SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES

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1 SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces, material ot icluded i the text. Cocept Developmet 1. I Sectio 1.5, we itroduce the cocept of summatio. I would like to begi with a story about Carl Freidrich Gauss ( ), sometimes rated as oe of the greatest mathematicias who ever lived. The story goes somethig like this: As a youth, Gauss of course atteded class with oly boys uder the tutelage of a headmaster. Oe day the boys were beig particularly uruly, so the master assiged them a task to keep them occupied: Add the umbers (that is ). The boys bet to their task, all but Carl Freidrich Gauss, who put his had i the air ad said Sir! The aswer is ad he was right! How did he do it? Oe way is to simply strig the umbers out as see below: If you look at the first ad last you see a sum of 101. If you look at the d ad the ext to last you see + 99 = 101. If you look at the ext pairig, , you see a sum of 101! There are 50 pairs i this overall sum ad each has a total of 101. The result: = 5050! Why is this process valid? If you start at the lower ed ad move i oe etry i the sum, that is, move from 1 to, you add 1. If you start at the upper ed ad move back oe i the sum, that is, move from 100 to 99, you lose 1. The loss o the upper ed compesates for the icrease o the lower ed. This process repeats itself as you move from oe pairig to aother throughout the overall sum. Result: 50 pairs of 101 ad total of This process works best whe you have a eve umber of terms i your sum. A alterative approach is to write the sum dow twice, oce frot to back ad the secod time back to frot as see below: This time we pair the umbers vertically, agai obtaiig sums of 101. How may such sums do we have? (Aswer: 100) Thus, i all the idividual sums we have a total of 100(101) = 10,100. However, we have couted each umber twice, so, to compesate, we divide by. Result? 5050! This process ca be geeralized with a sum of the form ( 1) + ( ) + where you have terms. Whe you write the sum dow twice, oce frot to back ad the secod time back to frot, the add vertically, you have pairs of ( + 1) as see below.

2 ( ) + ( 1) + + ( + 1) + ( + ) Total? ( + 1). However we couted each term i our iitial sum twice, so, to compesate, we divide by, leadig to the formula: ( 1) + ( ) + = ( + 1) Thus, ay time we see a sum of this form, we may use the formula, the first of those preseted o page 8.. We ofte see sums represeted with summatio otatio. For example the sum of Gauss story may be represeted as: where the symbol the bouds of summatio = i stads for summatio, i is kow as a idex, ad the umbers 1 ad 100 are 10 You try oe! What sum is represeted by j? (Aswer: ) i = See Examples 1 ad, pages ad 7 for further work with these cocepts. Note also properties that apply illustrated o page 7 umbered to the right of the page, 7 ad Additioal formulas, preseted without proof at the top of page 8, are as follows: 100 i = = ( 1) ( + 1) + ad i 3 = = ( + 1) 4 Please see additioal work with these formulas ad the summatio properties i Example 3, pages 8 ad Next, we iclude some additioal material ot preseted i the text. What if we looked at the terms of a summatio, but did ot add? Such a listig of terms is kow as a sequece or a progressio. Examples below: 1,, 3, 4, a ifiite sequece sice the dots idicate it cotiues forever or 1,, 3, 4, , 100 a sequece represetig the terms of the sum we discussed above

3 Obviously, terms of a sequece may be added. We may use summatio otatio or formulas ad properties discussed above whe appropriate. Such a summatio of the terms of a sequece is kow as a series. 5. We are ofte give a geeral term or th term for a sequece. For example, if I told you that you to list etries i a sequece with geeral term, you would respod with the listig, 4,, ad if I asked for etries i a sequece with geeral term, you would respod with 1, 4, 9, 1, 5,..... What about the sum of the first 10 etries of such a sequece? Use the formula preseted i step 3 above: ( 0 1) 10 (10 + 1) + = You would the evaluate usig the arithmetic tools of cacellatio ad calculatio to obtai There are two special types of sequeces, arithmetic ad geometric. (Note the ew use of familiar termiology) I a arithmetic sequece, we start with a iitial etry, the add a commo differece repeatedly obtaiig additioal etries i the sequece. If our iitial etry is ad our commo differece is 3, we would obtai the sequece, 5, 8, 11, 14, If our iitial etry was 3 ad our commo differece is, we would obtai the sequece 3, 5, 7, 9, 11, I a geometric sequece, we start with a iitial etry, the multiply by a commo ratio repeatedly obtaiig additioal etries i the sequece. If our iitial etry is ad our commo ratio is 3, we would obtai the sequece,, 18, 54, 1, If our iitial etry was 3 ad our commo ratio is, we would obtai the sequece 3,, 1, 4, 48, We are ofte give a arithmetic or geometric sequece ad asked to fid the geeral or th term. How do we go about this task?? I a very orgaized way! Let s start with the sequece 1, 3, 5, 7, 9, We have show the first terms i a ifiite sequece. The first term is 1, the d is 3, the 3 rd is 5 ad so o. Thus, you see that we may umber our terms. Let s put each term vertically ito a table, umber it ad aalyze it. Number of Etry () Term Aalysis Simplificatio () () () () () How ca we develop the geeral term from this iformatio? We ote, i the right had colum, that the iitial etry (1) ad the commo differece () are costat. The oly piece that chages is the coefficiet (multiplier) of. Ca we tie it to aythig? Yes! I each case the coefficiet is oe less

4 tha the umber of the term! It would appear that our terms are of the form 1 + ( 1). Simplified, our geeral term is 1 + or 1. Lets check this! Our th term above is ad () 1 = 11! It appears to work! Ca we carry this forward? What would the 100 th term i our sequece be? (100) 1 = 199. What would the 1000 th etry be? (1000) 1 = 1999! 7. What about the geeral term of a geometric sequece? Let s start with the sequece 3,, 1, 4, 48, We have show the first terms i a ifiite sequece. The first term is 3, the d is, the 3 rd is 1 ad so o. Thus, you see that we may umber our terms. Let s put each term vertically ito a table, umber it ad aalyze it. Number of Etry () Term Aalysis Simplificatio ( 0 ) 3 3 ( 1 ) ( ) ( 3 ) ( 4 ) How ca we develop the geeral term from this iformatio? We ote, i the right had colum, that the iitial etry (3) ad the commo ratio () are costat. The oly piece that chages is the expoet applied to. Ca we tie it to aythig? Yes! I each case the expoet is oe less tha the umber of the term! It would appear that our terms are of the form 3( 1 ). Lets check this! Our th term above is ad 3( 5 ) = 3 3 = 9! It appears to work! Ca we carry this forward? What would the 100 th term i our sequece be?. 3( 99 ) =?!? (Let s leave it i expoetial form!) What would the 1000 th etry be? 3( 999 ) Do you eed additioal help with these cocepts? Help optios: 1. Try the practice problems below. Solutios are preseted followig the assigmet. A. Use a formula preseted i Sectio 1.5 to fid B. Evaluate j j = 1 C. Fid the 4 th term i the sequece with geeral term 3( 5 ( 1) ) D. Fid the geeral term of the arithmetic sequece 4, 7, 10, 13, Summary I this sectio ad the additioal material, we leared work with summatio cocepts, formulae ad otatio ad with sequeces, usig a geeral term or format to develop a sequece ad fidig the geeral term of a give arithmetic or geometric sequece.

5 Assigmet: Complete the followig Sectio 1.5 problems (page 9): 1 15 (odd problems oly), 19, plus additioal problems below. 1. If a sequece is represeted as a i =, 3, 4, 5,, fid a 3.. Write the first 5 terms of a arithmetic sequece with first term of 3 ad commo differece. 3. Write the first 5 terms of a geometric sequece with first term 3 ad commo ratio. 4. What are the ext 3 etries i the ifiite sequece 5, 10, 17,, 37,....? 5. a.) Fid the geeral term of the ifiite sequece 3, 8, 13, 18, 3, b.) Fid the 100 th term of this sequece.. a.) Fid the geeral term of the ifiite sequece 3,, 1, 4, 48, b.) Fid the 100 th term of this sequece. (You may leave your aswer i expoetial form.) Solutio to Practice Problem A: The formula of choice is 45 45(4) i = = ( + 1) i =. I this applicatio, it becomes Solutio to Practice Problem B: The formula of choice is 5 i ( + 1)( = + 1). I this applicatio, we see 5()(51) 5(13)(17) i = = = 555. (Note the use of the basic math techique cacellatio. ) 1 Solutio to Practice Problem C: The 4 th term i the sequece with geeral term 3 5 ( 1) is 3 5 (4 1) = = 3 15 = 375

6 Solutio to Practice Problem D: The geeral term of the arithmetic sequece 4, 7, 10, 13, may be foud as show below. First umber the terms of the sequece with 4 as the 1 st, 7 as the d ad so o, where represets the umber of the geeral term. Term Aalysis of Term Simplificatio (3) (3) (3) (3) Thus we see that each term i the sequece is 4 + ( 1)(3) where is the umber of the term. Thus the geeral term is 4 + ( 1)(3) = = Check your work. Is 7 = 3() + 1? Yes! Is 10 = 3(3) + 1? Yes! ad so o.