Bồi dưỡng học sinh giỏi khảo sát hàm số

Cac bai toan lien quan den day so thiTcJng la nhufng bai t$p 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 vie

510.76

(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 d a y so {Xnlnri sao c h o x i = a va

N G U Y E N T A I C H U N G

B6i dvtdng hoc sinh gioi, chuyen toan

- 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 * '

L&i noi dau

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

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

sSp xep hpp li de giiip ban dpc tiep can tijfng bifdc, ttoig miic dp kien thiic va luyf n

t§.p giai toan

Phan 16n cac bai toan trong cuon sach nay du^c tuyen chpn t\t cac ki thi; Thi

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,

ho$c tren T^p chi toan hpc va tu6i tre Mpt so bai toan khac la do tac gia t\f bieii

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

Idi giai hay nhat va ngSn gpn nhat, hi vong ban dpc so c6 dxtac nhftng Idi giai hay

hdn

Vice phan chia cac chiroug, bai, myc dfin c6 tinh dpc lap tiWng doi va do do

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

bai kia Tuy vay, chiing toi van c6 gang xoay quanh chii de ciia bai iy de ban dpc

minh rut ra TiliuTng gi thudng gap va each giai quyet van de

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

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 l i " , cac ket luan, nhan xet va litu y tiT bai toan difa ra

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,

thi Olympic loan Sinh vien giiJta cac tritdng dai hpc

chiing minh menh de chiia bien A{7i) dung v6i moi so nguyen dUdng n

(bang phifdiig phap quy nap), ta tlittc hieu ba bitdc sau:

Bxidc 1 {hxidc c d s d , h a y bvTdc khdi d l u ) Kiein tra A{n) diing khi n = 1

Biidc 2 (birdc q u y n a p , h a y bi:rdc "di t r u y § n " ) Vdi fc e Z, fe > 1, gia sii A{n) diing khi n ^ fe, ta chiing minh A{n) cung diing khi n = fc + 1

Bifdc 3 Ket luan A{n) diing xcii moi so nguyen ditdng n

B a i t o a n 1 Cho day s6 nhu sau: | = ^ , ^ 2 + ^ , Vn = 1 , 2 ,

a) Tinh x\, X2, Xi , , b) Tim so hang tSng qudt {so hang thii n)

a) Ta CO x i = \/2 = 2 cos va

X 2 = V^2 + v / 2 = W2 1 + v/2' = ^ 2 ( 1 + COS J ) = 2cos J , X3 = V'2T1^ - ^ 2 ( 1 + COS J ) = 2 cos ^

3

b) Vdi e Z, k>l, gia sii xfc = 2 c o s K h i do

= V2T^k = y ^ l T c ^ ^ ^ = 2COS

Theo nguyen li quy n?ip suy ra x„ = 2 cos ^^^n = 1,2,

Bai toan 2. x„ = ^ 2 + \/2 + • • • + v^ Hay tinh

sa\ do sir di.ing bai toan 1 d trang 3

Bai toan 3 (Do thi OLYMPIC 30/04/2003) Cho day s6 (u„) djnh bdi

_ lan - + lan —

Do do = v/3 = tan ^ , ^ 2 = 3 L = tan f J + ^) va

1 - tan - tan - v 3 8 / tan - H — + tan -

tan 3 + ( " - ^ ) 8 , + t a n ^

1 - tan ^ 3 + ( n - l ) 3 ^ t a n -TT

= tan Tlieo nguyen ly quy nap suy ra

Bai toan 5 Tim so hq,ng tdng qudt cua day so (x„) cho nhu sau:

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

nhiJng thudng dung trong day so ma nhfmg phep doi bien nay con hay dimg

khi giai phUdng trinh, chuTng minh bat ding thiic ):

Chu y 1 Neu ta gap ham da thiic bQ.c hai f{x) = ax^+bx+c thi ta ddi goc tQa

do vi dinh A Pavabol TiCc la ta ddi bien ~

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

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

fix) TUc la ta doi bien X = x +

4a Bai todn 7. Tim day s6 {x„};^~i biet

Hi/dng dan Chii y 1 ci trang 6 gpi y cho ta each doi bien nhvl sau: Gpi

yn = x„ + —, thay vao (1) ta dupe:

Bai to6n 8 Tim day so {x„}^^ sac cho x\ a va

x„+i = a x 3 + 6 x 2 + c x „ 4 - d , V n = l , 2 , ^ a > 0 , c = ^ ^ =

HUdng dan Gpi y„ = x„ + thay vao (1) ta dupe

/ 3 , , 62 63 \ 2 262 63 62 6 62 6 ( c - 3 )

Suy ra j/„+i = ay^, Vn = 1,2, Suy ra

Vn = ayt, = a {ayt2f = a'^'vlU = a'^' {ay^.f = a'^'^'^t^

Bai toan 9 (De thi HSG quoc gia nam hpc 2000-2001, bang B) Cho day

^' = 5- ^"•^' = 2(2n/r)x„ + l ^ " ^ ' ' ' ' " " ' ^ '

Hay tinh tong cua 2001 so hqing dau tien cia day so {xn}n=i •

Gi&i. Dg t h i y x„ > 0, Vn = 1,2, do do tit i „ + i =

2(2n + l ) x „ + 1 ta c6 x„+i = p = 2 ( 2 n + l ) + — (n = 1 , 2 , )

2 ( 2 n + l ) + — ^"+1

D$,t — = u„ K h i do u i = 3 v&

x „

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

Nhu vay nioi cong thiJc lUdng giac se cho ta mot dong nhat thiJc dai so Tuy

nhien v6i so lUdng cac cong thiic bien doi lUdng giac qua nhieu, ban than cac

hp thiic ludiig giac da t ^ thanh mot chuycn dc c6 tinh doc l9,p tUdng doi

dan tach hkn cd sci dai so cua no, da lam cho chung ta qucn di mot lir^ng

Idn cac he thiic dai so c6 cung xuat xvi t\l mot he thiic lUdng giac quen biet

D i e b i f t trong chUdng trinh toan bac ph5 thong hien nay, cac ham s6 lir^ng

giac ngu^c, ham luqng giac hypebolic, khong nSm trong phan kien thiic

bat biioc t h i nhfmg bai toan lien quan den chung se la mot thach thiic Idn

doi vdi hoc sinh va ca giao vien Djnh l i 1 va dinh li 2 se giup ta thiet l^p

nen cac dong nhat thiic dai so d l t i i do di den 16i giai cho cac bai toan xdc

dinh so hang toiig quat cua day truy hoi phi tuycn bac nipt

D i n h l y 1 Gid sii cosnt = Pn (cost), vdi P„(x) la da thiic bac n Khi do

T

C h u y 4 Ta thay rdng cosnt duac hiiu dien thanh mot da thiic cua cost, 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

D i n h l y 2 Gid sii sin(2/!; + l)t = P2k+i (sint), 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

P2k+i{x) b&ng each giU nguyen nhUng hf so ling vdi luy thica chia 4 du I va

thay nhUng he so Hug vdi luy thiCa chia 4 diT 3 bang h$ so doi ddu Khi do

Sau day la mpt so he thiic dai so t h u dUdc tit cac hg thiic lien quan den ham hypebolic, vipc chiing minh cac ho thiic do xin danh cho ban doc j

• Tir he thiic cos 2t = 2 cos^ t - 1 ta thii dUdc h? thiic dgi so

• T i t h? thiic cos 3t = 4 cos^ f - 3 cos t ta thu dildc h f thiic dai so

(1)

1 /

Nghia la 4x^ " 2 a + - 7 - vdi X = - a + -

I l f thiic dai so ling vdi cos 5t = 16 cos^ t - 20 cos^ t + 5 cos t la

i (a^ + = 16m^ - 20m^ + 5m, trong d6m =

]-(a+-Tii h§ thiic sin3t = 4sin^ f - 3sint ta thu ditpc h§ thiic dai so

1.1.4 M o t so phep d6i bien dvtdc d i n h hMdng b d i cAc cong thiJc

lifdng giac

Trong muc nay ta so xct mot so bai toan diTdc giai bKng each di.ra tren ode

d^c trimg cua mpt s6 da thilc dai so sinh bcii ham so sinnx vk cosnx

Bai toan 10 Xdc dinh d&y so (j/„) thoa man dicu hien sam

j / i G R; Vn+i = 2yl - 1, Vn = 1, 2 ,

Giai

• Neu < 1 thl t6n tai 4> sao cho cos(^ = y^ Khi do

j/2 = 2 cos^ (j)-\ cos 20, y-i = cos 2^^,!/„ = cos 2"~^</>

• Khi I2/1I > 1 Xet so thi^c fj sao cho

Vay neu dat P = yi + i/j/jf - 1 thi

2/1 = 2 Taco

Cach t r l n h bay khdc cho trtrdng hdp > 1 Neu \yi\>\i ton tai

(P sao cho cosh = j/i Khi do

y2 = 2 cosh^ (i>-l = cosh 2(f), j/3 = cosh 2^ (A, , 2/„ = cosh 2"~ ^ (A

-6y„+i = at^y^ ^ ^ ^ ^^^^ = _aj,y2 _ 1 y^^^ = 2y2 - 1

Sau do Slit dung bai toan 10

LtTu y Phep dat x„ = - 6 y „ ditdc tim ra nhit sau: Cong thitc hwng giac

C O S 2 Q = 2cos2 Q - 1 gdi y cho ta c6 gang dua day so da cho ve day so

{yn)t=i thoa man

y „ + l = 2 y 2 - l , V n = l , 2 , (1) D§,t x„ = py„ Khi do

py„+i = ap^yl + 6 y„+i = apyl + | (2) Ttf (1) va (2) suy ra ta can tim p sao cho

B a i todn 12. Cho day so (x„) nhu sau: x i < va

4b

2 1

XiX2X3 X„ Tim so hang long qudt cua day so {u„)

G i a i Dat x„ = ^y„ Ta dUdc day so (y„) thoa man: j/i = 2xi < - 1 va

6i/„+i = a62y2 + 5 =^ = abyl + 1 y„+i = 2y2 + 1 (do a6 = 2)

Xet so th^tc /3 sao cho

Vay neu dat ^ = yi + y^yf+T thi

Tlieo nguyen If quy nap toan I I Q C suy ra

• Neu 12/11 < 1 thi ton t ^ i 0 sao cho cos(j> = j / i Khi do

2/1=4 cos^ (/) - 3 cos 0 = cos 3 0 , , 2/n = cos 3"0

• K h i I2/11 > 1 Xet so thyc l3 sao cho

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

do sii dung bai toan 14

Lifti y Phep doi bien trcnig bai toan 15 d traiig 15 dildc t h n ra nhu sau: T i t

h? thiic truy h6i ciia day ( x „ ) khien t a hen tucing den rong t h i k ludng giac

cos 3x = 4 CDS'* X — 3 cos x

Ta C O g^ng dyng cong thi'rc nay Gia sii x„ = 6y„ + c, k h i do

bVn+i + c = a {byn + c)^ - 3 (62/,, + c)

«>6y„+i + c = a {b'^yi + Sb^cy^ + 36c^2/„ + c^) - 3 (6y„ + c)

«>y„+i = abSi + 3o6cyf + Z{ac^ - l ) y „ + ^-'(ac^ - 4c)

Vay t a cho

a6^ = 4 3a6c = 0 3(ac2 - 1) = - 3 ' ( a c ^ - 4 c ) = 0

c = 0^

6 =

2

Do do t a dat x „ = —;=2/u- Tuy uhien c6 t h ^ t i m ra cong thiic doi bien nhanh

hon bSiig each dat x„ = byn, sau do t i m 6

B a i toan 16. Ttin {x,,},"^^, biet xi = a, x,i+i = axf, + 3x„, a > 0

-yi = 2 1 /

v

Bai toan 18 (De nghi thi OLYMPIC 30/04/1999) Xdc dmh so hang tong

qudt cm day so (u„) biet iting:

f u, = 2

\i = 9uf, + 3(t„,Vn = 1,2,

Hifdng dtn

2

Cach 1 Dat u„ = -Xu, khi do xi = 3 ; a;„+i = 4 ^ + 3x„, Vn = 1,2, Den

day, ta tien hanli luaug tu nlut bai toaii 16 d trang 15

each 2 Dat 3a„ ^ ^ { 1 ,,,, + 3,,^^ Chon xux^ sao cho { §| +

Bang quy nap, chiing niinh dUdc:

Vn = xf'~' + a f ,Vn = 1,2,

Lxiu y R6 rang each 1 se clio IcJi giai tu nhien va de nh6 hcJn ,16

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&

i;„+i - 1 = K - 1 ) ^ +3K - 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

Cach khac Xet phUdng trinh - 3x + 1 = 0 Phuong trinh n&,y e6 hai

nghiem xi, X2 va theo dinh ly Viet ta c6 xi +X2 = 3 vk X1X2 = 1 Ta se chiing

minh quy n?ip rang

Lifu y Phep dat t;„ = u„ + 1 (n = 1,2, ), d\X0c tim ra nhit sau: Xet h ^

so fix) = x^ + 3x2 _ 3 i^hi do u„+i = /(u„), Vn = 1, 2, Ta c6 /(x) la da thiic b§c 3 va

fix) = 3x2 ^ ^//(^) = 6x + 6 = 0 ^ X = -1

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

doi he true toa d6 theo cong thiie doi true sau: | Y = y + l ^

toan 19 ta phai d^it t;„ = u„ + 1 (n = 1,2, )

BM to4n 20 (De nghi thi OLYMPIC 30/04/2004) Cho day s6 (u„) nhu

sau:

ui = «„+i = 24u3 - 12V6u2 + i5u„ - 7 6 ( „ = 1, 2, )

Tin cong thiic so hg.ng tong qudt Un cua day so da cho

Hrfdng dan. D^t v„ = >/6u„ - 1 =i-vi = 2, v„+i = 4v^ + 3v„ (n = 1,2, )

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:

Hirdng dan 0^,1 x„ = y„ - 1 Ta thu diI0c day so (y„) nhu sau

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

Do cong thiic Ivr^ng gidc C O S 4 Q = 8 CDs'* Q - 8 cos^ a + 1 nen

2/2 = Scos" ^ - 8cos2 iL + 1 = cos ( 4 ^ )

y, = Scos" ( 4 ^ ) - 8cos2 (4.^^) + 1 = cos (42 .iL)

Gia sii y„ = cos Khi do

j/„,: = 8cos^ ( 4 - ^ ^ ) - 8cos2 ( 4 " - > ^ ) + 1 = cos ( 4 " ^ )

V^y theo nguyen h' quy n^p suy ra y„ = cos ( 4 " " ' • j ^ ) - Vn = 1,2, Do do

Theo nguySn l i quy n^p, suy ra y„ = ^ (a'^" ' + ^ 4 ^ ^ , Vn G N* Do do

B a i toan 24 (De t h i HSG T P Ho Chf M i n h nSni hoc 2011-2012) Cho day

so (u„) nhu sau : ui = ^ vd u„+i = — — , „,Vn e N* T i m so hmg

5 u * - 8 < -I- 8

long qudt cua day so da cho

G i 4 i D^t — =?;„ K h i d6 vi = 7 gia thiet suy ra

=>vn+i = 8t;^ - ivl -I-1, Vn = 1,2,

Xet so thi;c a sao cho

2a^ - 5a -I- 2 = 0

a = 2-1,

V$,y ngu dat a = 2 t h i J = i J^a + iY ^ = ^ Tudng t i ; n h u 15i giai bai

toan 23, t a chiing minh diMc

1 V - ^ ^ l = ^(2^"-"+2-^" ' ) , V n e N *

Do do un = ^ 4 n - i ^ " ^ 2 - 4 " - ' ' = 1' 2' • • •

L t f u y Ro rang, each giai dua tren cac hang d i n g thiic d ^ so (dU0c d}nh

hudng bcii cong thiic lit(?ng giac) n h u da trinh bay d tren la t u nhien, de hieu

hdn so v<3i vigc neu cong thiic roi chiing minh bang quy nap n h u d mpt so

ihi li?u kh4c

B a i toka 25 Cho day so (x„) nhu sau

XI = ^, x„+i = 16x1 - 20x1 + 5a;„, Vn = 1,2,

Tim so hQ.ng long qudt cua day so da cho. ^ - t i i ^ si>

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^ ^ + 5cos^ = c o s ^ ,

X3 = 16cos°-—- 20cos'^-—-|-5cos—= c o s — ' '

o 3 3 3

5"-17r Gia sur x„ = cos — - — , k h i do

= 16 cos^ ^ - 20 cos"-^ f + 5 c o s " - | = cos 5 ^

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

XI = -7, x„+i = 16x5 _ 20x^ -I- 5x„, Vn = 1 , 2 ,

Tim so hang tong qudt cua day so dd cho

G i a i Xet so thi/c a sao cho

1 f a + = -7 •«> a^ 4- 14a -H 1 = 0

2 V V?iy neu d^t a = v/48 - 7 t h i ^ = - v/48 - 7, ^ + ^ )

16 •x[a + -] = - a H 5a3 -MOa + — -I- -I- - r

Gia sur x„ = i (a^"-' + -l^r^, khi do

L 2 \

, ("1 / 5 „ - l 1 M 1 / gn 1 \

Theo nguyen h' quy n»p suy ra

Vay so hang tong quat cua day so da oho la

Bki todn 27 C/io ddj/ so (j/„) n/ii/ sau; j/i = a vd

y„+i = 180y^ - 48y^ + 5y„, Vn = 1,2,

71m so /i(in5 ton^ qudt cua day so da, cho

Htfdng d i n D^t y„ = - ^ x „ , Vn = 1,2, Thay vao (1) ta dvt0c

• Trudug h(?p |a| < Khi do |a\/3| < 1 D$,t a = arccos(av/3) Khi d6

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

26 d trang 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

2y„+i = 32y^ + 40^^ + iOx„, Vn = 1,2,

<^yn+i = 16y^ + 20y^ + 5y„, Vn = 1,2,

Xet so thyc a sao cho

• a = 3 + >/l0

L a = 3 - VIO

f a _ i ^ - 6a - 1 = 0 ^ V?iy neu d^t a = 3 + \/lO thi

l = - ( 3 - V Y o )

2 a j a Taco

1.1.5 P h i r c t a g p h a p h a m l a p

D6 t i m so hang t6ng qiiat cua day so (u„) bang phuong phap l^p ta thudng

tim cac ham so f{x) va h{x) sao cho

(*) / K ) = /i(/K_i))

Si'r dyng (•) lien tiep ta t h u dudc

/ ( U n ) = ft(/K_l))

= h{h (/(u„-2))) = h2 (/(«„_2)) = - - = K (/(«0)) (**) Tir {**) ta t i m dit0r Un- Ham so / dUdc gpi la ham so phu, con ham so h

dU0c gpi la ham lap Ta se bat dau bang hai bai toan rat ddn gian nhung da

the hicu kha day d i i phUdng phap

B k i t o 4 n 29 Tim so h(ing long qudt cua day so (x„) cho nhu sau:

^ 1 6

So - dU0c t i m ra n h u sau: Ta c6

6

Xn+\ fc = 7x„ - 1 - A: = 7 ( x „ - fc) + 6A; - 1

Ta can chpn /t sao cho 6A; - 1 = 0 A; = i V|iy t a se xet x„+i - ^ Co t h i

thay ngay r i n g day so da cho co dang x„+i = /(x„), trong do /(x) = 7x - 1,

va so - la nghi^m cua phudng trinh /(x) = x, hay noi each khac, so ^ Ik

diem bat dpng eiia ham so /

B a i t o d n 30 Cho day so (x„) nhu sau ; x j = 5; x„+i = 5f!L±i^ Vn 6 N '

x„ + 2

Chring minh ring vdi mgi n € N * thi Xnj^4 Tinh X2oi3

G i a i Ta CO x i = 5 ^ 4 Gia sur x„ jt 4, ta chiing minh x„+i ^ 4 Neu

Xn+i = 4 t h i ^"_^_2 = + 4 = 4x„ + 8 •«> x„ = 4, mau thuan vdi gia

thiet quy n?ip V^y x„ ^ 4, Vn e N * Ta c6

Vay nen trpng Icri giai tren ta da xet x„ - 4 va x„ + 1

B a i t o d n 3 1 T i m so h^ng tSng qudt cua day so (u„) cho nhU sau:

ui = 4 , tt„+i = i (u„ + 4 + 4^1 + 2u„) , Vn = 1,2,

G i a i Theo gia thiet ta c6 u„ > 0 vdi mpi n va:

9u„+i = u„ + 4 + 4v^l + 2un

<:»18u„+i = 2u„ + 8 + Sv'l + 2u„

•«>18u„+i + 9 = 2u„ + 1 + 8\/H- 2iin + 16

«.9(2u„+i + l ) = ( v T + 2 ^ + 4)^

<:»3V'2u„+i + 1 = v'2u„ + 1 + 4

D«Lt Vn = v/2u„ + 1 K h i do v i = 3 va ttt (*), ta c6:

1 4 3t;„+i = v„ + 4 i;„+i = 3^^" + 3

(*)

5ou «fdj/ <a si xdy difing mot so Idp ham, de tv[ dd tao ro cdc bai toan xdc dinh

so h(mg tSng qudt cua day so bdng phuang phdp ham l^p

V i 1 CApn c = 7, A;/ii rfd a = - 1 4 , b = 56 Ta dMc bai todn sou

B a i todn 32 CAo ddj/ 5 0 (x„) nhu sau: x i = « 6 R m

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,

«.u„+i = 3u2 + 4u„ + ^, Vn = 1,2,

Ta dxtdc bai todn sau

B a i toan 33 Cho day so (u„) nhu sau: ui = a eR vd

u„+i = 3u2 + 4ii„ + V n = 1,2,

o Lvru y Phep dat u„ = ^ ditdc tim ra nhif sau: T a d?it u„ = A;x„, vdi A; la hang so se tim sau Khi d6

I kxn+i = 3ifc2x2 + 4A;x„ + - =• xn+i = 3A;x2 + 4x„ + — 9 2

27

V$,y ta t i m k thoa man 3k = 1 =^ k = ^ Un = —

Ta tiip tuc lam kho hon bdng each dot bien x „ = 1 - 5u„ Khi do dUffc day

so (u„) thoa man diiu ki§n

1 - 5u„+i = (1 - 5w„)2 - 6 (1 - 5u„) + 12

«>1 - 5w„+i = 25ul + 20Un + 7^ Un+l = -5ul -4un

To dt^c'c oai toan sau *' ' - :!

B a i t o a n 34 Cho day so (u„) nhu sau: ui = a eR vd

It Tim so hang tdng quat cua day so da cho

G i a i D a t u„ = i x „ K h i do ' '

-5

- g X n+ i = - - x ^ + - x „ - - <^x„+i = x 2 - 4 x „ + 6 < ^ x „ + i - 2 = ( x „ - 2f

VHy x„ - 2 = (x„_i - 2f = (x„_2 - 2f = • • • = ( x i - 2)2'"' S6 h?ing t6ng

quat ciia day so da cho la

V i d u 4 C/ipn d = 1, khi do 6 = 3, a = - 3 , c = 0, to dU(7c day so (x„) t/ioo

mdn /ip thiic truy hoi

x„+i = x^ - 3x2 ^ 3^^^ Vn G N* ^

7\iy n/iien day so (x„) cho nhu vay "khd Id ", ta c6 the "ddu bdt di" bang each

doi bien x„ = 5u„ Khi do day (u„) thoa mdn h$ thiic truy hSi

5u„+i = 125u^ - 3.25u2 + 3.5u„ ^ u„+i = 25ul - 15^2 + 3 u „

Ta CO bai toan sau

29

Bki to&n 3 5 Cho day so (u„) nhu sau: ui = a e R ud

u„+i = 25u3 - \bul + 3u„, Vn e N*

Tim so hg.ng tong qudt cua day so da cho

Gi&i Dat Un = -^.Ta ditoc day so (x„) thoa man di6u kien xy = 5a

5

^X„+l = - + |x„ ^ Xn+1 =xl- 3x2 ^ g^,^

X „ + i - 1 = ( X „ - 1)^

W^y vdi mpi so nguyen dudng n ta c6

x„ - 1 = (x„_i - 1)3 = ( x „ _ 2 - 1 ) 3 ' = = (xi - if"' = (5Q - 1)3"

So hgng t6ng quat cua day so da cho la

Ta d?/(rc 6ai todn sau

Bai toan 36 Tim so h(ing tSng qudt cua day so (x„) cho nhu sau:

/(x„) ^ 2 - Xn _ /

2 + x„ V 22 + x„_iy - x„_i = [/(X„_,)]2 = [/(X„_2)]

|2>

(3) DP/3 = [ / ( a ) r " ' - T i i (3) ta c6

2 - x 2-2/3

= l3^2-x„ = 2P + 0Xn^x„ = y ^ V?iy neu a = -2 thi x„ = - 2 , Vn = 1,2, , neu Q ^ - 2 thl

2 n —1

x„ =

(2 + a )

Vn = l , 2 ,

Vf du 6 TVonp bai todn 36, <a do'i bien x„ = 3u„-l Khi do dUtfc day so

(un) thoa man

3un+i - 1 = 4 + ( 3 u r - l ) ^ 9 u 2 - 6 u „ + 5 8 (3u„ - 1) _ 24u„ - 8

24un - 8 , , 9ti2 + 18un - 3

Ta chpn a, b, c, d sao cho

- bdx^ + ax- cd= {x- df

=^x^ - bdx"^ + ax-cd = x^ - Sdx"^ + Zd^x - d^

6 = 3 ^ 3d2

( bd = Zd r 6 = 3

{ a^3d^ ^ a = 3d'

\ = d^ { C = (P

V i d u 7 I a = 3d2 chpn c = 2 ta duac 6 = 3 , a = 6, d = ±y/2 Ta c6

bai todn sau

B a i t o a n 37 Tim so h(ji.ng tong qudt cua day so (x„) cho nhu sau:

bai todn sau

B a i t o a n 38 Tim so hg,ng tong qudt cua day so (x„) cho nhu sau:

V i d u 9 Trong bai todn 38, doi bien x„ = u„ - 1 ia duac day so (u„) </ida

man dicu kien

I s _ _ X* + ax^ + 6 _ a:'* - cex^ + Q X^ - dex + b

Ta can chpn a, 6, c, d, e sao cho

x"* - cex^ + ax^ - dex + 6 = (x - e)"*

=»x'* - cex^ + ax^ - rfex + 6 = x'' - 4ex3 + Ge^x^ - 4e3x + e''

B k i todn 40 71m s6 hang tdng qudt cua day so (u„) cho nhu sau:

G i a i De thay, vdi mpi so nguyen duong n t h i luon ton tai u„ Neu a = thi u„ = - i Vn = 1,2, Tiep theo xet a / Ta c6

32u3 + Sun 16u^ + 32u3 + 24u^ + 8 U n + 1

ZUn+l + - ^g^4 ^ 24u2 + 1 + ^ 1 6 < + 24u2 + 1

{2Un + 1) 4 16u4 + 24u2 + 1 •

2x + 1 ^"

= - - = - ( / ( « i ) r " = - [ / ( « ) r - '

Tff (3) ta dUdc

^"" + 1 2 + 2 [ / ( a ) ] ' ' " - ' V^y n l u a = - 1 t h i u„ = Vn = 1,2, , c6n n l u a / - 1 t h i

(3)

= \2a + iy

B k i t o d n 4 1 Cho day so ( i „ ) nAi/ sau a;„+i = ^ x ^ , V n = 1 , 2 , 71m s6

hQng long qudt cua day so da cho

H i r d n g d i n Vdi moi n = 1,2, ta c6

^ 2 l + 2 ' + 2 2 = • • • = 2 l + 2 ' + 2 2 + - + ^ " 2"^ ' ^ 1 •

B a i t o a n 42 Tzm {x„}+^i hiet

Xi = a,x„+i = x \ 2a^",Vn G N*,a > 0,a 7"^ 1

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

toan 11 6 trang 11

B a i t o a n 43 Ttm {xn)t=x biet

X I = a,xn+i = 2a2"x2 - «("+i)2",Vn e N % a > 0

H u d n g d i n Dat 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

B a i t o a n 4 4 Tim so h^ng ting qudt cua day so (x„) cho nhu sau

Den day t a lam tuong t u nhu bai toan 15 6 trang 15

C h u y 5 Trong bdi todn 45 d trang 37, phep ddt

Theo bdi todn 44 ta c6 u„ = 2^ ^"u^" Vdy dat x„ = 2^~^''u§"j/„, t/iay vdo

(1), ta duac 2 '-^"^\r\^+, = (2»-3"urj/„)' - 3a3"x„, suy ra

Tom /at, ta ddt x„ = 2^-3" (ly/E)^" y„ = 2 ( v ^ ^ " y„ iTAz do

Vn+i =4y3 -3y„,Vn = 0 , l , 2 ,

B a i todn 46 (Olympic toka Sinh vien tokn qu6c-2010) Cho { u „ } , { v n } ,

{wn} la cdc day so dUOc xdc dinh bdi UQ = VQ = WQ = \

t;„+i = - 2 u „ - 8i;„ + 6tz;„ (2)

u;„+i = - A u n - 16v„ + 12t/;„ (3)

Chiing minh rhng Vn -2 chia hit cho 2"

H i r d n g d i n TCr (3) va (2) ta dUdc u;„+i = 2vn+i. (4)

Lay (2) t r i l (1) theo ve ta dUdc i;„+i - u„+i = - u „ - V n + u»„ (5)

Thay (4) vao (5) dudc

Vn+l - " n + l = - U n - t^n + U^n = — " n - + 2Vn |

= ti„ - U„ = • • • = t;2 - 1^2 = - U l - Hi + IWl = - 1

=^ u„+i = i;„+i + 1 (6) Tir (6) va (4) suy ra u;„+i = 2i;„+i = 2u„+i - 2 (7)

Thay (6) va (7) vao (2) dUdc

v„+i = - 2 u „ - 8i;„ + 6t/;„ = - 2 (t;„ + 1) - 8t;„ + 12v„

Hay Un+i = 2vn - 2 <=> i)„+i - 2 = 2 (t;„ - 2) Sau do diing quy nap hoic t i m

cong thiic c y the cua t;„+i - 2, chiing minh diTdc yeu cau bai toan

B a i toan 47 (HSG Qu6c gia-2010) Cho day so thrfc (a„) xdc dinh bdi a i = 5

vd

a „ = ^alZ\ 2 " - i + 2 3 " - i , Vn = 2 , 3 , a) T i m so /lan^ tong qudt cua day so (an)

b) Chang minh rhng (on) Id day so gidm '

V^y a„ > a„+i, Vn = 2 , 3 , Hay (a„) la day so giam

N h a n xet 1 Vi^c xdy dxjcng nen nhOng day so c6 he thiic truy hoi nhu d bdi

toan vita roi khd dan gidn Ch&ng han xet day so x „ = v^5" + 7", Vn = 1,2,

Khi do

Xn+i = " V 5 " + i + 7"+i = " V S (5" + 7") + 2.7" = "^^^x^ + 2.7"

Ta diCdc bdi toan sau

B a i toan 48 Cho day so (x„) nhu sau:

x i = 12, a;„+i = "+y5x^T2J", Vn = 1,2,

Xdc dinh so hang tong qudt cua day so da cho

B a i toan 49 (Olympic toan Trung Qu6c-2005) Cho day so (a„) nhu sau:

d = a2 = 1 vd o„+2 = 1- On, Vn = 1,2, Tim

02004-G i a i Ta c6 a „ + 2 a „ + i = a „ + i a „ + 1, Vn = 1,2, Do do

0302 = 0201 + 1

0403 = 0302 + 1

=>a„+ia„ = 0201 + n - 1 On+ian = n => a„+i = — , Vn = 1,2,

Bdi v^y a„+2 = = ^ ^ O n , Vn = 1 , 2 , T i l do

B a i toan 50 Cho day so (a„) xdc dinh nhu sau: oo = 1, o i = 4 va

a„+2 = 4a„+i - a„, Vn = 0 , 1 , 2 , ^

a) Tim so hang tong qudt cua day so dd cho j

_6) Chiing minh + a^_i - 4a„a„_i = 1, Vn = 1,2,

Ta CO dieu phai chiing minh

Ltfu y Vi§c t i m so hang tdng quat ciia day t r u y hoi tuyen t i n h cap hai

aXn+2 + bXn+l + C X „ = 0

se dUdc giai quyet oi muc 1.3.2 (6 trang 65) T u y nhien, dung phiWng phap

lap thong qua cap so nlian cho ta Idi giai so cap hdn, va qua do ta con t i m

ra difcJc t m h chat cua day so de dang hdn

B a i t o a n 51 Cho day so (a„) xdc dinh bdi oo = 1, a i = - 1 va

an +2 = 7ar^+l - 6a„, Vn e N a) Tim so hang tSng quat cua day so da, cho

b) Chitng mink - 7a„+ia„ + 60^ = 14.6", Vn e N ,|

c) Chiing minh an /?10 vdi mgi n G N

\„ - 6 a „ _ i = - 7

(an+i - a„) (a„+i - 6a„) = 14.6" ^ al+i - 7a„+ia„ + 6al = 14.6"(n G N ) :

Ta CO dieu phai chiing m i n h

c) Ta CO Oo va oi la nhQng so le Gia svt an va o„+i le K h i do v i yUnO'

o-n+2 = 7a„+i - 6a„ ' '

nen an+2 cung la so le Theo nguyen 11 quy n9,p suy ra o„ la so le vdi mpi

V i V9,y, kot hdp v6i (2) t a diMc 5(02012 + 13):2011 M a (2011,5) = 1 nen

(a2oi2 +13)^2011

Lxiu y Dinh l i Fermat nho ditoc phat bidu nhit sau: N6u p la s6 nguy§n to

va a khong chia het cho p t h i o^"' = l(modp)

1.2 Mot so litig dung cua sai phan

Viec t i m so hang tong quat ciia day so, nhieu khi diTdc quy ve viec tinh mOt

toiig nao do Sau day ta se trinh bay mot so kien thiic ve sai phan de t i i do

CO mgt cai nhin t6ng quat k h i tinh t5ng

D i n h nghia 2 Cho Ux la mpt ham theo x Khi dd A u j : = Ux+i - Ux dMc

gpi Id sai phan cap 1 cua ham

Ux-D i n h n g h i a 3 Neu Eux = Ux+i thi E duac goi la todn til dich chuyin

D i n h n g h i a 4. A ^ ' u j = A {iST'-'^Ux), E'^Ux = E {E"'-'^Ux) Ta goi Ian liCfH

gQi A ^ U i , E'^Ux Id sai phan cap m, todn tv! dich chuyin cap m cua

Ux-N h a n xet 2. [ A = E vdi l u x = Ux Vay

A"^ux = { E - ir Ux = Y^ {-iyci,E^-'ux (vdi E° = I )

t = 0 i

1.2.2 T i n h c h a t

(1) A C = 0 (vdi C la h^ng so)

(2) A (ux ± Vx) = Aux ± Awx

(3) A [kux] = kAux (vdi k la hang so)

(4) A (UxVx) = UxAVx +

Vx+lAUx-(5) Sai phan cap i cua ham so Ux la mpt toan t i i tuyen tinh

(6) Sai phan moi cap dcu c6 the bifiu dien theo gia t r i cua ham

1 D i n h nghia. Gia sii Ai;x = Ux A (i;x + C) = Avx = Ux- K h i dovx + C

gpi la tich phan bat djnh cua Ux, ky hi?u la A " ^ U x V$.y

Vi du Ta CO i )(3) \) 3

2 Mot so tinh chat

(1) ( u x ± v^) = A - i u x ± A - i i ; ^ + C

(2) A - ' (fcux) = fcA-'ux + <7 (v6i k la hang so)

(3) A - i ( u x A t ; ^ ) = U x t ^ x - A - ^ ( i i x + i A u x j + C (giong nhu cong thiJc tich phSn

2.sin- \ y

(5) A-icos(a + 6x) =

2 sin

1 Sin a - - + 6x + C 2 ) (6) Ta CO

0(71 H~ 1) A-> (x(x + l)(x + 2)) = A-^x + 2)W = (£±2)W _^ ^

4

1.2.4 Phi/dng phap tich phan tfiftig ph5n

Ta sii dung cong thiic

A-i (uxAt/x) = U x U x - A-1 (ox+iAux) + C

Bai toan 52 Tern tich phdn bat dmh cua /(x) = x3^

Bai todn 54 Tim tich phdn bat dmh cua f{x) = (j ^ + 2)

Hirdng dan Ta can tinh A'^^ ^^-n^- ^^t =? di

(x + l)(x + 2)

x2- _ /(x)2^

8 (6)

14 (6)

22 12 26

N h a n xet 3 Tic bdi todn nay, ta di ddng tinh dU(?c cdc tong sau day

5„ = sinx + sin2x 4- h sinnx; Q„ = sinx + sin3x H h sin(2n +

1.3 X a c d i n h day so b a n g phLfdng p h a p sai p h a n

• Cho day so {x„ Xet phuong trinh ')

aoxn+k + aiXn+k-i + ••• + akXn = gin) (1)

tiong dog{n) la haniso theo n, vkao,ai, ,a* lacac h§ngso K h i do phudng trinh

aoXn+k + aiXn+k-i H + akXn = 0 (2)

dvrdc gpi la phitdng t r i n h thuan nhat tUdng ling vdi phifdng t r i n h (1) C6n phitong t r i n h (an la A)

aoA^ + a i A ' ' " ' + • • • + = 0 (3) gpi la phudng trinh d i e triTng cua (1), dong thdi (3) cung diidc gpi la phUdng trinh d i e t n t n g cua (2)

• Nghifm tong quat cua phudng trinh (1) se c6 d^ng:

2;n = ^ + < , V n = 1,2,

trong do x;;^ la nghi§m tong quat cua phUdng trinh (2), c6n x* la mpt nghi^m rieng bat ky ciia phUdng trinh (1)

• Dvta tren mpt so ket qua ciia ly thuyet phUdng trinh sai phan t a suy ra

dupe mpt so ket qua sau day:

[i) Neu (3) CO A; nghi^m thifc phan bi?t Ai, A 2 , , Afc t h i (2) c6 nghi^m tong

quat la

Xn = ciA? + C2AJ + • • • + CfcA^, Vn = 1,2,

(vdi ci,c2, ,ck la cac h k i g so ) K h i biet x i , X2, • •, Xfc t a se t u n dupc cy the cac hang so c i , C 2 , , Cfc. K h i do

x„ = ciA? + C2A5 + • • • + CfcA?, Vn = 1,2, '

49

(vdi c i , C 2 , , Cfc l a cac hling so vita, t i m dit0c 6 treu) goi la m o t nghiem rieug

cua (2)

( i t ) Neu (3) dU0c viet lai n h u sau

aoA* + a i A ' ^ - i + • • • + Cfc = OQ (A - Ai)* (A - A2)'' (A - A 3 ) (A - A,) = 0,

vdi cac A i , A2, A 3 , , A, la khac nhau doi mot Tiic la (3) c6 Ai la nghi^m bpi

s, va A2 la nghi^m bgi h, va A 3 , , A, la cac nghi^m ddn, va s+h+{q-2) = A:,

t h i (2) CO nghi§m tong quat 1^

Xn = C3A? + • • • + c,AJ + ( c i i + cian + • • + cun''^) A?

+ ( c 2 i + C22n + • • • + C2/,n''~^) AJ, Vn = 1 , 2 ,

(vdi c u , C 1 2 , • •, c i j , C 2 1 , C 2 2 , , 02h, C3, ,Cq Ik hang so)

(Hi) Neu (3) c6 s nghifm thi;c phan bi?t A i , A2, , A, v&, '

Xq = a + bi = r (cos (i> + i sin(/>) (vdi r = | A , | = y + b'^, 4> = ArgXq)

la nghi?m phiJc b p i h, t h i so phufc lien h(?p \ a - fci = r (cos0 - i s i n 0 )

cung l a n g h i f m phiic bpi h cua (3) K h i do (2) c6 nghiem tSng quat la

TVong d 6 c dU0c t i m tijt i i = Q

C h u y 6. Neu trong gid thiet khong c6 xi = a, nghia Id x i tuy y thi ta khong

can tim c, nghia la c Id mgt hdng so tuy y vd x „ = cA", Vn = 1 , 2 , ggi Id

nghi$m tong quat cua (*)

B a i t o a n 6 0 Tim tat cd cdc day so (x„) thoa man dieu ki$n:

• Neu X j t l t h i x ^ la da thiic cung bac vdi da thiic / „

• Neu A = 1 t h i x ^ = n.gn, vdi 5„ la da thiic cung bac vdi da thiic / „

Thay x ^ vao phUdng t r i n h , ( i ) roi dong nhat he so t a t i m dUdc cac he so cua

x ; T i r x „ = x;: + x ; , Vn G N * lay n = 1 roi sii dyng x i = a t a t i m dupe c

C h u y 7. Neu trong gid thiet khong c6 x i = a, nghia Id x i tiiy y thi ta khong cdn tirh c, nghia la c Id mgt hdng so tiiy yvdx„ = cA" + x ; , Vn = 1 , 2 , ggi

la nghiem tSng qudt cua (i) i

B M tokn 62 Ttm day so (x„) biet ( ^ ^ ~ ^ o , 2 _ L I W

^ ' I a;„+i = 3 i „ + n'' + 1, Vn = 1, 2 ,

G i a i PhUdng t r i n h d i e trUng A - 3 = 0 - « > A = 3 V^y x„ = c.3" + x ; , vdi

x ; = an^ + ;9n + 7 (a 0), x ; 111 nghi^m rieng cua phudng t r i n h

x„+i = 3 x „ + n2 + l

Suy ra

a{n + 1)2 + /3(„ + 1) + ^ = 3(an2 + + ^) + „2 ^ 1

<^an'^ + (2a + /?)n + a + /? + 7 = (3a + l)n2 + 3/3n + 3 7 + 1

Dong nhat he so t a duoc

la da thuTc bac k theo n

C a c h giai Giai phuong trinh dac trUng aA + 6 = 0 t i m dvfdc A Ta c6

Trong do x^T = cA", Vn = 1,2, (c la hSing so se t i m sau) Con x ; \h mOt

nghiem rieng bat k i ciia ax„+i + 6x„ = PA:(n).7",Vn = 1,2, va ducjc xac dinh nhir sau:

• Neu A # 7 t h i x ; = Qkin).'r", v6i Qfc(n) la da thiifc bac k theo n

• Neu A = 7 thi x ; = nQfe(n).7", vdi Qfc(n) la da thiic bac k theo n

Thay x^ vao ax„+i + 6x„ = Pfc(n).7", roi d6ng nhat he so t a t i m dUdc x ;

T i t x„ = x;r + x ; , Vn = 1,2, , lay n = 1 roi suT dyng x i = a, t i m du^c c

B a i toan 66 Ttm day { x n j + ^ i biet x i = 2010, x„+i = 7x„ + 7 " + \ V n e N*

G i a i T i t gia thiet t a c6 v '

x i - 2 0 1 0 , x„+i = 7 x „ + 7 7 " , V n = 1,2, ' ' ' ' ' ' Phuong t r i n h dac trimg A - 7 = 0 c6 nghiem la A = 7 Vay so h^ng t6ng quat ciia day so da cho co dang

V$,y x ; = n7", Vn € N* Suy ra a;„ = c7" + n7",Vn G N* Ma x i = 2010 nen

So hgiig t6Hg quat cua day so da cho la

Bai toan 68 Tim day so {x„}^^ thoa man dieu ki$n

Bai tokn 70 Tfn/i fon^ 5„ = 1 + 4a + Oa^ H + n^a"-!, t;(?i a ^ 1

Giai 5„+i = 1 + 4a + 9a2 + + n2a"-i + (n + l ) 2 a " = 5 „ + (n + 1)20"

V^y day so (Sn) thoa man: 5 i = 1 va „ ,i:

5 „+i = 5 „ + (n + l)2a", Vn = 1,2,

Do do

5 „ = C + (an2 + 6n + c)a", Vn = 1,2,

Vi 5 ; = (an^ + 6n + C)Q" la nghiem rieng cua ( 1 ) nen

a(n + if + 6(n + 1 ) + c] a"+i = (an^ + 6n + c)a" + (n + 1)20"

C a c h giai Giai phUdng trinh d?Lc tntng aA + 6 = 0 t i m dvt(?c A Ta c6

Trong do x^T = cA", Vn = 1,2, (c Ik hang so se t i m sau) Con x ' du<?c xac

5^^j = cosx + cos2x H h cosnx + cos(n + l ) x = Sn + cos(n + l ) x

Vay day so (5„) thoa man 5 i = cosx va

5„ = 5„_i + cosnx, Vn = 2 , 3 , Phirdng trinh dftc t n m g la A = 1 Do do

Sn = C ( l ) " + >4sinnx + Bcosnx = C + >lsinnx + Bcosnx, Vn = 1 , 2 ,

Yi s* = y l s i n n x + 5 c o s n x la nghi?m rieng ciia (1) nen v6i mpi n = 2,3,

ta CO ylsinnx + Bcosnx = i4sin(n — l ) x + Bcos{n — l ) x + cosnx, hay

A sin nx + B cos nx = J4 (sin nx cos x — cos n x sin x)

+ B (cos nx cos X + sin nx sin x) + cos nx

So sanh h$ so ciia sin nx, cos nx a hai ve t a dirpc

2sin-sin(nx + - ) Vay 5„ = C + — Vn = 1,2, Do 5 i = cosx nen

2 s i n 3x 3x sin — sin —

L i A i y Qua cac bai toan 63, 70, 73, ta thay rang phUdng trinh sai phan cho

ta mot phitdng phap nhat quan dS: tinh t6ng

Dang 5 T i m day so (x„) bilt: aji = a va \

1^ ngb' !in rieng ciia

aXn+l + bXn = fin, aXn+1 + 6x„ = / 2 „

Bai toan 74 Tim x„ biet

ui = 1, Un+i = 2u„ + + 2.2", n e N *

Giai* Phitdng trinh d?lc t n m g A - 2 = 0 c6 nghi^m A = 2 Ta c6

trong do

u;r = C.2", u^=an^ + bn + c, u" = A.n.2"

Thay < vao u„+i = 2u„ + n^, ta dUdc

V$,y u„ = 5.2"-i - n2 - 2n - 3 + n.2", Vn = 1,2,

Dang 6 T i m day so (x„) biet: xi = a va

1 -{U'f'

aXn+l + bXn = / i n + / 2 n H H fan, Vn = 1 , 2,

trong do a, a, b la cac hkng so cho trifdc

Cach giai Giai phUdng trinh dSc trifng aA + 6 = 0 t i m diI0c A Ta c6

a^n = a;„ + + X2„ + • • • + x*„

Trong do x;: = cA", Vn = 1,2, (c la hang so se t i m sau), xl^ (A; = 1,2, , s)

1^ nghi§m rieng ciia

ax„+i + 6x„ =

fkn-Sau day ta se trinh bay mot phitdng phap t5ng quat hdn, goi la phUdng phap '^ien thien hang so, phitdng phap nay ditdc the hien qua dang 7 sau day

rieng c i i a x „ + i = x „ + s i n n x , V n = 1 , 2 , , v a x * dU(?c t i m n h u sau: T a c6

the x e m c l a niQt h a m theo n va t i m x * = c„ T h a y x * = c„ vao (*) t a dildc

Ac„ = s i n n x c„ = sinna; = cos (~ + nx)

B a i t o a n 7 6 Ttm day so { x „ } + f ° i thoa man

ma Vn+i = At'n + Vn lien

Zn+i (Avn + Vn) - a„z„t;„ = 6„ <=> {Zn+l - a„z„) Vn + z„+i A D „ = 6„ (*)

Gia suf Zn la mot nghiom rieng cua phitrtng trinh thuan nhat, khi do theo

trinh thuan nhat Tiep theo ta t i m rapt nghiom rieng ciia phudng trinh khong

thuan nhat Ta xem x„ = 1 2 (n - 1)C„, Vn = 1 , 2 , , thay vao phUdng

trinh khong thuan nhat ta dUdc

1 2 n C „ + i = 1 2 n C „ + n.n!

^Cn+i -Cn = n<^ ACn=n<=>Cn = A'^n = A-^n^^)

Vay x ; = 1.2 ( n - 1 ) " ^ " ^ = n ! V n = 1 , 2 , la mpt nghi?m rieng

cvia phiwiig trinh khong thuan nhat Do do

Vay ta chi can chpn / ( n ) = a (a la hang so) Thay vao (*), ta ditdc

'2a - (n + l)a — ii-l=^a — an = n — l=>a = —1

trong do A vk B diWc xac dinh khi biet x i , X 2

• Ngu Ai, A2 la hai nghiem thuc kep, Ai = A2 = A thi vdi mpi n e N * ta c6

x„ = {A + Bn)\'',

trong do A vh B ditdc xac diiih t\t xi = a, X2 =

P-• Neu A la nghiem phiic, X = x + iy, t h i dua ve dang litdng giac

A = r(cos()fi + i sin <;/))

K h i do Xn = r " ( ^ cos n4>+B sin ncp), v6i A, B diTdc xac dinh tijt xi = a,X2 = (3

C h u y 8 Neu trong gid thiet khong cd xi = a,X2 = P, nghta Id xi,X2 tuy y

thi ta khong can tim A, B, nghla la A, B Id cdc hhng so tuy y vd

Xn = AX1 + BX^;Xn = {A + Bn)X'';Xn = r''{Acosn<f) + Bsmn<p)

goi Id nghiem tong qudt cua (i) Ung vdi cdc trudng hop noi tren

B a i toan 79 Tim x„ biet

a;i = 0, a;2 = 1, Xn+i - Xn - 2a;„_i = 0, Vn = 2,3,

G i a i Philrtng trinh dac trmig

\Un+l = OUn - 9u„_i

G i a i Phvtdng trinh dac tnrrig A;^ - 6fc + 9 = 0 c6 nghiem kep fc = 3 nen cong

thiic t6ng quat ciia day so c6 d^ng u„ = (a + 6n).3" Do U Q = u i = 1 nen

Vay so hang tong quat cua day so (u„) la u„ = 3" - 2n.3"~S Vn e N

B a i toan 81 Cho day so (u„) nhu sau:

= 4 (C,l+i9" + C^+i9"-2.80 + C^+i9"-^802 + • • •)Y • ,

V i vdi niQi so ttt nhien n t h i

4 (Ci+i9" + C^+i9"-2.80 + C^+i9"-^802 + • • •)

la so nguyeii nen suy ra dieu phai chuTng minh. h^iX^'"

B a i toan 82 Cho f :N* —>R thoa man dieu kien

fin + 2) = fin + 1) - fin), /(I) = 1, /(2) = 0

Oft?

Chiing minh rang \ \< — - , V n 6 N*

Giai Dat / ( n ) = t/„, ta duqc

Un+2 = Un+1 - U„, Ui = / ( I ) = 1, U2 = f{2) = 0

Phudng trinh dSc trung - A + 1 = 0 c6 nghi?m phiic la

Ta CO A, = cos ^ + i sin | Suy r a / ( n ) = ^ cos ^ + B sin ^ Ta c6

axj + bxi + c = 0, axj + bx2 + c = 0

^ax1+' + 6^'/+' + cx'i = 0, ax2"+2 + ^^n+i ^ ^ ^

- a ( x r ^ + x ^ ^ ) + 6 (xr' + ' ) + c (x? + x^")

= ( a x ^ ' + 6x^+1 + cx?) + ( a x ^ ' + 6x^+i + CX2") = 0

B a i toan 84 Chiing minh r&ng vdi moi n e N thi

chia hit cho 5 ( H hieu [x] c/it p/idn nguyen cua s6 x )

[ S2k = (-l)'=5o (mods) S2k = (-1)*.2 (modS)

\+i = ( - l ) ^ 5 i (mods) ^ \+i = (-l)*=.10 = 0 (modS)

Ma [(S + 2^6)"] = 5„ - 1 nen [(S + 2\/6)"] = 2.(-l)'= - 1 (modS) ho$c

(5 + 2V6y = - l ( m o d S )

suy ra [(s -,-2^6)''] khong chia het cho 5

Bai toan 85 Chiing minh [(2 + 73)"] la so le vdi mpi n e N Bai toan 86 ChUng minh hieu thiic 5„ = (9 + 4\/5)" + (9 - 4v/5)" nhan

gid tri nguyen vd khong chia hit cho 17 vdi mgi n e N

Hudng dan Ta chilfng minh dUcJc SQ = 2, S I = 18, 5„+2 = 185„+i - 5„

Suy ra 5„ € Z Lai c6

Sn+3 + Sn = 185„+2 - 5„+i + 5„ = 17(5„+2 + 5„+i);i7

Ta CO 5o = 2 khong chia het cho 17 nen cung khong chia het cho 17 Ta

CO Si = 18 khong chia het cho 17 nen Szk+i ciing khong chia het cho 17 Ta

CO S2 = 322 khong chia h i t cho 17 nen 53^+2 cung khong chia het cho 17

V§,y Sn khong chia het cho 17 vdi moi n 6 N

69

Bai toan 87 Tinh USCLN {16" + 10" - l|n € N ' } , tUc la so nguyen Idn

n/ia< (5 > 1 sao c/io Vn e N*, (5| (16" + 10" - 1)

Giai Vdi moi n G N*, dat u„ = 16" + 10" - 1, t;„ = u„ + 1 = 16" + 10" va

(J = USCLN{u„|n€N*}

De daiig kilm tra dit0c day so (t;„) thog, man h§ thiic truy hoi

Vn+2 - (16 + 10)t;„+i + 16.10v„ = 0, Vn G N ' TiJt do

u„+2 = 26u„+i - 160u„ - 135, Vn G N* (1)

Vi (5|u„, (J|u„+i, 6\un+2 nen tiit (1) suy ra <J|135 M^t khac 5\ui, tiic la (J|25,

suy ra

(5|(135,25)=*(J|5=i>(5 = 5

Bai todn 88 (Chpn dpi tuyen Vi?t Nam thi toan Quoc te nam 2012) Cho

day so (x„) gdm vd hq,n cdc so nguyen duang dU0c xdc dmh bdi

f x i = 1, X2 = 2011

1 Xn+2 = 4622x„+i - x„, Vn G N*

Chring minh rdng ^^2012^ ^ c/iin/i phuang

Giai De ddn gian lidn trong vigc dung ki hiqu va hieu ro ban cliat van de, ta

se phat bieu va chiing minh bai toan tpng quat nhu sau: Cho p Id so nguyen

duang le Idn han 1 Xet day so (x„) gdm v6 h(f,n cdc so nguyen duang dupc

xdc dinh bdi

I X i = x„+2 = 2px„+i - x„, Vn G 1, X2 = p

Chang minh rdng ^^"""^"^ ^ Id so chinh phuong

PhUdng trinh d$c trimg cua day so da chp la

t = h = p + y/p^ - 1 V$.y sp h?ing t6ng qu^t cua day (x„) cp d?ing

t=0

N h u v^y, ta c6 + q = Ny/2(J+1) Ta t h u dU0c

p + 1 2(p + l ) 2 ( p + l ) T6m l ^ i : ^^"^^"^ ^ = la so chi'nh phuoiig Bai todn chiing minh xong

Lmi y Dang thiic ^ ( - l ) ' t 2 ^ ^ = (titrtr chiing minh

nhu sau: dat j = (p - 1) - i, kh i do

va la mOt nghiem rieng tuy y ciia phiXdng trinh khong thuan nhat

ox„+i + bx„ + cxn-i = f„ (/„ la da thiic khac da thiic khong)

Theo dang 9 d trang 65 ta tim dudc x^T, trong do cac he s6A,Bse dUdc tim

sau Con x* dUdc xac dinh n h u sau:

- M ^ i i 1

, Meu 1 khong la nghiem ciia aX'^ + hX + c = 0 t h i x^ la da thi'tc cung bac /„

]• Neu 1 la nghiem ddn cua aA^ + 6A + c = 0 t h i x*„ = np„, vdi 5„ la da thiic

rung bac vdi da thiic /„

, Neu 1 la nghiem kep ciia aA^ + 6A + c = 0 t h i x ; = n^gn, vdi 5„ la da thiic

j^^diig bac vdi da tlnic/„ ^ ^ ^

Thay x ; vao rtx„+i + 6 x „ + cx„_i = /„ roi dong nhat cac ho so (hoac Ian htrtt

lliy n ="1, n = 2, n = 3) ta tinh dUdc cac he so ciia x^ B i l t x i , X2, tit h§ thiic

B a i toan 89 Tim tat cd cac day so {x„};^f°i thoa man xi = - 4 , X 2 = 3 va

6x„ + x„+i - x„ + 2 = n + 1, Vn = 1,2, (1)

Giai Phiidng t r i n h dac trilng -A^ + A + 6 = 0 A = 3.^ nghiem

t6ng quat ciia phudng t r i n h thuan nhat tUdng ling la

G i a i Phirong trinh dac triftig + A + 1 = 0 c6 hai nghiem 1^

2 2 2 2

Ta CO

1 \/3 271 2n , 2n7r „ 2nn „ , „

+ 1 — = cos — + z s m — =>- x „ = ACQS — + B sm —^,\/n = 0 , 1 , 2 ,

V i 1 khong la nghiem cua phucJng trinh d i e trimg nen nghiem rieng x* cua

phirong trinh da cho c6 dang x* = C, Vn = 0 , 1 , 2 , (vdi C la hkng so)

Thay vao phUdng trinh da cho d\rdc C = - 1 Tat ca cac day so can t i m deu

Theo d9,ng 9 d trang 65 t a tim d\r0c x^, trong do cac he so A , B se t i m sau

Con X * dudc xac dinh n h u sau:

• Neu 7 khong l a nghi?m c u a aA^ + 6A + c = 0 thi x ; = Qfc(n).7", vdi Qk{n)

la da Ihiic bac A; theo n

• Neu 7 l a nghiem ddn c u a a>? + 6A + c = 0 thi x* = nQfc(n)7"

• Neu 7 l a nghi§m kep c u a aA^ + 6A + c = 0 thi x ; = n^Qk{n)^''

Thay x* vao (•), dCing phUdng phap dong nhat cac h? so t a se t i m dit^c

Qkip) Biet x i , X 2 , t i i x „ = x;;' + x* tinh dUdc A, B

B a i l o a n 9 1 Tim so hg,ng tong qudt cua day so (x„) thoa man x i = X2 = 1,

2x„+2 + 5x„+i + 2x„ = (35n + 51)3", Vn = 1 , 2 , (1)

G i a i Phudng trinh dac t n m g 2A2 + 5 A + 2 = 0 -f* A € | - 2 , V?ly

x „ = > l ( - 2 ) " + B ^ - l ^ + ( a n + 6)3n v n = 1 , 2 ,

trong do {an + 6)3", Vn = 1 , 2 , thoa man (1), nghia la vdi mpi n 6 N* t h i

2 [a(n + 2 ) + 6) 3"+^ + 5 [a(n + 1) + 6] 3"+^ + 2(an + 6)3" = (35n + 51)3"

hay 35an + 51a + 356 = 35n + 51, Vn € N*, nghia la

thuTc bac m , b a c i theo n

C a c h giai. Gpi k = max { m , Z}

• Neu 2 = cose ± i s i n e (vdi = -\) khong la nghiem c u a phUdng trinh d$.c

trung aA^ + 6A + c = 0 thi t i m x ; dudi d^ng

X* = Tk(n) cos en + Rk{n) sin e n

trong do Tk{n), Rk{n) l a cac da thiic bac k ciia n

• Neu 2 = cose ± i sine (vdi = - 1 ) l a ughiOm c u a phUdng trtnh dSx; trUng + 6A + c = 0 thi t i m x ; dudi dgng

X* = nrjk(n) cos en + nRk{n) sin en

trong do Tk{n), Rk{n) l a cac da thiic b^c A; c u a n

B a i t o a n 9 3 Tim so hang tong qudt cua day so ( x „ ) cho nhu sau: xi = 0,

2

(n + l)7r + 2 ( a n + 6) cos ^ + ( c n + d) sin

nx ( n + l)7r fn-K 7r\

cos + - s i n

(ivn 7 r \T

= " n T + 2 J = ' = ° ^ T '

s i n — ^

thay vac (1) ta dudc

- [a(n + 2) + 6] cos ^ - [c(n + 2) + d] sin ^

x i = a , X 2 = P, X 3 = 7, a x „ + 3 + 6 x „ + 2 + c x „ + i + d x „ = / „ ,