Bồi dưỡng kiến thức học sinh giỏi chuyên khảo dãy số (Tái bản có sửa chữa bổ sung): Phần 1

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Phần 1 tài liệu Bồi dưỡng học sinh giỏi chuyên khảo dãy số do Nguyễn Tài Chung biên soạn cung cấp cho người đọc các kiến thức cơ bản về: Xác định các dãy số, xác định giới hạn của dãy số. Mời các bạn cùng tham khảo nội dung chi tiết.

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Nội dung Text: Bồi dưỡng kiến thức học sinh giỏi chuyên khảo dãy số (Tái bản có sửa chữa bổ sung): Phần 1

510.76 B452D NGUYEN TAI CHUNG (B6i dtedfig hoc sinh gioi Todn, chuyen Todn) T A I B A N C O StTA CHUA V A B O SUNG Phan loai toan day so Phuong phap giai toan day so ^ Cac de thi hoc sinh gioi Quoc gia, khu vac -if Cac de thi Olympic Sinh vien, Olympic 30/04 Tim day so {Xnlnri sao c h o x i = a va xn.i = a x ^ b x ^ c X n + d,VneN* f a > 0,c= d = ^ ^ - I f ^ ). (1) y ia z/a^ DVL.013494 Tim so hang t o n g q u a t cua day so da cho.
NGUYEN TAI CHUNG B6i dvtdng hoc sinh gioi, chuyen toan C H U Y E N D E D A Y s6 V A I ? N G D U N G - Phan loai toan day s6 - PhiWng phap giai toan day s6 - Cac de t h i HSG Quoc gia, K h u vuTc - Cac d l t h i Olympic Sinh vien, Olympic 30/04 J*' N H A X U A T H A \I I I Q C Q U O C G I A H A NOI
-1 ^I3 (ivj*-! 1" L&i noi dau Chu'dng 1 Day so la mot. chuycn de quan trpng timoc clutrtng trinh chiiycn toan trong c&c trUdng THPT chuyen. Cac bai toan lien quan den day so thiTcJng la nhufng bai t$p Xac dinh day so kho, thirdng g?ip trong cac ki thi hpc sinh gi6i m5n To&n cap qu6c gia, khu v\fc, quoc te, Olympic 30/04 va Olympic Sinh vien. Toan day so rat phong phu, da d^ng va cung rat phutc h^p nen kho phan lo^i va hg thong hoa thanh cac chuycn do ricng bigt. Tuy v^y, chuug toi c6 gang toi da 1.1 Xac dinh day so bang phu^dng phap quy nap, sSp xep hpp li de giiip ban dpc tiep can tijfng bifdc, ttoig miic dp kien thiic va luyf n phu'cfng phap doi bien t§.p giai toan. Phan 16n cac bai toan trong cuon sach nay du^c tuyen chpn t\t cac ki thi; Thi 1.1.1 X a c d i n h d a y so b a n g phifdng p h a p q u y n a p . hpc sinh gioi quoc gia, thi chpn dpi tuyen quoc gia dil thi toan quoc te, Olympic 30/04, Olympic toan Sinh vien toan qu6c, thi hpc sinh gioi ciia cac ti'nh thanh, chiing minh menh de chiia bien A{7i) dung v6i moi so nguyen dUdng n ho$c tren T^p chi toan hpc va tu6i tre. Mpt so bai toan khac la do tac gia t\f bieii (bang phifdiig phap quy nap), ta tlittc hieu ba bitdc sau: soan. Xiii cam dii tac gia cac bai toan nia chiing toi da tri'ch chpn. Nhiing Idi giai dua ra dua tren tieu chi tir nhien, de hieu. 'l\iy nhien, IcJi giai d day chua han la Bxidc 1 {hxidc cd sd, hay bvTdc khdi d l u ) . Kiein tra A{n) diing khi n = 1. Idi giai hay nhat va ngSn gpn nhat, hi vong ban dpc so c6 dxtac nhftng Idi giai hay Biidc 2 (birdc quy n a p , hay bi:rdc "di t r u y § n " ) . Vdi fc e Z, fe > 1, gia hdn. sii A{n) diing khi n ^ fe, ta chiing minh A{n) cung diing khi n = fc + 1. Vice phan chia cac chiroug, bai, myc dfin c6 tinh dpc lap tiWng doi va do do Bifdc 3. Ket luan A{n) diing xcii moi so nguyen ditdng n. khong nhat thiet luc nao cung phai dpc theo trinh t\i. Cung can noi them rSng, that kho nia i)han chia cac van d^ theo mot bien gidi rach roi nhit tieu dg ciia tCrng bai. Dau do trong mpt vai van de ciia bai nay da xult hien bong dang van dg ciia B a i t o a n 1. Cho day s6 nhu sau: | = ^ , ^ 2 + ^ , Vn = 1 , 2 , . . . bai kia. Tuy vay, chiing toi van c6 gang xoay quanh chii de ciia bai iy de ban dpc a) Tinh x\, X2, Xi. . , , minh rut ra TiliuTng gi thudng gap va each giai quyet van de. b) Tim so hang tSng qudt {so hang thii n). Tot nhat, dpc gia t\t minh giai cac bai tap co trong sach nay. Tuy nhien, dc thay va lam cliii nhiiiig ki xao tinh vi khac, cac bai tap deu dUcJc giai san (tham chi la nhilu each giai) v6i nhiJng mute dp chi tiet khac nhau. Npi dung sach da c6 a) Ta CO x i = \/2 = 2 cos va gang tuan theo y chii d^o xuyen suot: BiSt diftfc IcJi giai ciia bai toan chi la yeu cku dau ticn - ma hon the - lam the nao de giai ditdc no, each ta xii li no, nhulig suy lu^n nao to ra "c6 li", cac ket luan, nhan xet va litu y tiT bai toan difa ra... v/2' = ^ 2 ( 1 + COS J ) = 2cos J , X 2 = V^2 + v / 2 = W2 1 + Hi vpng quygn sach nay la tai lieu tham khao c6 ich cho hpc sinh cac 16p chuyen toan trung hpc ph6 thong, giao vien toan, sinh vien toan ciia cac trirdng DHSP, DIIKIITN ciing nhir la tai lieu phyc vu cho cac ki thi hpc sinh gioi toan TIIPT, X3 = V'2T1^ - ^ 2 ( 1 + COS J ) = 2 cos ^ . thi Olympic loan Sinh vien giiJta cac tritdng dai hpc. Tac gia Th^ic sy Nguyen Thi Chung 3
BSLng phudng pliap quy nap ta se chiing minh b) Vdi e Z, k>l, gia sii xfc = 2 c o s K h i do Un = tan ,Vn = l , 2 , . . . = V2T^k = y ^ l T c ^ ^ ^ = 2COS . TrUdng h^p n = 1 da kigm tra d trgn. Gia sil Un = tan. [ | + ( n - l ) ^ ] . K h i do Theo nguyen l i quy n?ip suy ra x„ = 2 cos ^^^n = 1,2,... tan + tan^ Bai toan 2. x„ = ^ 2 + \/2 + • • • + v ^ . Hay tinh xjoro- .3+("-^)8, TT N CO ^ n dau cSn ' 1 - tan ^ 3 + ( n - l ) 3 ^ . t a n - ID = tan = tan Htrdng dan. Ta c6 x i = V2, x„_i = V^2+ •\/2T-• • + v/2. Suy ra .3-^"8 c6 n - 1 d&u c&n Tlieo nguyen ly quy nap suy ra x„ = v/2 + x„_i, Vn = 2,3,... Un = tan ,Vn = l , 2 , . . . L3+("-^)8J sa\ do sir di.ing bai toan 1 d trang 3. Do do Bai toan 3 (Do thi OLYMPIC 30/04/2003). Cho day s6 (u„) djnh bdi U2003 = tan = tan ui = v/3 Un + y/2~ 1 Wn+I = ,Vn= 1,2,... 1 + (1 - V ^ ) Un SO J Tim U2003- V3 Ui = G i a i . Tit cong thitc tan(x + y)= tanx + tany ^ Bai toan 4. Cho day s6 (u„) tfm/i bdi: < - + 2 - v/3 "n 1 - tanx. tan?/ ,Vne N» " " • ' ^ " l + (y3-2)u„ 2 tan? 7^m U2010- 1 = tan - = tan ( - + - = o_ 4 ^8 l-tan2? 1 cos - I tan - = ~i + y/2 Hvrdng d i n . Ta c6 ^an — = ^ = JIZ^ = 2 - N/3. Bang quy
3 / 2 \i 2^-1 2^ 2"~'-l 2"-' 1.1.2 Xac dinh day so b^ng phu'dng phap d6i bien (dat an phu). Ban dpc nen chii y mot so phep d6i bien rat thitdng diing sau day (khong V,y =« - - » r •- i ( a + ^ ) " ' - | ; , V n - 1.2,... T h . nhiJng thudng dung trong day so ma nhfmg phep doi bien nay con hay dimg khi giai phUdng trinh, chuTng minh bat ding thiic...): lai bang phUdng phap quy nap thay diing. Chu y 1. Neu ta gap ham da thiic bQ.c hai f{x) = ax^+bx+c thi ta ddi goc tQa Bai to6n 8. Tim day so {x„}^^ sac cho x\ a va do vi dinh A Pavabol. TiCc la ta ddi bien ~ x„+i = a x 3 + 6 x 2 + c x „ 4 - d , V n = l , 2 , . . . ^ a > 0 , c = ^ . ^ = (1) Chii y 2. Neu ta g&p ham da thiic bac ba f{x) = ax^ + bx^ + ex + d tht ta ddi goc toa do vc diem uSn Al - — (-—]) cAa do thi cua fix). TUc la HUdng dan. Gpi y„ = x„ + thay vao (1) ta dupe ta ddi bien X = x + —. 3a Chu y 3. Neu ta gg.p ham da thiic bac bon f{x) = ax* + bx^ + cx'^ + dx + e thi ta ddi qoc toa do diem sieu uon A (-—; f (- —]] cHa dd thi ciia / 3 , , 62 63 \ 2 262 63 62 6 62 6(c-3) fix). TUc la ta doi bien X = x + 4a / 3 63 $c-3) ( , 63 $ 6 Bai todn 7. Tim day s6 {x„};^~i biet xi = a, Xn+i = ox^ + 6x„ + c, Vn = 1,2,... (1)1 Suy ra j/„+i = ay^, Vn = 1,2,... Suy ra 62-26 Trong do a^O va c = 4a Vn = ayt, = a {ayt2f = a'^'vlU = a'^' {ay^.f = a'^'^'^t^ I4.'»J.l2-|....J.1'>-2 t-?"'' ,n-l 3"~'-l / 6 \ Hi/dng dan. Chii y 1 ci trang 6 gpi y cho ta each doi bien nhvl sau: Gpi = . . . = a*+^+-* + yf = a ' - 3 y\ I" 3a J yn = x„ + —, thay vao (1) ta dupe: 3»-> V*yx„ = a . (^a + - j - - , V n = l,2,... y „ ^ , - ^ = a ( j , „ - A ) + 6 ( y „ - A ) + c , V n = 1,2,... Luu y. Chd y 2 d trang 6 da gpi y cho ta each doi bien J/n = + 3^- 6 ( 1 hin b'^\ f b \ Bai toan 9 (De thi HSG quoc gia nam hpc 2000-2001, bang B). Cho day 62 62 62 - 26
T Gi&i. Dg t h i y x „ > 0, Vn = 1 , 2 , . . . do do tit i „ + i = ta c6 C h u y 4. Ta thay rdng cosnt duac hiiu dien thanh mot da thiic cua cost, 2(2n + l ) x „ + 1 goi la da thiic Tre-bu-sep loai 1. Con sinnt b&ng tich cua sint mgt da thUc cua cost, goi la da thiic Tre-bU-sep loQi 2. x„+i = p = 2 ( 2 n + l ) + — (n = 1,2,...). 2(2n+l) + — ^"+1 D i n h l y 2. Gid sii sin(2/!; + l)t = P2k+i ( s i n t ) , trong do P2fc+i(x) la da thiic dai so bac 2k + I . Ki hteu Q2k+i{x) la da thiic dai so bac 2k + 1 sinh bdi D$,t — = u„. K h i do u i = 3 v& P2k+i{x) b&ng each giU nguyen nhUng hf so ling vdi luy thica chia 4 du I va x„ thay nhUng he so Hug vdi luy thiCa chia 4 diT 3 bang h$ so doi ddu. Khi do u„+i = 4 ( 2 n + l ) + u„,Vn = 1 , 2 , . . . (1) Tilf (1) d l dang suy ra u„ = (2n - 1) (2n + 1), Vn = 1 , 2 , . . . . V$y 2 1 1 i„ = — = ,Vn=l,2,... Sau day la mpt so he thiic dai so t h u dUdc tit cac hg thiic lien quan den ham u„ ( 2 n - l ) ( 2 n + l) 2n - 1 2n + l hypebolic, vipc chiing m i n h cac ho thiic do x i n danh cho ban doc. j Dodo • Tir he thiic cos 2t = 2 cos^ t - 1 t a t h i i dUdc h? thiic dgi so /I (J. l_\ x i + X2 H + ajaooi = 1-3' + \ Sy "'"'•••^ V4001 4003/ (1) 1 _ 4002
1.1.4 M o t so p h e p d 6 i b i e n dvtdc d i n h hMdng b d i cAc cong thiJc r Cach t r l n h bay khdc cho trtrdng hdp > 1. Neu \yi\>\i ton tai l i f d n g giac. (P sao cho cosh = j / i . Khi do Trong muc nay ta so xct mot so bai toan diTdc giai bKng each di.ra tren ode y2 = 2 cosh^ (i>-l = cosh 2(f), j/3 = cosh 2^ (A,..., 2/„ = cosh 2"~ ^ (A. d^c trimg cua mpt s6 da thilc dai so sinh bcii ham so sinnx vk cosnx. B a i toan 10. Xdc dinh d&y so (j/„) thoa man dicu hien sam Tif cosh
B a i todn 12. Cho day so (x„) nhu sau: x i < va Vi 2 „ = 2j/„ = 4x„ nen 4b 2 1 4x„+i 4'-"x„+i ZxZ2Zz--Zn 4".XlX2X3...X„ XiX2X3...X„ So hang t6ng quat ciia day so (w„) la . Tim so hang long qudt cua day so {u„). XiX2X3...X„ 1\ -, Vn G N*. G i a i . Dat x„ = ^y„. Ta dUdc day so (y„) thoa man: j / i = 2xi < - 1 va 1 9 1 B a i t o a n 13. Ttm day so {x„}+f°i biet X I = Q , x „ + i = a x ^ + 6 , V n 6 N * , a 6 = 2. Sijf dung bhi toan 10 ci trang 10 ta dildc G i a i . D$,t x„ = 6j/„, khi do = - va 1 / 1 \ + ,Vn=l,2,... 6i/„+i = a62y2 + 5 =^ = abyl + 1 y„+i = 2y2 + 1 (do a6 = 2). vdi /3 = 2/1 + v^y2 _ 1 ^ + ^4x2 - 1. Dat Xet so th^tc /3 sao cho jn-l I- Khi do Vay neu dat ^ = yi + y ^ y f + T thi Taco 1 Boi vay 1 1 \ 2/1 = 2 1 2 y2 = 2y? + 1 = 2 +1 = r m pj\ Gia sii Vn = ^ (^P'^'"' - ) . K h i do + 1 1 Pj _ 1 T2 1 \ 1 \ y„+i = 2yi + 1 = 2 P -— 12 13
Tlieo nguyen If quy nap toan IIQC suy ra B a i t o a n 15. Tim { x , , } ^ ^ , hiet 3 2/n = 2 x i = n, Xn+\ o x „ - 3 x „ , a > 0. Af.. „o Bdi v^y 2 2n-l H i r i n g d i n . Dfit x „ = - p 2 / n - K h i do y„+i = Ayl - 32/„, Vn = 1, 2 , . . . Sau a a ,Vn do sii dung bai toan 14. 6 + V6^ + ' Lifti y. Phep doi bien trcnig bai toan 15 d traiig 15 dildc t h n r a n h u sau: T i t h? thiic t r u y h6i ciia day ( x „ ) khien t a hen tucing den rong t h i k ludng giac B a i t o a n 14. Tim {yn}n=i' biet cos 3x = 4 C D S ' * X — 3 cos x. yi 6 R, y„+i = Ayi - 3j/„. Ta C O g^ng d y n g cong thi'rc nay. Gia sii x „ = 6y„ + c, k h i do Giai. bVn+i + c = a {byn + c)^ - 3 (62/,, + c) • Neu 12/11 < 1 t h i ton t ^ i 0 sao cho cos(j> = j / i . K h i do «>6y„+i + c = a {b'^yi + Sb^cy^ + 36c^2/„ + c^) - 3 (6y„ + c) 2 / 1 = 4 cos^ (/) - 3 cos 0 = cos 3 0 , . . . , 2/n = cos 3"0. «>y„+i = abSi + 3o6cyf. + Z{ac^ - l ) y „ + ^-'(ac^ - 4c). • K h i I2/11 > 1. X e t so thyc l3 sao cho Vay t a cho a6^ = 4 c = 0^ /? = 2/1 + L /? = y i - V2 3a6c = 0 3(ac2 - 1) = - 3 6 = '(ac^-4c) = 0 Vay neu dat (3 = -1 thi 2 Do do t a dat x „ = —;=2/u- T u y uhien c6 t h ^ t i m ra cong thiic doi bien nhanh )' -r'''- hon bSiig each dat x „ = byn, sau do t i m 6. Ta CO 1 B a i t o a n 16. Ttin {x,,},"^^, biet xi = a, x,i+i = axf, + 3x„, a > 0. y:-2 G i a i . Dat x „ = -7=2/11- K h i do 2/1 = va 2/2 = 4 - 3 ( 2 \ ^2/a+i =4yf.+3y„(Vn6N*). yn] +3 Vv/a / Giasu2/„ = - /fl-^ + : ? r : ^ , khi do Xet so thuc /:* sao clio • /3 = y i + / y ? + 1 1 ^ 3„-, _ 1 _ yi = 1 - " '^^'^ - 1 = 0 ^ J i = y i - x / y f + i - 2/n+l = 4 -3 Vay theo nguyen h' quy nap toan hoc suy ra Vay nou dat /J = yi + V y f T T thi ,Vn = 1 , 2 , . . . 1 / yi = 2 v 14 15
Ta CO Bai to&n 19. Cho day so (u„) nhusau: | '^^^ = + Su^ - 3, Vn = 1,2,, . 7^m cong thiic so hang tong qudt cua day so da cho. Giai. D^t ii„ = u„ + 1 (n = 1,2,...). Khi do Hi = 3 v& Do (16 i;„+i - 1 = K - 1 ) ^ + 3 K - 1)2 - 3 = - 3 t ; „ - 1. V$,y Vn+i = ^ n - 3i^n, Vn = 1,2,... Den day ta tien hanh tUdng tit nhit bki toan 15 d trang 15. =4 Cach khac. Xet phUdng trinh - 3x + 1 = 0. Phuong trinh n&,y e6 hai 1 \ nghiem xi, X2 va theo dinh ly Viet ta c6 xi + X 2 = 3 vk X1X2 = 1. Ta se chiing minh quy n?ip rang 03" ,Vn = 1,2,... Khi do T3 Vr, = x\ , V n = l , 2 , . . . yn+1 = 4 1 /^a— 1 \ +3 /33"-» 2\,^ ^3" /J 2 l,^ ^3"^- Ta e6 t;i = 3 = xi + X2 = x f + xf. Gia sii t;„ = x f + xf'\i do: » v.^, = v l - 3t;„ = ( x f - ' + x f " ) ' - 3{xr-' + x f " ) fi Theo nguyeu h' quy n^p suy ra y„ = 1 ^ / j a - _ Vn = 1,2,... V$y = xf + 3xf x\f + xj )+xi - Z{x\ x5 ) x„ = + +1 + - \" = x f + 3(xr-' + x f " ' ) + xf - 3(xr-' + xf") = x f + xf. +1 V$.y theo nguyen ly quy n9,p suy ra Vn = x?"~* + x^""', Vn = 1,2,... V$,y Bai toan 17. Tir/i day S(J {^ulit^ *'ao c7io i i = a «n = . „ - l/ 3= +( v^/ 5j \ ' " " + (/ 3- -2v ^/ 5j\ ' " " ' - l . V n = l , 2 i„+, = ax^+6x^+cx„+d,V7i € N* a > 0 , c = — ,d= .(1) Lifu y. Phep dat t;„ = u„ + 1 (n = 1,2,...), d\X0c tim ra nhit sau: Xet h ^ HUotng d t n . Gpi y,, = x„ + ^ , thay vao (1) ta dudc 2/„+i = ay^ + 3y„. so fix) = x^ + 3x2 _ 3 i^hi do u„+i = /(u„), Vn = 1,2,... Ta c6 /(x) la da o(l thiic b§c 3 va Bai toan 18 (De nghi thi OLYMPIC 30/04/1999). Xdc dmh so hang tong fix) = 3x2 ^ ^//(^) = 6x + 6 = 0 ^ X = - 1 . qudt cm day so (u„) biet iting: Vay diem uon cua do thi ciia ham so fix) Ik Ai-l, -1). Ta biet ring d6 thj ham so fix) nh$n diem uon ^ ( - 1 , -1) lam tam doi xiing. Do do ta thUcJng f u, = 2 doi he true toa d6 theo cong thiie doi true sau: | Y = y + l. ^ \i = 9uf, + 3(t„,Vn = 1,2,... toan 19 ta phai d^it t;„ = u„ + 1 (n = 1,2,...). Hifdng d t n . BM to4n 20 (De nghi thi OLYMPIC 30/04/2004). Cho day s6 (u„) nhu 2 sau: Cach 1. Dat u„ = - X u , khi do xi = 3 ; a;„+i = 4 ^ + 3x„, Vn = 1,2,... Den ui = «„+i = 24u3 - 12V6u2 + i5u„ - 7 6 ( „ = 1,2,...) day, ta tien hanli luaug tu nlut bai toaii 16 d trang 15. each 2. Dat 3a„ ^ ^ { 1 ,,,, + 3,,^^. Chon xux^ sao cho { §| + Tin cong thiic so hg.ng tong qudt Un cua day so da cho. 'jl.. Bang quy nap, chiing niinh dUdc: , r V
Hrfdng dan. D^t v„ = >/6u„ - 1 =i-vi = 2, v„+i = 4v^ + 3v„ (n = 1,2,...). Bki todn 23. Tim so hQ.ng tong qudt cua day so (x„) cho bdi Xet phuong trinh - 4x - 1 = 0. Phudng trinh nay c6 hsd nghi?m xi, X2 vk xi + X2 = 4, X1X2 = - 1 . Ta chunig minh ditdc: = 3^2^ ^^^^ ^ 1 6 ^ ^ 4 _ 8^2x2 + ^ , Vn = 1,2,... Vn = ^ ( x f + x f " ' ) , V n = 1,2,... Gilti.D&tx„ = ^ . K h i d 6 y i = 3 v a Suy ra (2 + Viy +(2-^5)^ + 2 ,Vn=l,2,... 1 ^ = 16^24 - 8V/24 + ^ , Vn = 1,2,... u„ = v/2 4 2 2 2y/E «B'yn+i=8y;t-8y2 + l,Vn = l,2,... Bhi toSn 2 1 . Tim so hang tong qudt cua day so (x„) cho nhu sau XI = 6, x„+i = x^ + 4x^ - 2x2 _ i2x„ - 7, Vn = 1,2,... Xet so thvtc a sao cho i .ii'ii* / Hirdng dan. 0^,1 x„ = y„ - 1. Ta thu diI0c day so (y„) nhu sau a = 3 + 2v^ a+ - a2 - 6a + 1= 0 a = 3 - 2V2. 2/1 = 7, 2/„+i =y*- 8y2 + 1, Vn = 1, 2 , . . . Bai to4n 22. Tim so h^ng tong qudt cua day so (x„) cho bdi xi = \/2 + \/3 V?tynludata = 3 + 2 v / 2 t h i 3 = i ( a + i ) , 1 = 3 - 2^2. Ta c6 x„+i = x ^ - 4 x 2 + 2 , V n = 1,2,... 4 o 4 1\ a^+4a2+6+^ + ^ j . Giai. Dat x„ = 2y„. Khi do a+- 2j/„+, = 162/^-16i/2+2,Vn = l , 2 , . . . 1> 2 a + - = 8j/^ - 8y2 + 1, Vn = 1,2,... Tac6 yi = v/2 +V5 1 + 2 _ '1 + cos^ ^ —2 — - = cos—. 12 Do do ri / i\ 4 -8 1 2 a+ - ' 1 u Do cong thiic Ivr^ng gidc COS4Q =8 8 cos^ a + 1 nen CDs'* Q - Bdi v^y = Scos" ^ - 8cos2 iL + 1 = cos ( 4 . ^ ) . 2/2 1/ A 1 4 " ^ 1 y, = Scos" ( 4 . ^ ) - 8cos2 ( 4 . ^ ^ ) + 1 = cos (42.iL). Gia sii y„ = cos Khi do j/„,: = 8cos^ ( 4 - ^ ^ ) - 8cos2 ( 4 " - > . ^ ) + 1 = cos ( 4 " . ^ ) . Gia s» !M = i (
Theo nguySn l i quy n^p, suy ra y„ = ^ (a'^" ' + ^ 4 ^ ^ , Vn G N*. Do do G i a i . T a c6 cos 5a = 16 cos^ Q — 20 cos^ a + 5 cos a. V$,y x i = cos — va 3 X2 = 16cos^ ^ - 20cos^ ^ + 5 c o s ^ = c o s ^ , o o 3 3 ,Vn = l , 2 , . . . ~ 2v/2 . X3 = 1 6 c o s ° - — - 20cos'^-—-|-5cos—= c o s — . ' ' o 3 3 3 B a i t o a n 24 (De t h i HSG T P Ho Chf M i n h nSni hoc 2011-2012). Cho day 5"-17r so (u„) nhu sau : ui = ^ vd u„+i = — . — , „,Vn e N*. T i m so hmg Gia sur x„ = cos — - — , k h i do 5 u * - 8 < -I- 8 long qudt cua day so da cho. = 16 cos^ ^ - 20 cos"-^ f + 5 cos"- | = cos 5^. G i 4 i . D^t — =?;„. K h i d6 v i = 7. gia thiet suy ra Theo nguyen l i quy n^ip, suy ra x„ = cos — - — , Vn = 1,2,... o B a i toan 26. Cho day so (x„) n/ii/ sau: , f =>vn+i = 8t;^ - ivl - I - 1 , Vn = 1,2,... X I = -7, x„+i = 16x5 _ 20x^ -I- 5x„, Vn = 1 , 2 , . . . X e t so thi;c a sao cho Tim so hang tong qudt cua day so dd cho. 5 1/ n 2a^ - 5a -I- 2 = 0 a = 2 G i a i . Xet so thi/c a sao cho 4=2r+a a= 2-1, a = -7-v/48 1 fa + = -7 •«> a^ 4- 14a -H 1 = 0 a = -7 + v/48. 2V V$,y ngu dat a = 2 t h i J = i J^a + i Y ^ = ^. Tudng t i ; n h u 15i giai bai V?iy neu d^t a = v/48 - 7 t h i ^ = - v/48 - 7, ^ + ^ ) = -7. Tac6 toan 23, t a chiing minh diMc 1 V - ^ ^ l = ^(2^"-"+2-^" ' ) , V n e N * . 16 •x[a + -] 2 \j 2 \ = - a H 5a3 -MOa + — -I- a^ J -I- - r (3) iM^ 5 / 3 0 3 1\ (4) 20 - a + - = -[a^ +3a+- + -^ Do do u n = ^4n-i ^"^2-4"-'' = 1' 2' • • • 2 \ J L t f u y. Ro rang, each giai dua tren cac hang d i n g thiic d ^ so (dU0c d}nh 1 l^" _ 5 ( l^ a4- - a-H - (5) hudng bcii cong thiic lit(?ng giac) n h u da t r i n h bay d tren la t u nhien, de hieu hdn so v 1 / 1> 1 B a i toka 25. Cho day so ( x „ ) nhu sau 16. - (a + - 2 V a; -20 0 2 V u + -A ; 4-5 2 1^ a; 2V (6) XI = ^, x„+i = 16x1 - 20x1 + 5a;„, Vn = 1,2,... If l\ Bdi vay x i = - a 4- - va T i m so hQ.ng long qudt cua day so da cho. ^ - t i i ^ si> 2V a) 3 ri / i\ 5 'l 1> 1 1\ 1 5^ ^\ - 20 a 4- - 4- 5 a4- - a" 4- -? X2 = 16. 2 ^ a^ —2 \/ 2 ^ aV 20 21
Bai todn 28. Cho day so (x„) nhu sau: xi = 6; x„+i = x^ + 5x1 + 5x„, Vn = 1,2,... Tim so hang tong qudt cua day so da cho. Giai. Vdi m6i n G N*, d$,t x„ = 2y„. Khi do yi = 3 vli Gia sur x„ = i (a^"-' + -l^r^, khi do 2y„+i = 32y^ + 40^^ + iOx„, Vn = 1,2,... /l0 , ("1 / 5„-l 1 M 1 / gn 1 \ fa _ i^ - 6a - 1 = 0 ^ La = 3 - VIO. Theo nguyen h' quy n»p suy ra V?iy neu d^t a = 3 + \/lO thi 1 l1 =_ - ( 3 - V Y o ) . / 2 aj a Vay so hang tong quat cua day so da oho la Taco ia[l(a-i)f = i ( a - - = . . o . - ^ . A . ^ x„ = - ,Vn=l,2,... 1/ _1\ Bki todn 27. C/io ddj/ so (j/„) n/ii/ sau; j/i = a vd 20 L2V «/J y„+i = 180y^ - 48y^ + 5y„, Vn = 1,2,... 5/ n (1) 5 [1 1^ [2 V 2 V ay 71m so /i(in5 ton^ qudt cua day so da, cho. Tlit (1), (2), (3) ta CO Htfdng d i n . D^t y„ = - ^ x „ , Vn = 1,2,... Thay vao (1) ta dvt0c v3 16 1 ( ^\ T3 + 5' 1 ( A 1 2V ayj 2 V a/ :x„+i = 180.-^x^ - 48.-^x^ + — x „ , Vn = 1,2,... Dodo ^Xn+i = 20x1 - 16x1 + 5x„, Vn = 1,2,... ' 0 1 N a V^y ta thu dilpc day so (x„) nhu sau - a'" J J^=2r 'a^'J X , = aV3, x„+i = 20x^ - 16x^ + 5x„, Vn = 1,2,... Gi& sut y,. = i ^a*"-' - - i - ; - ^ . Theo (4) suy ra y„+i = ^ • Trudug h(?p |a| < Khi do |a\/3| < 1. D$,t a = arccos(av/3). Khi d6 Theo nguyen h' quy n^p todn hpc ta c6 xi = cos a, sau do tien hanh tUdng t\f nhu bai to4n 25 d trang 20. • TVudng Irijp |a| > 4=- Khi do ay/3 > 1. Tien hanh tUdng ti; nhu bai to4n v3 ,Vn = 1,2,... m v^y x„ = [(yiO + 3)^""' - (v/TO - 3) 26 d trang 21. 23
1.1.5 Phirctag p h a p h a m lap. G i a i . T a CO x i = 5 ^ 4. G i a sur x „ jt 4, ta chiing m i n h x„+i ^ 4. Neu D6 t i m .so hang t6ng qiiat cua day so (u„) bang phuong phap l^p t a thudng X n + i = 4 t h i ^"_^_2 = + 4 = 4x„ + 8 •«> x „ = 4, mau thuan v d i gia t i m cac ham so f{x) va h{x) sao cho thiet quy n?ip. V^y x „ ^ 4, Vn e N * . T a c6 / K ) = /i(/K_i)). (*) 5Xn-i4-4 ^ _ X n _ i - 4 ^glLlLli.i 6(x„_i + l ) Si'r d y n g (•) lien tiep t a t h u dudc. Xn ^ 1 + 2 - x„_i + 2 ' ~ " ' - X n _ i + 2 ' - x„-i + 2 / ( U n ) = ft(/K_l)) Suy ra = h{h (/(u„-2))) = h2 (/(«„_2)) = - - = K (/(«0)). (**) X„ + 1 _ .gn-JL+l p2^n-2 + 1 ^1 + 1 ^ fin ^ - % „ _ l - 4 X„_2-4 XI - 4 Tir {**) t a t i m dit0r Un- H a m so / dUdc gpi la ham so phu, con ham so h dU0c gpi l a ham lap. T a se bat dau bang hai bai toan rat ddn gian nhung da 4 6" + l 4.62°13 + 1 D o d 6 x „ + l = 6 " x „ - 4 . 6 " 18u„+i + 9 = 2u„ + 1 + 8\/H- 2iin + 16 va so - la nghi^m cua phudng t r i n h /(x) = x, hay noi each khac, so ^ Ik «.9(2u„+i + l ) = ( v T + 2 ^ + 4)^ .
•**t;„+i - 2 = -vn - - v„+i - 2 = -(v„ - 2). V§Ly so h9iig tong quat cua day so da cho la: ' x„ = 7 + { a - 7 ) 2 " - ' , V n 6 N * . Nhu v$.y: livhi y. Trong bai toan nay thi ham so phy la /(x) = x - 7, c6n hhm l?Lp la '^n - 2 = 1 {Vn-l -2) = {Vn-2 " 2) = ^ ( f n - 2 " 2 ) h{x) = V i d u 2 . C/ipn c = -2, khi do a = 4, 6 = 2. Th (ft/tfc day so (x„) n/irr sau; = -=3;;^(-i-2) = 3 ; ^ ( 3 - 2 ) = ^ . x„+i = x 2 + 4 x „ + 2,Vn = 1,2,... Do d6: Vn = + 2. Suy ra: Ta tiep tuc lam kho han bhng each doi bien x„ = 3u„. Khi do dU0c day so (un) thoa man dieu ki$n 3u„+i = 9u2 + 12u„ + 2, Vn = 1,2,... So hgng t6ng qu4t cua day so (u„) da cho la: «.u„+i = 3u2 + 4u„ + ^, Vn = 1,2,... 1 1/ 1 V " " = - 2 + 2 ( 3 ^ + 2 j ,Vn = l , 2 , . . . Ta dxtdc bai todn sau. B a i t o a n 33. Cho day so (u„) nhu sau: ui = a eR vd 5ou «fdj/
V$,y t a t i m k thoa man 3k = 1 =^ k = ^ Un = —. «>1 - 5w„+i = 25ul + 20Un + 7^ Un+l = -5ul -4un--. C d c h 2. Ta c6 2 To dt^c'c oai toan sau. *' ' - :! . 3 B a i t o a n 3 4 . Cho day so (u„) nhu sau: ui = a eR vd 2 2' 2 2' / 2\ = 3 = 3 3 U n - 2 + ^^ V 3y V 3y It 2» Tim so hang tdng quat cua day so da cho. = 3^+2 2V _ 31+2+22 2V' 3 + X "n-3 + ^ 3/ G i a i . D a t u„ = - i x „ . K h i do ' ' - 5 _ . . . _ 3l+2+-+2"-2 / 2\ 3/ -gXn+i = --x^+ - x „ - -
Bki to&n 3 5 . Cho day so (u„) nhu sau: u i = a e R ud Giai. N i u a = - 2 t h i x„ = - 2 , Vn = 1,2,... T i l p theo x6t a ^ - 2 . Ta c6 u„+i = 25u3 - \bul + 3 u „ , Vn e N * . ^ 8x„-i 2xti-8xn-i+8 2(2-x„-i)2 Tim so hg.ng tong qudt cua day so da cho. 8X„-1 2x2_i 8xn-i + 8 _ 2(2 + X n - i ) 2 2 + x„ = 2 + 4 + x 2 _ j (2) Gi&i. Dat Un = -^.Ta. ditoc day so (x„) thoa man di6u kien xy = 5a 4+ x2_i 4 + x2_i 5 2-x Xet ham s6 / ( x ) = — . Tit (1) v i (2) ta c6 ^X „+l = - + |x„ ^ Xn+1 =xl- 3x2 ^ g^,^ ^ 2 - Xn _ / 2 - x„_i = [/(X„_,)]2 = [/(X„_2)] |2> X „ + i - 1 = ( X „ - 1)^ . /(x„) 2 + x„ V 2 + x„_iy W^y vdi mpi so nguyen dudng n ta c6 (3) x„ - 1 = (x„_i - 1)3 = (x„_2 - 1)3' = . . . = (xi - if"' = (5Q - 1)3" DP/3 = [ / ( a ) r " ' - T i i (3) ta c6 So hgng t6ng quat cua day so da cho la 2-x 2-2/3 = l3^2-x„ = 2P + 0Xn^x„ = y ^ . V?iy neu a = -2 t h i x„ = - 2 , Vn = 1,2,..., neu Q ^ - 2 t h l OX 2n —1 3. Ldp ham g{x) b + c^x^' Tac6 (2 + a ) , / \ ax bd- ax + dc^x^ Vn = l , 2 , . . . d-g{x) = d- = „ , . x„ = Ta can chpn a, 6, c, d sao cho 6d - ax + dc^x^ = (d - xf = d2 _ 2dx + x^ Vf du 6. TVonp bai todn 36, «n+l 5(a;) = ~ 9 u 2 - 6 u „ + 5" Nhu v^y ta dupe bai todn 48 d trang 142. Ta d?/(rc 6ai todn sau. + ax 4. Ldp ham g(x) = Bai toan 36. Tim so h(ing tSng qudt cua day so (x„) cho nhu sau: 6x2 + c* 8x Tac6 , , , x^ + ax x^ - 6dx2 + Qx - cd x i = a e R vd x„+i = — ~ , Vn = 1,2,... 4 + x2 31 30
Ta chpn a, b, c, d sao cho r 6=3 V f d u 8 . Tt)f ^ a = 33d2 c/ipn c = A ta dU0c 6 = 3 , o = 12, d = ± 2 . To cd - bdx^ + ax- cd= {x- df I c = dd2 =^x^ - bdx"^ + ax-cd = x^ - Sdx"^ + Zd^x - d^ bai todn sau. ( bd = Zd r 66 = 3 ^ { a^3d^ ^ a = 3d2 3d' B a i t o a n 3 8 . Tim so hg,ng tong qudt cua day so (x„) cho nhu sau: \ = d^ { C = (P. 3.3 I 12x xi = a>Ovd x„+i = ^ ^ 2 + 4 " ' ^ " = 1,2,... V i d u 7. I a = 3d2 chpn c = 2 ta duac 6 = 3 , a = 6 , d = ±y/2. Ta c6 G i d i . T a c6 bai todn sau. o _ 4 - 1 + 12Xn ^ _ X t i - 6 x 2 _ ^ + 12Xn - 8 _ ( X n - i - 2)^ B a i t o a n 3 7 . Tim so h(ji.ng tong qudt cua day so (x„) cho nhu sau: ~ 3x2_, + 4 3x2_, + 4 - 3x2_,+4- .S! 4-1 + 12xn ^^ x3_^ + 6 x 2 _ ^ + 12x„ + 8 (x„_i+2f xl + 6Xn x„ + 2 = - r - 0 —:- + •i = r-o — xi = a > 0 vd Xn+i = •,Vn=l,2,., = -r—o - • (2) 3x2+2 3 ^ - 1 + 4 34-1+4 3 4 - 1 + 4 x-2 G i a i . Ta c6 X6t h k n s6 fix) = T i i (1) va (2) suy ra X *T* ^ _ / - _ 4 + 6x^ /7;_xl- 3 v / 2 x 2 + 6 x n - 2y/2 _ (x,, - \ x„-2 /x„_i-2\ 3x2+2 3x2+2 3x2+2 Xn + 2 \Xn-l+2j = [/(x„_i)P = [/(x„_2)] i3» I ^/o ^ + , ^/o ^ 4 + 3 v ^ x 2 + 6 X n + 2 v ^ ^ (x^ + s/2f • • • = [ / ( x i ) r ' = [/(a)r' 0) 3x2 + 2 3x2+2 3x2+2 T i t (1) ta CO Xet ham so / ( x ) = ^ - — K h i do Xn-2 x +v/2' = [/ { a ) ] 3 " ' ^ x„ - 2 = [/ x „ + 2. [/ (a)] Xn + 2 30-1 :„ - /x„_i - v/2\ 3* /a-2\ /(x„) = = [/(x„_i)]' = [/(x„_2)] 2 +2 x„ + \/2 \x„_i + v ^ / ^x _ 2 + 2. [ / ( a ) r _ [a + 2) = . . . = [ / ( x O P " - ' = [/(a)r^ (1) 3"-i • /a-2\ 1 - [a + 2j T i l (1) t a c6 V i d u 9 . Trong bai todn 38, doi bien x„ = u„ - 1 ia duac day so (u„) Un - 1 V a + v/2/ 3 4 _ i - 6u„-i + 7 4_1 - 3 ^ 2 _ i + I5un-1 - 13 =>u„ - 1 3 u 2 _ j - 6u„-i + 7 32 33
= " n - 1+ 9 " n - l -6 Xit ham s6 f{x) = Vx > 0. Tit (1) (2) t a suy r a 3n2_j-6u„_i+7- Nhu v^y, ta dU(fc bai todn 49 d trang 144. 5. L o p h a m g{x) = cx^ + dx . = ---=[/(xi)r' = [/Hr". Taco D ? t t = TCt (3) t a c o I s _ _ X* + ax^ + 6 _ a:'* - cex^ + QX^ - dex + b = / 3 „ ^ x „ - v/2 = / 3 „ x „ + V2/3„ ^ x „ = 4^t^. X„ + ^ 1 - Ta can chpn a, 6, c, d, e sao cho So hang tSng quat ciia day so da cho la x"* - cex^ + ax^ - dex + 6 = (x - e)"* =»x'* - cex^ + ax^ - rfex + 6 = x'' - 4ex3 + Ge^x^ - 4 e 3 x + e'' /' C 6 = 4 e f c= 4 xn = . VnGN*. a = 6e2 ^ I a = 6e2 de = 4e3 =^ S d = 4e2 6 = 6" t 0, Vn e N*. Ta c6 4 {2Un 1)+ 16u4 + 24u2 + 1 • Mat khac ^ 4 + 4^2x3 + 12x^ + 8v/2x„ + 4 ^ ( X n + y ^ ) ^ _ 32u3 + 8Un _ ^ -16uf, + 32^3 - 24^2 ^ g U n ^ 4x3 + 8x„ ~ 4x3 + 8x„ ' (1) 16< + 24u2+l 1 6 < +24^2 + 1 Mat khac - (2Un - 1 ) ' 1 6 < + 24u2 + 1 • IT — 1 4 x 3 + 82;^^ Xet ham so / ( x ) = . T i t (1) va (2) t a c6 2x + 1 ^" ^ x^ - 4v/2x3 + 12x^ - 8\/2x„ + 4 ^ (xn - 4x3+Sx-„ ~ 4x3+8x„ • (2) 34 35
= - - = - ( / ( « i ) r " = -[/(«)r-'. Hi' 0, n e N . = \2a + iy H i r d n g d i n . D a t x„ = 2i-3". (2^0)^" y„ = 2 ( V a ) ^ " y „ . T h a y v&o (1) t a dU0c y„+, = 4 y ^ - 3 y „ , V n = 0 , l , 2 , . . . B k i t o d n 4 1 . Cho day so ( i „ ) nAi/ sau a;„+i = ^ x ^ , V n = 1 , 2 , . . . 71m s6 Den day t a lam tuong t u n h u bai toan 15 6 trang 15. hQng long qudt cua day so da cho. C h u y 5. Trong bdi todn 45 d trang 37, phep ddt H i r d n g d i n . Vdi moi n = 1,2,... ta c6 x„ = 2 ' - 3 " . ( 2 y ^ ' \ = 2(v^)'\ diMc tlm ra nhu sau. Tic {!) ta thay rhng c & n d d t x„ =?i/n, di du 1 1 2""* y„+i = 4 y 3 - 3 y „ , V n = 0 , l , 2 , . . . ^ 2l+2'+22 = ••• = 2l+2'+22+-+^" 2"^ ' ^1 • Tit (1), ta CO x„+i = - ( 4 x ^ ) - 3a^"x„. TrUdc hit ta tim Un thoa man B a i t o a n 4 2 . Tzm { x „ } + ^ i hiet " n + i = j u ^ V n = 0,1,2,... Xi = a , x „ + i = x \ 2a^",Vn G N * , a > 0,a 7"^ 1. Theo bdi todn 4 4 ta c6 u„ = 2^ ^ " u ^ " . Vdy dat x„ = 2^~^''u§"j/„, t/iay vdo H i f d n g d a n . Dat x „ = 2a2""'y„ (vi sao 1^ dat nhu vay, hay xem chu y 5 6 trang 37), khi do y^+i = 22/2 - i , Vn = :, 2,... Sau do lam tuong t u n h u bai (1), ta duac 2'-^"^\r\^+, = (2»-3"urj/„)' - 3a3"x„, suy ra toan 1 1 6 trang 11. = 2'-'"''4"'\l - 3a3"2»-^"-3 M 2 — < - y „ + i "0 yn B a i t o a n 4 3 . Ttm {xn)t=x biet =^yn+l = 22y3 _ 3 a 3 " 2 - 3 " + 3 - ' „ 3 " - 3 " - XI = a,xn+i = 2a2"x2 - «("+i)2",Vn e N % a > 0. ^J/n+1 = 4y3 - 3 a 3 " 2 " • ) ^- : chqn UQ sao cho i H u d n g d i n . D a t x„ = a"^"~'y„, khi do y„+i = 2yl - l , V n = 1 , 2 , . . . Sau do lam tUdng t u nhu bai toan 1 1 6 trai 11. 3a3"22 3 \ - 2 . 3 " ^ 3 ^ y 2 . 3 " ^ ^3'.22.3'' ^ ^ = 2 v ^ . B a i t o a n 4 4 . Tim so h^ng ting qudt cua day so (x„) cho nhu sau Tom /at, ta ddt x„ = 2^-3". (ly/E)^" y„ = 2 ( v ^ ^ " y„. iTAz do u„+i = ^ u ^ , V n = 0 , 1 , 2 , . . . Vn+i =4y3 - 3 y „ , V n = 0 , l , 2 , . . . 36 37