intTypePromotion=1

Bồi dưỡng kiến thức học sinh giỏi chuyên khảo dãy số (Tái bản có sửa chữa bổ sung): Phần 2

Chia sẻ: Cô đơn | Ngày: | Loại File: PDF | Số trang:163

0
392
lượt xem
123
download

Bồi dưỡng kiến thức học sinh giỏi chuyên khảo dãy số (Tái bản có sửa chữa bổ sung): Phần 2

Mô tả tài liệu
  Download Vui lòng tải xuống để xem tài liệu đầy đủ

Nối tiếp nội dung phần 1 tài liệu Bồi dưỡng học sinh giỏi chuyên khảo dãy số, phần 2 cung cấp cho người đọc các kiến thức xác định giới hạn của dãy số, một số ứng dụng của dãy số. Mời các bạn cùng tham khảo nội dung chi tiết.

Chủ đề:
Lưu

Nội dung Text: Bồi dưỡng kiến thức học sinh giỏi chuyên khảo dãy số (Tái bản có sửa chữa bổ sung): Phần 2

  1. 2,9 H a i day tiep c a n n h a u Giai. Xet/„ = j (sinx)"dx,tac6: 2 9.1 Mot s6 kien thufc thifdng dung. plnh n g h i a 2 0 . Hai day so ( x „ ) vd (2/„) gQi la tiep c^n nhau niu chung /„ = - y ( s i n x ) " - ^ d(cosx) = - ( s i n x ) " " * .cosx|| + Jcosxd(sinx)"~\ cldng thdi thoa man ba dieu kien sau: 0 0 (1) (xn) la day so tdng. (2) (2/„) Id day so gidm. (3) l i m | x „ - 2 / „ | = 0 . /„ = ( n - l ) y " c o s 2 x ( s i n x ) " - 2 d x = ( n - l ) J {1 - sin^ x) ( s i n x ) " - ^ dx, 0 0 D i n h l y 18. Hai day tiep cq,n nhau thi hoi tu den cung mpt gidi han. f f Cbufng m i n h . Gia sii ( x „ ) , (j/„) la hai day so t i l p nhau ( x „ ) la day /„ = ( n - 1) y (sinx)"-=^dx - ( n - 1) J(smx)''dx = (n - l)/„_2 - (n - 1)4. tang, (yn) la day so giam. Ta cut cho i&ng ba dieu kien (1), (2), (3) dung ngay tif n = 1 (neu khong t h i t a load di mot so so hang). Xet day so (zn) dinh bcti 0 0 = x „ - j / „ , V n = 1,2,... V?Ly n/„ = (n - l ) / „ _ 2 /„ = ^ ^ / „ _ 2 . V i /Q = | , / i = 1 nen: Ta CO 2n - 1 2n - 1 2n - 3 , 2n • 2- ; r1- 22n n -- 23- " - 43- 21- TT 2' 2TI ' 2 n - 2 - Zn+i - 2n = (x„+i - J/n+i) " (x„ - y„) = (x„+i - x „ ) + (y„ - j/„+i),Vn e N * . 2n 2n-2 4 2^ "'2n ^ _2n_ 2n-2 2n + l ' 2 n - l " 5 ' 3 ' Ma x„+i - x„ > 0, Vn = 1,2,... va 2/„ - y„+i > 0, Vn = 1,2,... nen 2n + l 2n + r 2 n -l Zn+i -z„>0,Vn=l,2,... Do (16: (*) Do do ( 2 „ ) la day so tang. T i t (3) suy ra l i m \zn\ 0 hay lim z„ = 0. n—»+oo n—•-j-oo "7^" [3.5...(2n-1)1^2x1+1) V$y (z„) la day tang va hpi t u ve 0. Do do Zk < l i m 2 „ , VA; = 1,2,... hay n—+oo /2n+i 2n ^fc+oo yn- 340 341
  2. 2.9.2 C a c bai toan. , ^, 2 _ _]_ ^ 2 - (n + 1) _ 1- n (n+1)! n! (n+1)! " ( n + 1 ) ! ^ 0 ' ^ " = L 2 , • •. B a i t o a n 234. Cho hai day so { a „ ) va (6„) nhu sau: V$.y day ( i ' n ) la day giam. Ta cd 0 < 6 , < a , ; a „ + i = ^^4^, 6„+, = (Vn = 1,2,...). dn + On 2 lim \un-Vn\= lim ^ = 0 . Chiing minh rang hai day so da, cho hqi tu va gidi hg,n cua hai day so do bduy n—>+oo n — ' + 0 0 n! • i 'I nhau. Po do (Ufi) v a (t;n) l a hai day ticp can nhau. Suy r a hai day nay h o i t u den G i a i . R6 rang a„ > 0, 6„ > 0 (Vn = 1,2,...). T i t 2 {al + bl) > (a„ + b„f cung mPt so, t a k i hieu so do l a e. Ta cd suy r a a„ > 6„, Vn = 1,2,... T a c6 " n < e < t^n, Vn = 1,2,... On+i = -T- < — r — = a„, Vn = 1,2,... Ta gia suf phaii chiing rang e la so hOXi t i , e = — , vdi p v a no la hai so t u a„ + 0„ a„ + &„ nhien, no khac khdng. Ta cd Suy r a day (a„) la day so giam. T a c6 1 1 1 p , 1 1 1 1 l + 7r + ;77 + -- - + — r < — < l + 77 + r T + -- - + — 7 + — : . 1! 2! no! no l ! 2! no! no! Nhan t a t ca cho no! t a dudc Suy r a day (6„) la day so tang. Vay > l < p ( n o - l ) ! < A + 1, (*) 6i < 62 < •• < 6„ < a„ < ••• < a i . Do do day so (a„) giain va bi chan dudi (bdi so 61), day so (&„) tang va bi trong do A la so t u nhien, >l = no! + ^ + ^ H 1- M a (*) khong the chan tren (bdi so o i ) . Vay ca hai day so da cho cung hpi t u . D a t lim a„=a 1! 2! no! xay ra dUdc (vi A\k{A-\-1) la hai so t u nhien lien tiep, p(no - 1)! cung la so va l i m b„ = b, k h i do tijf t u nhien). Dieu vo If nay cluing to e khong the la so hiiu t i . V$,y e la so vd t i . n-»+oo Bai toan 2 3 6 . Cho day so (a;„) nhii sau : 6„4.i = ^4^ (Vn = l , 2 , . . . ) cho n —• +00 t a dildc b = —^—, hay a = b. Vay t a cd dieu phai chiing minh. Chiing minh r&ng day so nay cd gidi han la mQt so vd ti. B a i toan 235 (De nghi t h i O L Y M P I C 30/04/2003). Cho hai day so (u„) va (vn) xdc dinh bdi: H i f d n g dan. Xet day so (?/„) n h u sau 2/„ = x„ + J ^ ' ^ ' ' ^ - = 1.2,. • • K h i dd Wn = 1 + Tf + ;7f + •• • + ^ , i;„ = u„ + - ^ , V n = 1,2,... ^^cng t u n h u bai toan 235 t a chiing m i n h dir0c ( x „ ) va (y„) la hai day tiep 1! 2! n! TV. •^ta nhau va hai day so nay cd gidi h^n chung la so vd t i . Chiing minh (u„) va (u„) cd gidi hQ,n chung la mqt so v6 ti. ^ a i toan 237. Cho day so {x„}+f°i nhu sau: G i a i . u„+i -Un = -.—^—rr-. > 0, Vn = 1,2,... Suy r a day (ti„) la day tang- (n + 1)! x„ = V ^ + - ^ , V n= l,2,... ^ k\! k=o Ch ' ''^^9 minh r&ng day so nay cd gidi hfin la mgt so vd ti. 342 343
  3. HUdng d i n . Xet day so {yn}n=i " h u sau: V^y day (xn) tang va bi chan tren (bdi yi chang h^n) nen hOi tu, day (y„) giain va bi chan dudi (bdi x i chang han) nen hgi ty. Ta c6 " 1 fc=0 lim y„ = lim 1+ - n—+00 n—+00 \ ny Khi do titcmg tit nhit bai toan 235 ta chi'mg minh diroc (a;„) va {y„) la hai day tiep c§n nhau va hai day so nay c6 gidi hgn chung la so v6 t i . = lim ( l + - ) . lim ( l + i)= Um x„. n—+00 \ 71/ n-.+oo \ nj n-.+oo V^y ta CO difiu phai chi'mg minh. Bai toan 238. Xet hai day so {xn}n=i. {yn}t^i nhu sau: isjhan xet 23. Tic (*) ta thay x„=(l + i)", iM=(l + i ) " ' (V„=.,,2,...) n+l ,Vn=l,2,... ChUng minh r&ng: Mgitdi ta chiing minh dUcfc rang e Id so vd ti vde = 2,7182818284... Logarit a) Xn < 2/n,Vn = 1 , 2 , . . . c(j so e goi la logarit tu nhien vd ky hieu Inx = log^ x. So e dong mot vai trd b) Day (x„) tdng thuc si/ vd ddy (2/„) gidm thtfc stf. rat quan trQng trong todn hgc. SvC dung {**) ta gidi dugc hai bdi todn kho {bai c) Hai ddy so {x„}+~i, {yn}t=i cimg gidi han, duac goi Id so e. todn 239 vd bdi todn 240). Giai. a) Hien nhien. Bai toan 239. Cho hai ddy so (x„) vd (y„) nhu sau: b) Sur dung bat ding thiic Cauchy cho n + 1 so, gom mpt so 1 va n so 1 + - x„ = l + i + ...+ - l ^ - l n n , y„ = l + i + ...+ _ l _ + i _ i n n ( n = l , 2 , . . . ) . ta dildc: 1 + - < 1 +n n + l a.,.a n + l 1+ - 1 + n+l Chatng minh rdng hai ddy so da cho hoi tu. Hi/dng d i n . Tfir (**) 6 n h ^ xet 23 t a c6: n\n^^^^ < l n e < (n + 1) In Vn = 1,2,... Vay x„ < x„+i, Vn = 1,2,..., nghia 1^ day (x„) tang thvfc sy. L^i diing b a t n n dang thiic Cauchy cho (n + l ) so, gom mgt so 1 va n so 1 ta dUdc Hay vdi mgi n = 1,2,... ta c6: In ^ < 1 va ^ < I n ^ . Do do vdi '^Wn = i , 2 , . . . t a c o ^n+i - x„ = in - ln(n + 1) + Inn = in - In n > 0, 2/n+i - y„ = - ln(n + 1) + Inn = — In ^ ^ - ^ < 0. '- \ J \ n - \ \ \ n+\l n ta .se chiing m i n h diMc hai day so (x„) va (y„) la h a i d a y tiep c a n n h a u Vay y„ < y„_i, Vn = 2,3,..., nghla la day {y„) giam thi^c s\f. 1 Cling tien den ciing mot gidi han. Tit do ta c6 dieu p h a i chiing m i n h . c) Theo tren ta c6: toan 240. Cho hai ddy s6 {x„} vd {y„} nhu sau: X I < X 2 < • • • < x„-i < x„ < y„ < yn-i < • • • < y 2 < yi, Vn = 1,2,... 344 345
  4. g l i i t o a n 2 4 1 . Xet day so { u „ } duac xdc dfnh bdi u„ = n^" (n g N*). DCit w„ = ^ 4- ^ = + • •• + , ^ =,Vn = 1 , 2 , . . . ^" ^ n ( n + 1) v / { n + l ) ( n + 2) v^2n(2n+l) 1 , 1 1 . Hay tlm l i m x„ ud lim y„. n—"+00 n—•+00 Chihig minh rdng day so { x „ } + f ^ c6 gidi han hQu h(in khi n tdng len vo h^n G i a i . TCr 16i giiii bai toan 239 t a c6: t,a gi(ii hO'^ '^^ so vd ti. ^ < l n ^ < i , V n = l , 2 , . . . n + 1 n n G i a i . Ta c6 > 0 , V n = 1,2,... Do do: 1 1 1 , n , n + 1 , , 2n , 2n - + + •••+ — < In + In + • • • + In = In -, Y|ty {xn}t=\y ^ ^ ^ t khac v6i m p i n = 1 , 2 , . . . t a c6: n n + 1 2n n - 1 n 2n-l n - 1 1 1 1 , n + l , n + 2 ,2n + l , 2 n + l - + + •••+ — > In + In + • • • + In — = In . / I I 1 , , n n + 1 2n n n+ 1 2n n — + — + ••• + — 1 = 1 + n2"; V"2 "3 "n Vay / I 1 l „ ? ! ! d l l < x „ < l n - ^ , V n = 2,3,... (1) g (do a > 1). Vdi moi so t\S q C h u y 3 0 . Ta duac phep sii dung cong thilc sau: Cho {xn}t=\ '''"^n m > 2 ta c6: so nhdn c6 cong boi q thoa man dieu kien - I < q
  5. Do do WlU2..Wr-l < , "^"^w +i.Vj = 0,1,2,... gdi vay nen vdi mpi j = 0 , 1 , 2 , . . . t a c6 " 1 + +— ^ + lim y —• ' Ihn V J - V l i m y^^^-^-'-^r-x {U\U2••-Ur-l) Suy ra hm > " 1 p r i +^ .•••. + — .^_J_l>o. < lim y n—+00 ^-^ k=m }^hi A: nhan cac gia t r i 0 , 1 , 2 , . . . , n t a cho j = k, t a se c6: Suy ra U1U2 u - i . q l i m E — la mot so nguyen dudng, do do n J (U1U2 • • - " r - l ) lim n—•+(» (2) U1U2 fc=m Hay (1) (2) suy ra ton t a i = r > 2 di fc=m 1 (urU2 . . lim — Tiep theo t a se chilng m i n h < -. 7 V- 1 1 Dieu nay mau thuan vdi (1). Vay a la so v6 t i . 3m>2: "i«2 • •-Wm-l- lim 2^ - < fc=m L\iu y. Sau day la bai toan 242, t6ng quat cua bai toan 2 4 1 , day la bai toan rat sau sic va kho nen doi hoi ban doc phai c6 nang luc va sU kien t r i dang d6 daii d i u mot diSu mau thuan. Goi r la s6 nguyen dUdng nho n h i t sao cho ke. Tuy nhien c6 mgt thuan Idi la trong bai toan nay c6 nhieu l?lp luan da r +j > 9 + l , V j =0,1,2,... Khidotaco c6 6 bai toan trade. i2' < r 2 \ V i = l , 2 , . . . , r - l . Bai t o a n 2 4 2 . Cho {ukj^^i la day dan di^u tang cdc so nguyen dMng sao J»n = + 0 0 . Chy:ng minh rdng day so (a;„)+f°i, vdi V i vay " 1 .2'+22+-+2'- .2'-2 U1U2 ••-Ur-l ^ ^ x„ = (n=l,2,...) Ur+j Ur+j fc=i vd gidi han cua day (a;„)+^i la rnqt s6 v6 ti. '^•ai. T a.,CO Suy ra x„+i - x „ = > 0 , V n = 1,2,... "n+l < 349
  6. m {x,y^i l a day s6 tang. T a d n g t u nhit bai toan 2 4 1 , t a cln'tng n i i n h cUtgc f hVfc v$.y neu u i < 02 t h i clipn vi - 2. Neu o i > a2 t h i do J i m a*: = + 0 0 nen .^i fc du Idn t h i ak > a i , do do chon r i la so nguyen duong nho nhat Idn hdn day c6 gidi h a n hCtu han. Jiin = +00 nen J i m ^ = 0, 2 sao cho a i < a^,. Thi^c ra ton t ^ i day v6 h^n cac so nguyen dUdng {Tk)k=\ D o do ton tai fci, fca,..., fcn sao cho CO tinh chat nay, tiJc la O j < a^^, Vj = 1,2,..., - 1. De t i m t a d p dyng 1 each tren cho day so {an\X=n^ • Gpi r la so nguyen duOng nho nhat sao cho ar+j > 9 + l , V j = : 0 , l , 2 , . . . v a a j < a^, Vj = 1,2,..., r - 1. Dat lim = a, k l i i do t f l x„ > 0,Vn = 1,2,... va day (x„) tang suy ra a > TTIIP theo t a chimg minh a la mpt s6 v6 t i . Gia ngUOc 1^, a la .5 Vi {w*}fc:^ 1^ '^'^'^ ''•'^^S faf^ •''o nguyen duong nen Ur < " r + j , do do hQu t i : a = ^, vdi p,g 6 N% = 1- Vdi m6i s6 tU nhien m > 2 t a c6: ar = ( U r ) ^ < { U r + j ) ^ = ( « ' + / ) ^ = " r ^ ^ V j = 1 , 2 , . . . , r - 1. , Q Viv?ly t a c o : = lim V — = I™ yUl Urn V-2 n-.+oo ^ Uk n->+oo fc=l r.2'-2 lim > — Ur+j < : lim n—+00 — + ••• + "»n-l n->+oo *:=Tn Ufc a2;y (9 + iP^' Do do " 1 U1U2 • • - U r - l .}_,...+-J— + lim y —. "r+j {q + 1)2^+' = 0' ^' 2. • • • S u y ra 1 Bdi v^y nen vdi uipi j = 0 , 1 , 2 , . . . t a c6 lim E - = -- fc=m > 0. / " 1 \ ("i«2...a.-,) lim y^-!- = lim V ^1^2••-^r-l N h u v^y u,u, .. - n . - . . 7 „ l i m £ ^ la mot s6 nguyen duong, do do k=0 71—•TJu k=m " lira " 1 *=0| (g + 1 ) ^ + 1 u,+jt KiU2...u„.-i.q^^lim^2->l Khi it nh?Ln cac gia t r i 0 , 1 , 2 , . . . , n t a cho j = fc, t a se c6: (1) fc=ni («.U2...,.._0(^„hm,l.—y lim y \- (2) fc=m "•^^ (2) suy r a t o n t ^ m = r > 2 d ^ T i 6 p theo t a chumg minh 3m > 2 : u i u j .. " m - i - „hm^ 2^ g' fc=m ^. dgn d i l u man t h u l n . Dat = ^V^, t i t gia t h i l t suy ra ^ lirn^ ak = + 0 0 / " 1 \ se chi ra rang ton t?ii so nguyen dUdng r j sao cho (uiU2...u„_i) lim y"—
  7. B a i t o a n 2 4 3 . Cho day cdc so nguyen {an}t=i tf^oa man diiu hen ^ " - ;V r iliim m ± ' - ' ^ 1 < a„ < max { 1 , n - 1}, Vn € N * . i ni ni! n->+oo fe=rn+l 2 " Tirn diiu kien can va du di so a = lim ^ la mot so hvCu ti = 1 — + ( n i - 1)! l i m V ni n — + 0 0 .it—' fc! A:=ni+1 Giai. 1 - + (ni-l)! lim y ( ^ D i g u k i e n ckn. Gia sH a = lim • — e Q. T a chiing m i n h rling k h i n—+00 Kl 3no e N* sao cho a„ = n - l , V n > UQ. Gia s i i phan chiing r i n g vdi mpi no € N * , t o n t a i n > no sao cho a„ / n - 1. D o a € Q nen a = —, vol = 1 - rii + (ni-l)! lim n-.+oo \^ni! f - ^ - An ! yl 2 1 •/ 2 1 1 K, M eN*. Suy ra vdi moi n> M t h i n!a € N * . Theo gia thiet phan chiing, = l - -+ (ni-l)!—= 1 - - + = 1 < 1. ton t a i ni > 2M sao cho a „ , / n i - 1, tiic la 1 < a„j < n i - 2 . Ta c6 ni ni nj! ni ni Vav 0 < ( n i - 1)! l i m 5^ T T < l i d i i u nay mau t h u i n vdi k=l fc=ni n (n, - 1)! l i m T ^ e N\ fc=ni Man thuan nhan ditpc rhi'mg t o rKng 3no 6 N* sao cho a„ = n - 1, Vn > no. Ta CO ( n i - l ) ! a 6 N* va vdi mpi n e ( 1 , 2 , . . . , n i - 1} t h i ( n i - 1)! chia het Dieu k i e n d u . Gia sii 3no e N* sao cho a„ = n - 1, Vn > no. K h i do cho n ! . Do do n (no + l ) ! a = (no + 1)! l i m ^ (ni - 1)! >i—+CX) ^ kl fc=ni V fc=i /J "0 n m-l = (no + 1)! l i m (ni-l)!a-(ni-l)!5;^^ , , fc! ^ 2^ fc! • no M a t khac do 1 < o„i < n i - 2 nen t a c6: = lim T»-« + 00 fc=ni+l ~ 1^ Ti + (no + l ) ! hm > —- k=ni V fc=ni+l n fc=no+l = ( n i - 1)! + hm > , TT _v^a.(no + l)! ^ / l i I n-.+oo K! nm > .— m! fe=ni+l / k=l n \ fc=no+l no a„i + 1 1 - + hm > T7 = ( n i - 1)! ni m n-+oo, fc! , E rn + ("0 + 1)! h m —r - -r Kl n-*+oo \no'. n! ^ ^ M m! n->+oo fc! fc! , ni ^ fc^ + fc=ni+l / 352 353
  8. To c & i t i m A sao cho ^ ^ ajAno + ! ) ' • ^ + 1 ) 6 iV* C T» 6 + Ad = A (a + Ac). (i) Z-^ fc! k=i j no- " 1 yn+2 = ayn+1 + KcVn + dz„) = a2/„+i + 6cy„ + d{y„+i - ay„) N h a n xet 24. TO kit qua bdi todn 243 suy ra e = J u n ^ E ^ ^ "Q- = (a + d)2/„+i + (6c - od)y„. Vay p h i T d n g t r i n h dac triTiig c i i a day so {y„} la 2.10 D a y s o phan tuyen tinh A^ - (a + d)A + (ad - 6c) = 0. 2.10.1 D i n h nghia day p h a n tuyen t i n h . Giai p h i r d n g t r i n h dac; t n m g t r e n t a t i m d U d c so h a n g t o n g q u a t y„ ciia day so {i/n}i v a tiir do t a t i m duoc so h a n g t Q n g q u a t 2„ ciia day so {zn}- Cho a, b, c, d thupc R sao cho ad - 6c 7^ 0 va c 7^ 0. Xet day so (x„) nhu sau: C h u y 31. Neu bd = 0 thi ta c6 the tuyen tinh hoa day so phdn tuyen tinh (xn) dd cho, rSi sau do tim dU0c so hang tSng qudt cua day so (xn), lam cdch my nhanh gon hdn. Ngodi ra ta cung c6 the tim dUdc so hang tong qudt cua 1 Vn € N*, x„+i = — — ^ , neu 116 ton tai. day so phdn tuyen tinh bang nhftn.g cdch khdc hay han nila, ban doc co thi CXn + d xem them bdi todn 30 d trang 2 4 {phudng phdp ham lap). Khi do day so (x„)+rj goi la day so phan tuyen tinh. Q u i \idc. Neu clio (x„)^^j la day so phan tuyen tinh thi ta higu rang vdi 2.10.3 M o t so t m h c h a t c i i a d a y so p h a n t u y i n t i n h . nipi n = 1 , 2 , . . . luon ton x-„. D j n h ly 19. Cho a, 6, c, d € M sao cho ad - 6c ^ 0, c ^ 0. Cho xi eR va vdi 2.10.2 X a c d i n h so h a n g t S n g q u a t c u a d a y so p h a n t u y e n t i n h . mqi n - 1,2,..., ddt , = x „ + i , neu no ton tai. Xet ham so fix) nhU Xet day so phan tuyen ti'nh { x „ } xac dinh bdi: sau: • cx„ + d ax,i + 6 ox + 6 CX + d trong do a, 6, c, d va p la cax. hang s6 cho tnfdc. Bang phudng phap quy na, ta c L g nunh duoc x „ ^ £ , V n = 1 , 2 , . . t . o n g do thoa man Chiing minh f Id song dnh. I Cho day so ((„) duoc dinh nghia bdi: 1 / yn+i = a y n + | ; 2 n , V n = l , 2 , . . . 2/1 = p , zi = 1,| 2,^^, = c y „ + d2„,Vn = 1 , 2 , . . . tn+i = rHtn),yn = i,2,... Vay van d^ con lai la t i n i s6 hang tfing quat cua hai day s6 (y„):ri (^n)»= thoa man ( * ) . T a t h u d n g dung haicax^hnha sau: 'tdj/ CO the khong xdc dinh ki tit mgt thU tU ndo do. Con f'Htn) la tg.p SQ,U: e a c h 1 (ap d u n g k h i (i) c6 h a i nghiem p h a n b i e t ) . T a co y„+i + A2„+i = (a + A c ) y n + (6 + Ad) 2„,Vn = 1 , 2 , . . . rVM = | x e R \ | ^ } : / ( x ) = („}. 355 354
  9. V^y vdi f la song anh thi dang tMc 2) ditdc xac dinh. K h i do vdi mpi fc = 1,2,..., n - 1 ^•^'J^g m i n h . T I T X „ + I = " ' ^ " " ^ ^ , V n = 1 , 2 , . . . cho n + 0 0 ta dUdc cx„ + d Xk dudc xac dinh va Xk^h. T a c6 xi 9^ ti. Neu x i = t2 thi: aL + b ^2 L = —— ^ cL' + {d-a)L-b = Q. Xl= r\h)=> f{xi)=^ti^X2 = ti cL + d ^ ' 356 357
  10. D i n h l y 2 1 . Neu A = ( d - a)^ + 46c < 0 thi day phan ki {khong hqi ty). (ad - 6c)(xn - 0) cQ + d D j n h l y 2 2 . Gid sU A = {d - af + 46c > 0 . GQI a,P Id hai nghi$m CUQ, (•0 + d ' {ad - bc){xn ~ a) phuang tfinh {dn Id x) cx^ + {d - a)x - 6 = 0. Khi do: cn + d x „ - B a) £i = a khi vd chi khi x „ = Q , V n = 1 , 2 , . . . = AX„,Vn=l,2,... 6) Old thiit X I 7^ a, dat Xn = p—^yn G N*, A = ^ . Khi do: x„ - a cP + d c) T h e o k i t q u a c a n 6), ,suy r a A",, = A " - ' X i , V n = 1 , 2 , . . . • N e u |A| < 1 t h i l i m A " " ' = 0. D o do Xn+i = AX„, Vn = 1 , 2 , . . . n—'+oo c) Gid thiet xi ^ a lim Xn = l i m X'^'^Xi = 0. n—.+00 n—+OC CQ + d • Neu |A| = < 1 thi l i m x „ = (3. n—•+00 _ -J- ~~ 0. TCr Xn = t a CO ca + d X„ - Q Neu |A| = > 1 vd x i 7^ /3 lim x„ = a. c.(i + d • Neu A = - 1 vd xi — P thi l i m x „ = (3. X,J:n--V„.Q = x „ - , 3 ^ x „ = ^ ^ ^ ^ ^ lim x „ = lim ~ ^ = /? A „ - 1 n — + 0 0 n-.+oo Xn — 1 • yVeu A = - 1 vd xi ^ f3 thi day { x „ } phan ky. • Trudng hap A = 1 A;/iong 1 t h i lim = 0 ( d o Xi 0). D o do C h u f n g m i n h . V i o, 0 la iighi^ni ciia phirong t r i n h L = " f ' ^ nen cL + d a - i - , lim Xn = lim ~ ^ = lim _ Q- 0 aa + b , a0 + b a = —„0 = ca + d cp + d' a) T a c h i c a n c h i i n g m i n h n e u xi = a t h i x „ = a , V n = 1 , 2 , . . . v i c h i l u ngUdc • N e u A = - 1 v a x i =^9 t h i x „ = / 3 , V n G N * s u y r a l i m x „ = ^fl. ' n—'+00 i, l a i l a h i e n n h i e n . T a d i m g p h u o n g p h a p q u y n a p . G i a s i i x i = a. K h i d o • Neu A = - 1 \h,xxi- 0 thi 7^ 0 v a axi + 6 aa + b X2 = : = J = a. X„+i =(-l)"Xi,Vn=l,2,... cxi + d ca + d Ta se c l u i n g m i n h d a y so ( y „ ) v d i y „ = ( - l ) " , V n = 1 , 2 , . . . , k h o n g h o i t u G i a siif x „ = a. K h i d o (phan k y ) . axn + b an + b Cach 1 . T a c6 lim y i n - x = h m (-1) = - 1 / 1 = l i m 2/2n- V a y day n—>+oo n—'+00 n—'+oo cx„ + a ca + d . phan ky. Vay theo nguyen l y q u y n a p suy r a neu x i = a t h i x „ = a, V n = 1 , 2 , . . . C a c h 2 . V d i m o i a thuQC K t a c l u i n g m i n h d a y (?/„) k h o n g h p i t u v e a. N e u 6) T a CO " 7^ 1 t h i |a - 1| = (5 > 0. C h o n e = ^ |a - k h i do Vn = 1,2,... ton tai Xn+i - 0 (aXn +b a0 + b\ +b aa + b ''0 > n sao c h o y„u = 1, n g h i a l a Xn+l = Xn+i - a \cxn + d c0 + dJ ' \cxn + d ca + d l2/no -a\ \\-a\>z. ac0Xn + adxn + bc0 + bd- ac/3x„ - ad0 - 6cx„ - 6d {cxn + d){c0 + d) " o n e n d a y n a y k h o n g t h e h o i t u ve a. N e u a- \i a 7^ - 1 . L y l u a n t i t o n g V u h u t r e n t a c u n g t h a y d a y (i/,,) k h o n g t h e h o i t u v e a. V a y d a y ( y „ ) k h o n g / acaxn + adxn + baa + bd- a c a x „ - ada - 6cx„ - 6d^ V icxn + d){ca + d) , (2/n) k h o n g h o i t y m a A : „ + I = 2/„Xi, V n = 1, 2 , . . . v a 7^ 0 n e n d a y 358 359
  11. X —3 _ 2 c ( x n - g) + 2cff + 2d ^ 2 c ( x n - g) ^ 2{cg + d) (a + d)(x„ - g) (a + d ) ( x „ - g ) (a + d ) ( x „ - 5 ) {Xn} cung khong hoi t u . Vay t i l X „ = suy r a day { x „ } kh6ng hpi 2c (a + d ) ' 2c 1 (vi neu h m x „ = G R t h i l i m X„ = , dieu nay mau thuan v6i d a v a + d'^ ia + d){x„-g) a+ d+ '^x„-g {X„} khong hoi t u ) . I2(c9 + d) = a + d VI 2cg + 2d = a - d + 2d = a + d). • Tritdng hop A = 1 khong the xay r a bdi v i n^u A = 1 t h i \u x i = g t h i theo a) suy ra x„ = g, Vn = 1,2,... do do l i m x„ = g. C) ^ n—•+00 ^ = 1 ca + d =' ciJ + d -t^ ca = cf3 a = (i. cp + d fjlu X I 7^ g t h i theo b) t a c6 Xn+i = Xn + Vn = 1,2,... suy r a {X^} Ih c^p so cpng C O cong sai la f.i va so hang dau la Xi. Do do Dion nay khong thfi xay r a dUdc do A = (d - 6)^ + 46c > 0. X „ = X i + ( n - l ) / y , V n = 1,2,... D i n h l y 2 3 . Gid thiit A = {d - a)^ + 46c = 0 ua d(it g = Khi do 2c a) x\ g khi vd chi khi x„ = 5, Vn = 1,2,... 6) Gia thiet xx ^ g, ddt X,, = — - — , Vn = 1,2,dat = Khi do Xn- g a+d lim (x„ - g) = li iium i — = = liim mi — ; = u -r— = 0 liiim i u x„ = g. X „ + i = X „ + M-Vn = l , 2 , . . . n->+c» n->+oo A „ n-.+oo A i + (U — l)/i n—+00 c) Um Xn = 9- V^y trong moi trUtJng hup t a dcu c6 l i m x„ = g. Chufng m i n h . 2.10.4 Cac b a i toan. a ) V i A = 0 n e u phUdng trinh cL^ + {d - a)L - b = 0 (tutc l a phuong trinh Nhir vay t a da khao sat day phan tuyen t i n h , da t r a Idi diXdc cau hoi v6i dieu L = ^ ) C O nghi^ni kep l a y = ^ T T - ^ . Tiep theo t a l a m tUrtng tit nhir da ki?ii nao ciiia a, 6, c, d t h i day phan tuyen t i n h hoi t y , phan ky. T u y nhien ket lam d dinh ly 22a) d trang 358. qua cua dinh l i 2 2 ci trang 358 va 23 ct trang 3G0 k h i d i t h i neu s\t dung t h i 6) Vdi moi n = 1, 2 , . . . , t a c6 phai trinh bay luon ca phan chiing m i n h cac ket qua do. Do vay neu t a nhd 1 _ faxn + b a -d\ dudc cac dinh ly va cac chiing m i n h ciia cac dinh do t h i t a c6 thg giai dufdc Xn+l — Xn+l -9 — i^ ' \cXn + d 2C J '• hau het cac bai toan ve xet sir hoi t u ciia day phan tuyen t i n h . Sau day la cfir bai toan ve xac dinh day phan tuyen t i n h , xet sir hoi t u cua day phan _ 2 c ( c X n + d) tuyen tinh. D l xet sii hoi t u t a chi can ap dung dinh ly 2 2 , 23 la r a ket qua. 2acx„ + 26c - acx„ + cdx„ - ad + Tuy nhien t a ciing c6 the giai b a n g each khac, mac d u giai b a n g each khac ^ 2c{cx„ + d) " l i H i g cac kot qua t l cac d i n h ly do c6 tac dvmg d i n h hucing, giup t a t i m dugc c(a + d)xn + 26c - ad + d'^' ' ^ i giai nhanh chong hdn. V\A = {d- a f + 46c = 0, nen 26c = Do do fiai t o a n 2 4 4 . Ttm so hQng tdng qudt cua day so (u„) cho nhu sau 26c - ad + d2 = - ^±1^ _ ad + d2 = i + 2ad - - 2ad + 2d^) l ^ - ' = 2 ^ ' " ^ ' - J-{d^ - a-") =-\{a - d){a + d) = - i . 2 5 c ( a + d) = - c ( a + d)g. ^^ 1 6„+i = 2a„ + 3 6 „ , n > 1. ,1^ ""^^ c{a + d)x„ - c(a + d)
  12. C a c h 2. Ta chutng niinh V c = I vk d = 3 {ad - be = I 0, c 0). Phuong t r i n h + 3x + 1 = 0 c6 I 2 j "'4-. nghi?m plian bi^t la x\ p va X2 = a — . T a co Do do A / f A \ l^n+i - L| < - |t;„ - L| < / - ) |t;„_i - L| < • • • < ( - \v^-L\. " $ 1^. - L| < ( - j |wo - L| , V n = 1, 2 , . . . cho n + 0 0 va sU dung Vay theo diuh ly 22c), day d a cho hoi t u va - 3 + v/5 ' ^ y ^ ' ^ ly kep t a d U d c lim ^„ = L = hm v„^a= . n—+00 2 n—*-\-oo 2.
  13. Bai t o a n 246. Cho day { r „ } nhu sau: vi = a va • Ta CO 6 + cvn > b + cv = > 0. Yi v„ > V, vdi V la nghi^m Idn ciia tam thtJc fix) = cx^ + bx + a nen cv^ + bvn + a > 0. Do do v„ - v„+i > 0, vdi . ^, = - ,Vn 6 (a, 6, c > 0, A = 62 - 4 a c > 0, a > (1) iiipi n e N * . Hay (i;„) la day giam. V?iy (t;„) la day giam va bj chan dudi nen -6+ V/A C/i»?np m i n / i rdng day s6 da. cho hoi tu va tinh ^}^rn^^n- hpi tu. Tir r i > v 2 > V 3 > - > v „ > v„+i > — , Vn = 1, 2 , . . . , suy ra lim Vn > — — V ^ y theo \i lu(ln d phan d i u Idi giai suy ra Giai. Ckch 1. 1. N§u Neu lim i;„ = L tlii tit (1) cho n - +oo ta dUdc -b + ^/A n->+oo lim Vn = - 6 ± n—»+oo 2c -a C a c h 2. T a c6 6 + cZ, 2c K y hi§u = — — T a cluing minh ' 2c 2c 2c Taco - 6 + v/A (2) - 6 - v/A b + c 2c 6 - v/A 6 - < 1. b + 6 + v/A 6+v/A Tir gia thiet Q > ^ s u y ra (2) dung khi n = 1. G i a sijf (2) dung tdi 2c 2c n = A; - 1, tiic la Vk-i > v. K h i do t;(i;jfe_i - f ) < 0 (vi i; < 0). Do do -b+ v/A v.Vk-i < cv.Vk-i < cv^ ^ cv.Vk-i + bv + a < cv^ + bv + a. Vay theo (hull Iv 22c) suy ra lim t;„ = . n—'+00 ZC Bai toan 247 (De thi vo dich sinh vien Moskva-1982). Cho day {x„} nhu Vi V la nghi?ni ciia piiUOng triuh cx^ + bx + a = 0 nen cv'^ + bv + a = 0. Do do sau: xo = 1982, x,.+i = . \n = 0 , 1 , . . . ) . Hay tlm lim x „ . cv.Vk-i + bv + a < 0. 4 —0 X „ n->+oo —a Giai. T a CO 6 + cvk-i >b + cv, ma v = ueu CSch 1. T a CO ( x „ ) la day phan tuyen tinh, x „ + i = , vdi a = 0, 6 = 1, cx„ + a b + cvk-i >b + cv = — >0. c ^ - 3 va d = 4 (od - 6c = 3 0, c ^ 0). Phudng triuh S i ^ - 4x + 1 = 0 c6 ''ai nghi^m phan bi?t la xi = ;3 = ^ va X 2 = Q = 1. T a c6 Boi vay c.v.Vk—i + bv + a a —a i b + c.Vk-i b + c.Vk-i b + c.Vk-i Vay (2) diing khi n = k. Do do theo nguyen ly quy nap suy ra (1) dung ^''^ theo djuh ly 22c) suy ra day da cho hpi tu va ^ h ^ ^ v„ = 0 = ^• nioi n = 1 , 2 , . . . Vay ta da dn'nig minh dttrir day so da cho bi chan difcii T-'* CO 2. Xet ham s6 / ( x ) = khi do fix) lien tyc tren R \| va a cvf. + bvn + a Vn - V„+i = Vn + -r-, = 7- • x„+i=/(x„),Vn = 0,l,... b + CVn b+ CVn 365 364
  14. T a CO fix) = (4 - 3x) phirffiig t r i n h dac t n m g cua day so {y„} la - 4A + 3 = 0 A= 1 1 rx = i A = 3. fix) =x^ = X 3x^ - 4x + 1 = 0 So h * " g tSng c"^* 1^- 4-3x "=3- y„ = >1 + S.3", Vn = 1 , 2 , . . . (^, B la cac liKng so se t i m sau). Taco 1 -1 Vi zn = 2n = >1 + B . 3 " + ' , Vu =: 1 , 2 , . . . V i 21 = 1 nen 1 = + 9B. x i = / { x o ) = /(1982) = 4 - 3.1982 5942 3 x ^ A = Vi j/i = XI nen xi= A+ 35. V§y: | ^ + f f ^ X 3 Vay theo nguyen l y quy nap suy ra O = /(X2) > ^ = 0 , 2 5 > X 2 . 3 x ^ -.3"] 2 + 6 Do do thoo kot qua bai toan 50a) 6 trang 148, suy ra ke t i t so hang t l n i hai 3x1 - 1 T^I^T ' Vn = 1, 2 , . . . t r d d i , day (x„) la day so tang. - ^ ± . - ^ . 3 - ^ De thay x„ e ^ ^ \n = 3 , 4 , . . . Suy ra day (x„) hoi t u . D a t l i m x „ = x, 4'3 n—>+oo Sl^r^^3T(r31^;)3^'^" = l ' 2 , . . ^ 1 1 khi do X 6 va X G Vay t a lo9.i tritdng h0p x = 1 . Do do x = 'Jo Xn khong xac: dinh khi va chi k h i : 4'3 hay l i m x„ = - . - 3 + ( 1 - ^ J 3 n + i = 0 ^ ( 9 - 3 " + i ) x i = 3 - 3 " + ' ^x, = LJ1_ n-'+oo i 3-3« B a i t o a n 248'. Cho day so { x „ } dinh nyhta tr~uy hoi bdi: % ta CO k^t qua nlur sau: • Khi X ^ 1 - 3 " (n=l,2,...). ' 3 - "3 ^ ' 3 ^ day khong xac dinh. 4-3x, . xrl: " ^ ^ " = 1. Vn = 1, 2 , . . . , do do l i m x „ = 1. Hay tlm cdc gid tri cua x i dc day trcn hoi tji vd trong cdc trUdng help do * ^"^i c a c ' 1 ' tinh l i m x „ . k; ciia x i t h i x„ xac din] gia t r , khac dinh vdi moi n = 1 , 2 , . . . v^ n—>+oo ^ 9x1-3.+ (l-xi)3" GTiai. Day so da cho c6 dang x„+i = ^, vdi a = 0, 6 = 1 , c = - 3 , d^^' CXn + d c ^ 0, ad - be = 3 0. Xet hai day so (y„) va {z„) thoa man dieu kien saU- „lim x„= Ihn -g^-3+(l-xi)3" yi "-+00 n^+^ 9a;i - 3 + (1 - a:i )3"+i 366 367
  15. 911 - 3 „ 3x„+i - 1 •3a:„ _ 1 " „„_i Xi - 1 = lim '3x„_j - 1 „"+oo 9 £ | ^ ^ _ ^^^3 (l-xi)3 3" '3x1 - r 3 " - ' ( x , - 1) , Ltfu y. ^-1 De cho IcJi giai dUdc ngin gpn va hdp logic thi vi§c tim ra cong thiic 3x, - 1 t6ng quat x„ cua day {a;„} diWc lam d ngoai giay nhap, con khi trinh g a i toan 249 (Do thi vo dich Kicp). Cho day so (a„) dU(fc xac dinh nhu bay 16i giai ta tien hanh theo trinh tif sau: sau: 3 - Bang quy nap (theo n G N*), hay chuTng minh: u„ dvr0c xac dinh ai = 2 , a„+i = 4 (Vn = 1 , 2 , . . . ) . neu va chi neu: (3* - 3)xi 3* - 1 khi 1 < A: < n; (1) a„ hPn nuta, vcii dieu ki^n (1), thi Chttriy minh rdny day ad da cho c6 gidi h(f.n hHu hg.n va tinh gidi h^n cua day so (i^- ( 3 " - ! _ 1) _ (3"-! _ 3)xi (2) Dap so. lim a„ = 3. (3" - 1) - (3" - 3)xi - V^y, Hipi so h ^ g cua day deu dM0c xac djnh neu chi n6u: Bai toan 250 (De thi hpc sinh gioi qupc gia, bang B, n i m hoc 2002-2003). Cho so thuc a ^ 0 va day no thxjtc { x „ } , n = 1 , 2 , 3 , . . . , xac dinh bdi: 3 ^ , ^ > 2 } . (3) xi = 0, x„+i (x„ + a) = a + 1 (Vn = 1 , 2 , . . . ) . Ngoai ra, vdi digu kien (3), thi (2) diing vdi mpi n e N*. I a) Ildy lim so hung long qudl cua day so dd cho. - Tijr nay, gia suf (3) dit0c thoa man. Khi do, neu a 7^ 1, de thay: (2) b) Chiing minh day s6 (x„) co gidi han hHu hQ,n khi n —> +00. Hay tim gidi ban do. keo theo ton tai ciia gidi han: Giai. 1 1 a) m-l in-2 1 -a lim x,i — lim --, ^ IViidng hdp 1: a = - 1 . K h i do x „ = 0, Vn = 1 , 2 , . . . n—'+00 I _ 1 \ 1 3-3a 3 3-i - 3 - m-2 IVUcJng hdp 2: a - 1 . K h i do x„ ^ - a , Vn = 1 , 2 , . . . Do do ta c6 - Con neu o = 1, thi x„ = 1 (Vn € N') x„+i = ^ ^ , V n = l , 2 , . . . (1) lim Xn = 1. x„ + a . Ngoai ra ta c6 the tim dUdc so hang t6ng quat cua day bang pli'^'^"^ (1) CO dang x,.+i = vdi a = 0, 6 = a + 1, c = 1, d = a, c ^ 0 va ad I cxn + d phap ham l^p (xem myc 1.1.5 ci trang 24) nhu sau: Neu x i = 1 "'^ = a + 1 ^ 0 ( a 7^ - 1 ) . Xet hai day so (j/„) va {z„) thoa man dieu kien x„ = 1, Vn = 1, 2 , . . . , nlu x, = ^ thi x„ = \ Vn = 1 , 2 , . . . T i l p th^'"' 1 xet xi 7.^ 1 v a x i 7^ - . T a c6 '^J'i do 1 3(:^»-l) x„+i - 1 = - 1 = J/„+2 = (a + 1) zn+i = (a + 1) (y„ + az„) 4-3x„ 4-3x„ 3 3x„-l 3x„+i - 1 = - 1 = 4-3x„ 4-3x„ 368 369
  16. f lidng t u tritcJng hop 2a, bang quy nap ta clnbig niiuh difdc: Vay plntdiig t r i n h d i e t n m g c u a day so {yn} l a y^_{a+l) ( - 1 ) " + (g + 1)" _ (g + 1) [ ( - ! ) " + (g + I f i ] A= -1 - QA - (a + 1) = 0 la cac hang so se t i m sau). , Neu a = - 2 thi x„ = Vn = 1, 2 , . . . Do do '••'ax Vi 2/1 = 0 nen C + £> = 0. Vi j/2 = (a + 1) zi = Q + 1 = - 1 nen C + 2£) = -1 lim Xn - lim —- = lim ( 1 - i ) V a y { g t S f i _ i ^{g=i:^ 1. Dodo n—'+00 n-*+oo n n->+oo \ , Neu g 7^ - 2 thi y„ = (l-n)(-l)^Vn=l,2,...,^„=|^=n(-lr^Vn=l,2,... Q + 1) [ ( - l ) " + ( a + l ) n-l ,Vn= 1,2,... Ta CO — = x i . Gia Slit — = x„, khi do Taco y„+i ia+l)zn a+l _ Q+ 1 _ „ , Zn V$,y theo nguyen ly q u y nap s u y ra ;\ (g + 1) x„ = — = ,Vn= 1,2,... \2n- +1 Zn n - + 1 + (» + 1)^" _ (Q + 1) 2^271 = -,Vn- 1,2,... Trtrdng hdp 2b: n ^ - 2 . Khi do so hang tong quat ciia day so {y„} la: (g + l)'"-l 1 - ,2n (g+1) yn = A ( - 1 ) " + B.{a + 1)" , Vn = 1,2,...{A,B la cac hang so se t i m sau). •Do do ngu | Q + 1| > 1 thi Vi 2/1 - 0 nen -A + B. {a + I) = 0. Vi y2 = {a + 1) zi = a + I nen t a c6 lim X2n = lim X2„-i = 1 => lim x„ = 1. A + B. (a + lf =n + l. Vay n->+oo n—+00 n-»+oo •Neu | Q + I| < 1 thi -A + B.{a + l) = 0 A - { Q + 2 A + B.{a'+lf = a+ l lim x 2 „ = - ( g + l ) = lim X2„-i lim x„ = - ( a + l ) . Trong so hang tong quat ciia day so d bai toan 250 0 trang 369 lay V$,y vdi mpi n = 1 , 2 , . . . ta c6 ~3 ta dUdc so hang t5ng quat cua day so trong bai toan 247 ci trang 365 ^ ^ ^ g khac gia tri ban dau. Ta cung c6 the tim dUdc so hang t6ng quat ciia a + 2 a+2 y so nhanh hdn bang phudng phap lap, diia tren cac ket qua sau: ^n+i - 1 = _ 1 = -^" + 1 ^ - {Xn - 1) Xn + a x„ +a x„+a 370 371
  17. ia + l)x„ + (a + lf ^ ( g + 1) (a:,. + a + 1) X n + l + « + 1 = Xn +—a + " + 1 = x„ + a x„ + a p a l t o a n 2 5 3 . Cho day so (u„) n/ii/ sau UQ = 0 vd x„+i+a+l a + lXn + a+ 1 ( a + 1 ) " ' x i + a + 1' - u „ -H2010' B a i t o a n 2 5 1 . Cho day so {x„} nhu sau: fl) Chiing minh day {un) c6 gidi han hHu han vd tinh l i m u„ 1 '')^^^^" = £^;;^2oo8- ^^''^^ n-^+oo n + 2009 • Chiing minh rdng day {xn} hqi ty. vd tim Urn Xn- G i j ii aa ii .. ^ 2008 a) Ta CO uo < 1. Gia siif u„ < 1, k h i do do ham / ( x ) = dong bien H\Xdng d a n . D ^ t (\/2 + l ) " " ^ ' x„ = u„. K h i do do ^ X ~T~ ^UXU tren khoang ( - 0 0 ; 2010) nen UQ = 1, u„+i = ^ ^, Vn = 0 , 1 , 2 , . . . u„ + 3 ti„+i = / K ) < / ( ! ) = ! . 2008 Ta chi'mg m i n h ditdc u„ = — ^ — — r ; i + r , Vn = 0 , 1 , 2 , . . . Vay (V2 + 1) - ( v 2 - 1) Theo nguyen l i quy nap suy ra u„ < 1, Vn € N . L ^ i c6 u j = ^QJQ > UQ. G i a sii hm Xn = Hm = \/2 - 1. u„ > u„_i =^ / ( u „ ) > / ( u „ - i ) =^ u„+i > u „ . Theo nguyen l i quy nap suy r a u„+i > u„, Vn e N . Vay day (u„) tang va b i B a i t o a n 252 {D^ t h i HSG Gia Lai, nam hoc 2008-2009). Cho day so (x,,) chan tren nen c6 gicii han hfm han. Dat l i m u „ = L , k h i do L < 1. Tilt (1) (n 0 , 1 , 2 , . . . ) thoa man n-.+oo cho n -+ + 0 0 t a dUdc 2x„ + l xo = 2,x„+i = ^ : ^ , V n = 0 , l , 2 , . . . L = i l ± ^ ^ L 2 - 2009L + 2 0 0 8 - 0 ' ° ^ ^ ' L = l ^ l i m u„ = 1. —L + 2010 n—>+oo a) Tim lim x„. 6) Ta CO n->+oo b) Chiing minh x i + X2 + • • • + X2008 < 2009. _ 2008 = - 2008 = ^009 -2008) H t f d n g d i n . Sii d u n g phiXdng p h a p h a m l a p ( x e m m u c 1.1.5 6 t r a n g 24) ta -Uit-i+2010 -Ufc_i+2010 3"+i ^ I 1 ^ -Ufc-i -h2010 2 1_ de d a n g t i m d u o c so h a n g tong q u a t c i i a d a y so ( x „ ) l a x „ = y^+rZi' Ufc - 2008 ~ 2009 - 2008) ~ 2009 {uk-i - 2008) 2009 = 1+ 3„+i 2 _ i = l + i + 3+ 1 32 + . . . + 3 n < l + 3;r< 1 , J_ +2^- Uk - 2008, 2009 E l ^ ^ j U f c _ i - 2008 n 2009' Suy ra He quala n
  18. -1 2 n o„-i - -K Tn 2008 + 2009 Un - 2008 2009' cos a (a„_i cos 2a + 1) (cos a - cos 2a) cos^a' \u if) V^y TifOng tir 1 2007 _ 2 n 2009 " ~ 2008 2009(u„ - 2008) 2009 a„ + cosa (2) cos a (a„_i cos 2a + 1) (cos a + cos 2a) cos'^a' -2009 ' 2 n ^Tn = rp^ (1) va (2) suy ra 2007.2008 2007(w„ - 2008) 2007' ( c o s a ) a n - 1 _ [ ( c o s a ) a n - i - 1] (cosa - cos2a) Vi (cos a ) a „ + 1 [(cos a ) a „ _ i + 1] (cos a + cos2a) -2009 n 1 = 0, l i m _ [(cosa)a„_2 - 1] (cosa - cos2a)^ n^+oo 2007.2008(71 + 2009) n " + 0 O 2007(n + 2009) 2007 [(cos a ) a „ _ 2 + 1] (cos a + cos 2a)^ 2 = 0 _ _ [(cos(t)ao - 1] (cosa - cos2a)" n "lini +oo 2007(u„ - 2008) (n + 2009) [(cos a)ao + 1] (cos a + cos 2 a ) " (cosa - cos2a)"+^ (cos a + cos 2a) n+i- L v f u y. Ta cou c6 the giai bang each t i m so hang tong quat n \ 2 ^n - 1 — - 1 2008 V2009y 2.11 D a y tong V2009; dc siiy ra kct qua. 2.11.1 M o t s 6 \\iu y. B a i t o a n 2 5 4 (De nghi Olympic 30/04/2011). Cho day (a„) xac dinh bdi: Dinh n g h i a 21. Cho day so (x„)+ri- Xet day s6 (5„)+^i nhu sau: n , - 1 Un-i + 1 - tan^Q °"~cos2a'''" a„_iCos2a +r ( - 2 < " < 2 ) - E i=l Xi, tlic Id Sn = Xi + X2 l' vivn; 1- Xn- Tim so hang tong quat cua day va xac dinh gidi han cua day. do day SO {Sn)l=i diMc got Id day tSng. . Hirdng d i n . Ta eo Ltfu y. M o t so (lay so eon dime xay diftig thong qua nhfmg tong thoa man aieu kien nao do. Chang han cac bai toan 31 (6 trang 127), 33 (d trang 128), 1 a„_i + 1 - t a n ^ Q 1 34 (6 trang 129), 35 {d trang 130), 36 (a trang 130)... a„" - cos a a„-icos2Q + l cos a ^ h a n x e t 2 5 . Doi vdi nhxtng day so {xn) cddangxn+i = / ( x i + X 2 + -• •+x„), sm^Q O n - 1 cos a + cos a - - 1 - o„-icos2a hang each dat Sn = xi + X2 + • • • + Xn, ta dua nc day so { 5 „ } + ^ i nhu sau COSQ ( a „ _ i c o s 2 n + l)coso: Si = Xi, Sn+i = Sn + fiSn). Ch' - ' V cos a (cos ct - cos 2 a ) O n - i + (cos^a - sin^a - cos a ) ^^.V y 32. Van de xet sif hoi tu hay phan ki cua mot day tong diMc dc cap (a„_i cos 2 a + l ) c o s 2 a di^^ chuang trinh toan cao cap, tuy nhien trong bdi nay ta chi xet van cos a (cos Q - cos 2 a ) a „ - i + (cos 2 a - cos a) . ^ ^0 cap han do Id tim gidi han cua mot so day tong thudng gap trong cac ( a „ _ i cos2a + l ) c o s 2 a hoc sinh gidi trung hoc pho thong. 374 375
  19. 2.11.2 P h i f d n g p h a p t i m gidi h a n c u a day t6ng. Bdi vay T a tlnrrJiig d u n g c a r phUrtng p h a p sau d a y : • R i i t g p n hoSc t h n so h a n g t 6 n g q u a t c i i a day so ( 5 „ ) (day l a d a n g t o a n r§,t „ii!?«, ("1 + "2 + •• + u„) = lim 1 - = 1. (n+ 1)2 t h u d n g g a p t r o n g cac k j ' t h i H S G ) . • So s a n i i d a y so (sit d u n g n g u y e n l y k ^ p ) . g a i t o a n 2 5 6 . Xet ddy ( x „ ) nhu sau: x „ = w , . ''' • Sit ( i i i n g cac d a y c o n ke n h a u . (2n-l)2(2n+l)2'^"=l-2,... • C h u y e n ve d a y t i c h . fim lim Yl • Sit d u n g k e t q u a c u a b a i t o d n 7 2 6 t r a n g 163, 7 3 6 t r a n g 163, nh|,n x e t 5 d t r a n g 163, nh§,n x e t 6 d t r a n g 164, b a i t o a n 2 0 0 a t r a n g 289. G i a i . T a c6 C h u y 3 3 . Neu day ( x „ ) cho bdi he thiic truy hoi {ching han bdi toan 260 n 1 d trany 379, bdi toan 2 6 1 0 trang 380...) thi ta thudng Idrn nhit sau: ^" (2ri-l)2(2«+l)2 - § ,Vn = l,2,... • Chung minh lim x„ = +oo hofic lira x„ = -oo bdng each chi ra day (2n-l)2 (2n + l ) 2 _ {x„) tang vd khong bi chdn tren hodc day ( x „ ) gidm vd khong bi ch^n dudi. Do do • Rut gon tong 5 „ , tii do tim lim 5„. ( X l + X 2 + -- -f-x-„) = i 1 - n->+oo 8 (2n+l)2 n ^ Bdi va\ C h u y 3 4 . De' rut gon tSng S„ = thudng bien doi i=l n n ^ l i m ^ ( x i + x 2 + ... + x„) = i lim 1 - 1 8 n->+(xi (2n + l ) 2 _ 8' 1=1 i=l n - 1 B a i t o a n 2 5 7 . Tinh l i m V x^. biet: x„ = I n ,Vn = 2,3,... ji—+30 n-* + 1 2.11.3 C a c bai toan. G i a i . T a c6 B a i t o a n 2 5 5 . Xet ddy (u„) nhu sau: u„ = ! , „ , V n 6 N*. Tim n^{n + 1)2 ^ 2^-1 , 3 ^ - 1 n ^ - l , lim (lii +1X2 H Hu„). u->+oo G i a i . T a c6 2^ + i A 3 ^ + i y li^TT + ly ^ ^ (iiilK^lt^ _ - 1) [(^ + 1)^ - (fc + 1) + 1 ^•2(A.•+l)2 - Jt2 (fc4-l)2'^*""~ ^ + 1 (^•+1)(A:2-A:+1) (RTJOt^TI^T) 1^0(16 „, , . , D o do / 1 Ul +U2 + h 1 = 1 - 376 377
  20. 1 1 u„ = i / a 2 + 4ay„, u„ > 0. T i f (2) suy ra: 2 n2 2 Suyra lim ft fe^ = ^ . ^ - U i = ^. V i ham s 6 / ( x ) = i . + =K+2a)2, VneN*. lien tuc tren (0; + 0 0 ) nen t a c6: Do Un > 0' Vn e N ' va a > 0, t a c6 (3) u„+i = u„ + 2a, Vn s N*, hay (u„) / n la niot cap so cpng c6 cong sai d = 2a va u i = 3a. Vay In = ln = ln2 lim y Xk = lim n->+oo fc3 AA; +l 3' u„ - 2an + a => 2/„ = a(n^ + n), Vn € N* ' n->+oo ^—^ fc=2 n—•+00 /J fc=2 1 1 / 1 1 1 an(n + 1 ) a \n n+ 1 •+ + ••• + B a i t o a n 258. Tinh Jjm^ ( ^ + I T 2 " r + 2 + 3 =^5„ = V a ; * : = - f l 7-7)=^ h m Sn = -. ^ 1 /I 1 a \ +1 y n-^+oo a G i a i . T a c6 = 2 . V i vay 1 + 2 + --- + A: k{k + l) k+l Bai toan 260. Cho so ihuc a > 1. Xet day so (u„) n/iU sau: u\ a vd 1 1+ 2 1+ 2+ 3 ' ' l+2+--- +n u„+i = u ^ - u „ + l , V n = l , 2 , . . . (1) =1.2 ^ - ^ ) + . V3 4; + - - + ' ' n - n + i y Haytm lim | — + — H + — )• , = 2 - n +1 Giai. V i a > 1 va u„+i = + (u„ - 1)^ , Vn e N* nen de dang suy r a Do do 1 < a = U i < U2 < • • • < u,, < Un+l < ••• (2) / 1 2 \ Td (1) va (2) t a c6: lim 1 1^ 1 = lim 2- = 2. + ,^ n+ 1 n-.+oo \+ 2 r + --' + l +2+ --- + n u„+i - 1 = u„ (u„ - 1)
ADSENSE
ADSENSE

CÓ THỂ BẠN MUỐN DOWNLOAD

 

Đồng bộ tài khoản
2=>2