5 40k 4 70k 3 080k 80k 43. Pascal s triagle is a easy ad useful tool i fidig the coefficiets of the biomial expasio for relatively small values of. It is ot very efficiet doig that for large values of. Imagie you wat to evaluate (x y) 0. Usig Pascal s triagle, you will eed the terms i the 9th row ad the 8th row ad so o. This makes the process tedious ad ot practical. Luckily, we have a formula that ca fid the coefficiets of ay Pascal s triagle row. This formula is the biomial formula, whose proof is beyod the scope of this book. Every etry i Pascal s triagle is deoted by ( r ), which is also kow as the biomial coefficiet. I ( r ), is the row umber ad r is the colum umber. The factorial otatio makes may formulae ivolvig the multiplicatio of cosecutive positive itegers shorter ad easier to write. That icludes the biomial coefficiet. The biomial coefficiet With ad r as o-egative itegers such that > r, the biomial coefficiet ( r ) is defied by ( r ) 5! r!( r)! Example 3 Fid the value of a) ( 7 3 ) b) ( 7 4 ) c) ( 7 0 ) d) ( 7 7 ) Solutio a) ( 3 7 ) 5 7! 5 7! 3!(7 3)! 3!4! ( 3)( 3 4) b) ( 4 7 ) 5 7! 5 7! 3!(7 4)! 4!3! ( 3 4)( 3) c) ( 0 7 ) 5 7! 5! 7/ 0!(7 0)! 0!7/! 5 5 d) ( 7 7 ) 5 7! 5! 7/ 7!(7 7)! 7/!0! 5 5 Hit: Your calculator ca do the tedious work of evaluatig the biomial coefficiet. If you have a TI, the biomial coefficiet appears as C r, which is aother otatio frequetly used i mathematical literature. 7 Cr 3 7 Cr 4 7 Cr

36 4 Sequeces ad Series Although the biomial coefficiet ( r ) appears as a fractio, all its results where ad r are o-egative itegers are positive itegers. Also, otice the symmetry of the coefficiet i the previous examples. This is a property that you are asked to prove i the exercises: Example 3 Calculate the followig: ( r ) 5 ( r ) ( 6 0 ), ( 6 ), ( 6 ), ( 6 3 ), ( 6 4 ), ( 6 5 ), ( 6 6 ) Solutio ( 6 0 ) 5, ( 6 ) 5 6, ( 6 ) 5 5, ( 6 3 ) 5 0, ( 6 4 ) 5 5, ( 6 5 ) 5 6, ( 6 6 ) 5 The values we calculated above are precisely the etries i the sixth row of Pascal s triagle. We ca write Pascal s triagle i the followig maer: ( 0 0 ) ( 0 ) ( ) ( 0 ) ( ) ( ) ( 0 3 ) ( 3 ) ( 3 ) ( 3 3 ) ( 0 ) ( ) ( ) Example 33 Calculate ( r ) ( r ). This is called Pascal s rule. Hit: You will be able to provide reasos for the steps after you do the exercises! Solutio ( r ) ( r ) 5! (r )!( r )!! r!( r)! 5! r r (r )!( r )!! ( r ) r!( r)! ( r ) 5! r! ( r ) r!( r )! r!( r )!! r! ( r ) 5 5!(r r ) r!( r )! r!( r )!!( ) 5 r!( r )! 5 ( )! 5 ( r!( r)! r ) 86

37 If we read the result above carefully, it says that the sum of the terms i the th row (r )th ad rth colums is equal to the etry i the ( )th row ad rth colum. That is, the two etries o the left are adjacet etries i the th row of Pascal s triagle ad the etry o the right is the etry i the ( )th row directly below the rightmost etry. This is precisely the priciple behid Pascal s triagle! Usig the biomial theorem We are ow prepared to state the biomial theorem. The proof of the theorem is optioal ad will require mathematical iductio. We will develop the proof i Sectio 4.7. (x y) 5 ( 0 ) x ( ) x y ( ) x y ( 3 ) x 3 y 3 ( ) xy ( ) y I a compact form, we ca use sigma otatio to express the theorem as follows: (x y) 5 i 5 0 ( i ) x i y i Example 34 Use the biomial theorem to expad (x y) 7. Solutio (x y) 7 5 ( 7 0 ) x 7 ( 7 ) x 7 y ( 7 ) x 7 y ( 7 3 ) x 7 3 y 3 ( 7 4 ) x 7 4 y 4 Example 35 ( 7 5 ) x 7 5 y 5 ( 7 6 ) xy 6 ( 7 7 ) y 7 5 x 7 7x 6 y x 5 y 35x 4 y 3 35x 3 y 4 x y 5 7xy 6 y 7 Fid the expasio for (k 3) 5. Solutio (k 3) 5 5 ( 5 0 ) (k) 5 ( 5 ) (k) 4 (3) ( 5 ) (k) 3 (3) ( 5 3 ) (k) (3) 3 ( 5 4 ) (k)(3) 4 ( 5 5 ) (3) 5 5 3k 5 40k 4 70k 3 080k 80k 43 Note: Why is the biomial theorem related to the umber of combiatios of elemets take r at a time? Cosider evaluatig (x y). I doig so, you have to multiply (x y) times by itself. As you kow, oe term has to be x. How to get this term? x is the result of multiplyig x i each of the factors (x y) ad that ca oly happe i oe way. However, cosider the term cotaiig x r. To have a power of r over the x, meas that the x i each of r factors has to be multiplied, ad the rest will be the r y-terms. This ca happe i ( r ) ways. Hece, the coefficiet of the term x r y r is ( r ). Example 36 Fid the term cotaiig a 3 i the expasio (a 3b) 9. 87

38 4 Sequeces ad Series Solutio To fid the term, we do ot eed to expad the whole expressio. Sice (x y) 5 i 5 0 ( i ) x i y i, the term cotaiig a 3 is the term where i 5 3, i.e. whe i 5 6. So, the required term is ( 9 6 ) (a) 9 6 (3b) a 3 79b a 3 b 6. Example 37 Fid the term idepedet of x i ( 4x 3 x ) 5. Solutio The phrase idepedet of x meas the term with o x variable, i.e. the costat term. A costat is equivalet to the product of a umber ad x 0, sice x 0 =. We are lookig for the term i the expasio such that the resultig power is zero. I terms of i, each term i the expasio is give by ( 5 i ) (4x 3 ) 5 i ( x ) i Thus, for the costat term: 3(5 i) i = 0 5 5i = 0 i = 3 Therefore, the term idepedet of x is: ( 5 3 ) (4x 3 ) ( x ) 3 = 0 6x 6 ( 8x 6 ) = 80 Example 38 Fid the coefficiet of b 6 i the expasio of ( b b ). Solutio The geeral term is ( i ) (b ) i ( b ) i 5 ( i ) () i (b ) i ( b ) i 5 ( i ) () i b 4 i b i () i 5 ( i ) () i b 4 3i () i 4 3i 5 6 i 5 6. So, the coefficiet i questio is ( 6 ) () 6 () Exercise 4.6 Use Pascal s triagle to expad each biomial. a) (x y) 5 b) (a b) 4 c) (x 3) 6 d) ( x 3 ) 4 e) (x 3b) 7 f ) ( ) 6 g) ( 3 x x ) 4 88

39 Evaluate each expressio. a) ( 3 8 ) b) ( 8 5 ) ( 3 8 ) c) ( 4 7 ) ( 3 7 ) d) ( 5 0 ) ( 5 ) ( 5 ) ( 5 3 ) ( 5 4 ) ( 5 5 ) e) ( 6 0 ) ( 6 ) ( 6 ) ( 6 3 ) ( 6 4 ) ( 6 5 ) ( 6 6 ) 3 Use the biomial theorem to expad each of the followig. a) (x y) 7 b) (a b) 6 c) (x 3) 5 d) ( x 3 ) 6 e) (x 3b) 7 f ) ( ) 6 g) ( 3 x x ) 4 h) ( 5 ) 4 ( 5 ) 4 i) ( 3 ) 8 ( 3 ) 8 j) ( i ) 8, where i 5 k) ( i ) 6, where i 5 4 Cosider the expressio ( x x ) 45. a) Fid the first three terms of this expasio. b) Fid the costat term if it exists or justify why it does ot exist. c) Fid the last three terms of the expasio. d) Fid the term cotaiig x 3 if it exists or justify why it does ot exist. 5 Prove that ( k ) 5 ( k ) for all, k [ N ad > k. 6 Prove that for ay positive iteger, ( ) ( ) ( ) ( ) 5 Hit: 5 ( ) 7 Cosider all, k [ N ad > k. a) Verify that k! 5 k(k )! b) Verify that ( k )! 5 ( k ) ( k)! c) Justify the steps give i the proof of ( r ) ( r ) 5 ( r ) i the examples. 8 Fid the value of the expressio: ( 6 0 ) ( 3 ) 6 ( 6 ) ( 3 ) 5 ( 3 ) ( 6 ) ( 3 ) 4 ( 3 ) ( 6 6 ) ( 3 ) 6 9 Fid the value of the expressio: ( 8 0 ) ( 5 ) 8 ( 8 ) ( 5 ) 7 ( 3 5 ) ( 8 ) ( 5 ) 6 ( 3 5 ) ( 8 8 ) ( 3 5 ) 8 0 Fid the value of the expressio: ( 0 ) ( 7 ) ( ) ( 7 ) ( 6 7 ) ( ) ( 7 ) ( 6 7 ) ( ) ( 6 7 ) Fid the term idepedet of x i the expasio of ( x x ) 6. Fid the term idepedet of x i the expasio of ( 3x x ) 8. 3 Fid the term idepedet of x i the expasio of ( x 3 x 3 ) 8. 4 Fid the first three terms of the expasio of ( x) 0 ad use them to fid a approximatio to a).0 0 b)

40 4 Sequeces ad Series ( ) r ) ( ) 5 ad iterpret your result o the 5 Show that ( ) ( r r r etries i Pascal s triagle. 6 Express each repeatig decimal as a fractio: a) 0. 7 b) c) Fid the coefficiet of x 6 i the expasio of (x 3)9. 8 Fid the coefficiet of x 3b 4 i (ax b) Fid the costat term of z. z 0 Expad (3 m)5. ( ) Fid the coefficiet of r 0 i (4 3r ) Mathematical iductio Domio effect I additio to playig games of strategy, aother familiar activity usig domioes is to place them o edge i lies, the topple the first tile, which falls o ad topples the secod, which topples the third, etc., resultig i all of the tiles fallig. Arragemets of millios of tiles have bee made that have take may miutes to fall. The Netherlads has hosted a aual domio topplig competitio called Domio Day sice 986. The record, achieved i 006, is domioes. Similar pheomea of chais of small evets each causig similar evets leadig to a evetual grad result, by aalogy, are called domio effects. The pheomeo also has some theoretical bearig to familiar applicatios like the amplifier, digital sigals, or iformatio processig. 90