Mu Sequences/Series Solutions National Convention 2014

Transcription

1 Mu Sequeces/Seres Solutos Natoal Coveto 04 C 6 E A 6C A 6 B B 7 A D 7 D C 7 A B 8 A B 8 A C 8 E 4 B 9 B 4 E 9 B 4 C 9 E C 0 A A 0 D B 0 C C Usg basc propertes of arthmetc sequeces, we fd a ad bm m We eed to fd teger solutos to m Takg both sdes mod, we obta that m mod, or that mkfor tegers k The the umbers both bm ad a are of the form k 6k 7 Fally, 6k , so there are 6 commo elemets B Recall that T So the sum we seek s the Let S deote The S Subtractg these two values yelds S T S Let T deote The Subtractg the two yelds T T Dvdg ths by two results Subtractg these two gves 4 T T 6 Hece, the sum s ST B Wrte out the sum as Notce we ca rewrte the sum, backwards, as Takg the lmt to fty, we e recogze each term after the frst as a power of e: e e e e e 4 B Sce the harmoc seres dverges, we kow by defto that the sequece of partal sum dverges Sce the alteratg harmoc seres s codtoally coverget, by the Rema Seres Theorem, ts terms ca be arraged a permutato so that the sum coverges to ay gve value D ca be show true by cosderg how may terms o each terval 0,0 are left over after terms are removed, fdg a upper boud for the sum as a result, ad by comparso test, t coverges Whle the harmoc seres dverges, the rato test s coclusve C If r s the proporto of reboud o each bouce, the the total dstace the ball travels s r Solvg, r, so t rebouds % ts prevous heght r 6 E By Bomal Theorem, ths s Page of

2 Mu Sequeces/Seres Solutos Natoal Coveto 04 7 A Notce we ca rewrte ths sum as By the p-seres test, ths coverges so log as, or 0 The roots of ths are,, ad Testg tervals for postvty, we fd that the postve area s,, a 8 A Let the roots be, a, ad ar 7 The the product of the roots, a a Takg the r sum of the roots two at a tme, a r Lettg a ad solvg, we get r 8 4 r, So the roots are,, ad 9, whch has a sum of p p 8 9 B Wrtg ths out, 4 0 l l l l Usg propertes of logs, ths s l 4 04 Cacellg, we obta l l A Lettg shows that f () f () f () f ( ) f ( ) Lettg yelds f () f () f () f ( ) ( ) Subtractg the two, f ( ), so the sum of the coeffcets s A The Rock s umber s of the form What we eed to fd, the, s 4 ca be factored out of 4 7 4, ad CM Puk s umber s of the form both ust fte geometrc seres, we obta to yeld Workg from the sde out, we realze that 4 7 Realzg these are ad 4 8 D Note that E E, so E E S The, E E E E E E E, so the aswer s 7 6 Page of

3 Mu Sequeces/Seres Solutos Natoal Coveto 04 B Note that we ca splt the sum up as The secod sum s clearly e For the 0! 0! frst sum, ote that, so Also recogze that 0! 0! 0! e Smlarly, e Thus,!!!!!! 0 the etre sum s e e e 0 4 E Ths s smply the teth Fboacc umber A Note that C, whch from Bomal Theorem, s To fd the last three dgts, we eed 9 0 to fd ths sum modulo 000 Notce that we oly eed to cosder the last three terms, sce ay term whch cludes a multple of 0 or hgher wll be a multple of 000, mod000, ad mod000 Hece, the 0 last three dgts are 0, 0, ad, ad the sum s a lm lm lm a coverge ths must be less tha For the seres to 7 D By the P-Seres Test, A coverges By the Alteratg Seres Test, B coverges By the Root Test, C coverges By the Dvergece Test, D dverges 8 A Sce 0 e! 9 B Recall that, e 0! yelds l l l Multplyg by!! 0 0 yelds Dvdg both sdes by ad tegratg wth respect to Our desred sum occurs whe we let to obta Page of

4 Mu Sequeces/Seres Solutos Natoal Coveto 04 0 D The form provded hts at epoetal growth So we look for solutos to the recurrece of the form c for some costat c Substtutg, we get c c 6c Dvdg through by c ad gettg everythg o oe sde, we obta our characterstc equato of c c 6 0, whch factorg yelds solutos of c, Thus, a, b Now, pluggg our tal codtos results, ad From the frst equato, we ow kow a b 0 If we wated to solve the system, 6 we d obta, geerates our recurrece A Frstly, sdes, result, so that 6 a It s easly verfable that ths s the geeratg fucto for,,,, Dfferetatg both wll thus result the geeratg fucto for (,,,4,) Multplyg by wll, the geeratg fucto for 0,,,, Dfferetatg oe last tme wll hece gve coeffcets that are eactly the postve perfect squares, ad a fal aswer of C I the deomator, dvde uder the radcal by to obta lm We k 4k recogze ths as a Rema Sum To covert to a tegral, let k so that d The tegral, the, s d sec 0 4 d We ca evaluate ths by lettg d The tegral would hece become ta 0 ta so that secd Recall that sec d l sec ta Thus, we obta l secta l C f ( ) f ( a) f '( a)( a) f ''( a)( a), gve a 4 ad f l So f '( ) ad f ''( ) Pluggg the values gves the desred result Page 4 of

5 Mu Sequeces/Seres Solutos Natoal Coveto C Note that ta ta ta cos The, ta sec 0 cos s s s s s 0 Factorg, s s 0 s,s O 0,, the solutos are,, ad for a total of 6 6 B Usg learty, we ca splt ths up as 60 ad 6 Hece, D Let S deote the orgal sum, ad ote the symmetry: Addg S to tself, we obta S S A Note that s, ad so coverget s By Comparso Test, ths seres s absolutely!! 8 E Whe the product s epaded, we ed up wth a product of coses over a product of ses Due to the cofucto detty cos s, the umerator ad deomator have precsely the same etres, so the product s 9 E Noe of the aswers must ecessarly be true 0 C Recall that 0! cos So,!! 0 0 cos Page of