Bồi dưỡng học sinh giỏi khảo sát hàm số

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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=
y

DVL.013494

Tim so hang t o n g q u a t cua day so da cho.

d = ^ ^ - I f ^ ).
ia

z/a^

(1)

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"

Chu'dng 1

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 li", 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.

Xac dinh day so
1.1
1.1.1

Xac dinh day so bang phu^dng phap quy nap,
phu'cfng phap doi bien
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 .

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 cd sd, hay bvTdc khdi d l u ) . Kiein tra A{n) diing khi n = 1.
Biidc

2 (birdc quy n a p , hay 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

X 2 = V^2 + v / 2 = W2

. ,

va

1 +

v/2' = ^ 2 ( 1 + COS J ) = 2cos J ,

X3 = V'2T1^ - ^ 2 ( 1 + COS J ) = 2 cos ^ .

Tac gia
Th^ic sy Nguyen Thi Chung
3

,

b) Vdi

e Z, k>l,

BSLng phudng pliap quy nap ta se chiing minh

gia sii xfc = 2 c o s K h i do

Un

= V2T^k = y ^ l T c ^ ^ ^ = 2COS
Theo nguyen l i quy n?ip suy ra x„ = 2 cos ^^^n

. TrUdng h^p n = 1 da kigm tra d trgn.
Gia sil Un = tan. [ | + ( n - l ) ^ ] . K h i do

= 1,2,...

tan

x„ = ^ 2 + \/2 + • • • + v ^ . Hay tinh xjoro-

Bai toan 2.

N

^
CO

'

n dau

1

cSn

x„ = v/2 + x„_i, Vn = 2,3,...

Un

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

U2003 =

Tim

Un +

y/2~ 1
V ^ ) Un

,Vn=

=

tan

L3+("-^)8J

1

= tan

tan

1,2,...

SO J

tanx + tany ^
1 - tanx. tan?/

= tan - = tan ( - + 4
^8

=

<^ tan^ - + 2 tan - - 1 = Q <^
8

TT

0

^

TT

Vi tan - > 0 nen tan - =

Do do

= v/3 = tan

_
^,^2

tan -

Bai toan 4. Cho day s6 (u„) tfm/i bdi: <

2 tan?

7^m

o_

1

tan - = ~i + y/2
tan - = - 1 - v ^ .
+ tan ^ , Vn = 1,2,.
1 - u„.tan
Un

- 1. Ta c6 u„+, =

tan -- +
+ lan
tan —
lan
=
3__

L . = tan f J
1 - tan - tan v3

H—

8/

+ tan -

l-tan(- + -)tan4

^3

8/

va

cos -

Hvrdng d i n . Ta c6 ^an — = ^

V3

+ 2 - v/3
,Vne N»
" " • ' ^ " l + (y3-2)u„
-

"n

I
= JIZ^

uap ta clnhig miuh du^c u„ = tan

= 2 - N/3. Bang quy

,Vn=l,2,...

L6+^^"-')l2

Bai toan 5. Tim so hq,ng tdng qudt cua day so (x„) cho nhu sau:
x„ + \/2 - 1
Xn+l

+ ^)

=

U2010-

l-tan2?

^

.3-^"8

,Vn = l , 2 , . . .

Ui

G i a i . Tit cong thitc tan(x + y)=

ID

= tan

U2003-

, ^

TT

- tan ^ 3 + ( n - l ) 3 ^ . t a n -

Do do

ui = v/3
1 + (1 -

+ tan^

Tlieo nguyen ly quy nap suy ra

c6 n - 1 d&u c&n

=

.3+("-^)8,

= tan

Htrdng dan. Ta c6 x i = V2, x„_i = V^2+ •\/2T-• • + v/2. Suy ra

Wn+I

,Vn = l , 2 , . . .

= tan

=

-,VnG

(l-v/2)x„ + l '

Bai todn 6. Tim so hq,ng long qudt cua day so (u„) cho nhxt sau:

{

uo = l
" " + 1 = ^;

_ _ , V n = 0,l,...
5

3 /

1.1.2

2

\i

2^-1

2^

2"~'-l

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...):

V,y

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

~

=« - - » r •- i

(

a + ^ ) " ' - | ; , V n - 1.2,... T h .

lai bang phUdng phap quy nap thay diing.

(1)

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

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
diem sieu uon A (-—; f (- —]]

thi ta ddi qoc toa do

fix).

TUc la ta doi bien X = x +

cHa dd thi

ciia

Trong do a^O va c =

Xn+i

3

/

3

62

, ,

63 \ 2

262

(

63 \)

62

63

6 62

6(c-3)

4a

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

/

= ox^ + 6x„ + c, Vn = 1,2,...

(1)1

63 \(c-3)

,

6

Suy ra j/„+i = ay^, Vn = 1,2,... Suy ra

62-26

Vn = ayt,

4a

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:

= a {ayt2f

= a'^'vlU = a'^' {ay^.f

I4.'»J.l2-|....J.1'>-2

= . . . = a*+^+-* +

t-?"''

yf

,n-l

= a ' - 3 y\

= a'^'^'^t^
3"~'-l

I"

/

3a J

6

\

3»->

y „ ^ , - ^ = a ( j , „ - A ) + 6 ( y „ - A ) + c , V n = 1,2,...
6

(

V*yx„ = a .

(^a + - j

- - , V n = l,2,...

Luu y. Chd y 2 d trang 6 da gpi y cho ta each doi bien J/n =
1

hin

b'^\ f

+ 3^-

b \

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

62
62
62 - 26
<*J/n+i-^=ay„-6i,„ + - +6 j , „ - - +
,Vn

4a

" <*J/n+i = a y 2 , V n = 1,2,...

= l,2,...

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

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

Vn = aj/2_i = a {ayl_2f

=

6

7

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

p
2(2n+l) + —

2(2n + l ) x „ + 1

ta c6

= 2 ( 2 n + l ) + — (n = 1,2,...).

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 ( 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
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

^"+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
2
i„ = — =
u„
( 2 n - l ) ( 2 n + l)

1

1

2n - 1

2n + l

,Vn=l,2,...

Sau day la mpt so he thiic dai so t h u dUdc tit cac hg thiic lien quan den ham
hypebolic, vipc chiing m i n h cac ho thiic do x i n danh cho ban doc.
j
• Tir he thiic cos 2t = 2 cos^ t - 1 t a t h i i dUdc h? thiic dgi so

Dodo
x i + X2 H
<*Xl + X2 H

+ ajaooi =

(J.

/I

1-3'

h X2001 = 1

+ \
_ 4002

1

Sy "'"'•••^ V4001

l_\
(1)

4003/

4003 ~ 4003'

• T i t h? thiic cos 3t = 4 cos^ f - 3 cos t t a t h u dildc h f thiic dai so
1 \

1.1.3

C ^ c d o n g n h a t thtfc b o s u n g

TCt cong thiic sin^ t + cos^ < = 1,

e R, n l u t a thay x = tan - t h i dit0c

Nghia la 4x^ "

N h u vay nioi cong thiJc lUdng giac se cho t a 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 t i n h doc l9,p tUdng doi

dan tach hkn cd sci dai so cua no, da lam cho chung t a 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 t r i n h 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 l i 2 se giup t a 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.
vdi P„(x) la da thiic bac n. Khi

8

1 /
2 a + -7

1

- vdi X = -

a+ -

.

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

(2x)2 + ( l - x 2 ) 2 = ( l + x 2 ) 2 , V x e R .

D i n h l y 1 . Gid sii cosnt = Pn (cost),

1 f

do

i (a^ +

= 16m^ - 20m^ + 5m, trong d6m

=

]-(a+-

Tii h§ thiic sin3t = 4sin^ f - 3 s i n t t a t h u ditpc h§ thiic dai so
= 3 fl /
J
.2 V

0

+ 4

a

Nghla la 4x3 ^ 3 ^ ^ ^ ^^3 _ i . ^

1
2

3; = 1

(5)

a-

-

_

, a ^ 0.

• He thiic dai so ling vdi sin 5x = 16 sin^ x - 20 sin^ x + 5 sin x Ik
1 (^a' - ^ )

= 16m^ + 20m^ + 5m, trong

9

= ^

" ^) •

1.1.4

M o t so p h e p d 6 i b i e n dvtdc d i n h hMdng
l i f d n g giac.

b d i cAc cong thiJc

r
Cach t r l n h bay khdc cho trtrdng hdp

(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
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
j / i G R;

Giai.
• Neu

Vn+i

= 2yl

-

1,

Vn =

1,

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

y2 = 2 cosh^ (i>-l = cosh 2(f), j/3 = cosh 2^ (A,..., 2/„ = cosh 2"~ ^ (A.
Tif cosh <;/) = y i , ta C O

2,...

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

= cos 2"~^</>.

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

> 1. Neu \yi\>\i ton tai

T

,
= yi

li'

=

2/1 + ^^y\ 1.

Do do
_ 2" — 1

,
j/n =

+

+ (yi +
2

y/yl

-1)

— , V n = 1,2,...

B a i toAn 11. Tim day so {x„}+~i
x i = Q, x„+i = ox^ + 6, Vn e N ' , at = - 2 .
Vay neu dat P = yi + i/j/jf - 1 thi

2/1 = 2

Hvtdng d i n . Dat x„ = - 6 y „ , khi do yi =

/

I

1

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

y^^^ = 2y2 - 1.

Taco

Sau do Slit dung bai toan 10.

Dodo

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=2y2-l,Vn=l,2,...
(1)
D§,t x„ = py„. Khi do

1 /
2

py„+i = ap^yl + 6

y„+i = apyl + |.

Ttf (1) va (2) suy ra ta can tim p sao cho
f ap = 2
r
\a man a6 = - 2 ) .

2

V^y ta se dat

Theo nguyen h' quy n^p suy ra
1 /^2"->

1

10

^

= -y„
a

x„ = - 6 y „ (do a6 = - 2 ) .

,V7i = l , 2 , . . .

14

(2)

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

va

Vi

4b

2 „

= 2j/„ = 4x„ nen

ZxZ2Zz--Zn

XiX2X3...X„

4'-"x„+i

4x„+i

1

2

4".XlX2X3...X„

So hang t6ng quat ciia day so (w„) la

. Tim so hang long qudt cua day so {u„).

1\

-, Vn G N*.

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

B a i t o a n 13. Ttm day so {x„}+f°i biet

1

9

XiX2X3...X„

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

1 /

vdi /3 = 2/1 + v^y2 _ 1 ^

1

+

\

G i a i . D$,t x„ = 6j/„, khi do
,Vn=l,2,...

= - va
= abyl + 1

6i/„+i = a62y2 + 5 =^

+ ^4x2 - 1. Dat

y„+i = 2y2 + 1 (do a6 = 2).

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
2/1 = 2

1 \

1

1 2
y2

1

Pj

+ 1
_ 1

= 2y? + 1 = 2

Gia sii Vn = ^ (^P'^'"' -

y„+i = 2yi + 1 = 2
12

pj\

m

+1 = r

) . K h i do
1

P
13

\

T2

1 \

-—

Tlieo nguyen If quy nap toan

IIQC

suy ra

B a i t o a n 15. Tim { x , , } ^ ^ , hiet
3

x i = n,

2/n = 2

Xn+\ o x „ - 3 x „ , a > 0.
Af..

„o

2

Bdi v^y

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

2n-l

6 +

a

V6^ + '

B a i t o a n 14. Tim {yn}n=i'

a

,Vn

do sii dung bai toan 14.
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

biet

cos

Ta C O g^ng

yi 6 R, y„+i = Ayi - 3j/„.

3x

=

4 C D S ' * X — 3 cos

x.

d y n g cong thi'rc nay. Gia sii x „ = 6y„ + c, k h i do

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

Giai.

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

• Neu 12/11 < 1 t h i ton t ^ i 0 sao cho cos(j> = j / i . K h i do
2 / 1 = 4 cos^ (/) - 3 cos 0 = cos 3 0 , . . . , 2/n = cos 3"0.
• K h i I2/11 > 1. X e t so thyc l3 sao cho

Vay t a cho

/? =

2/1 +

L /? = y i

Vay neu dat (3 =

-1

«>y„+i = abSi

-

+ 3o6cyf. + Z{ac^ - l ) y „ + ^-'(ac^ - 4c).

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

V2

thi

c = 0^
6 =

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

)'
Ta

-r'''-

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

CO

1

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

y:-2
2/2 = 4

Giasu2/„ = -

2/n+l = 4

G i a i . Dat x „ = -7=2/11- K h i do 2/1 =
( 2
yn]
+3
Vv/a

- 3

/fl-^

+ : ? r : ^ , khi do

-3

1 ^

Xet so thuc /:* sao clio
3„-,

_ 1 _

Vay theo nguyen h' quy nap toan hoc suy ra
,Vn = 1 , 2 , . . .

14

va
\ ^2/a+i =4yf.+3y„(Vn6N*).
/

yi =

1

-

"

Vay nou dat /J = yi + V y f T T

• /3 = y i + / y ? + 1
'^^'^ - 1 = 0 ^ J i = y i - x / y f + i thi

1 /

yi = 2

v

15

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.

Ta CO

Giai. D^t ii„ = u„ + 1 (n = 1,2,...). Khi do Hi = 3 v&

Do (16
=4

yn+1

= 4 1 /^a—

1 \

03" ,Vn = 1,2,... Khi do
T3
1 \ +3

^3"

2 l,^

/J

Theo nguyeu h' quy n^p suy ra y„ = 1 ^ / j a - _

x„ =

+

+1

2\,^

/33"-»

^3"^-

Vn = 1,2,... V$y
- \"

+

+1

,d=

.(1)

HUotng d t n . Gpi y,, = x„ + ^ , thay vao (1) ta dudc 2/„+i = ay^ + 3y„.
o(l

Bai toan 18 (De nghi thi OLYMPIC 30/04/1999). Xdc dmh so hang tong
qudt cm day so (u„) biet iting:

Hifdng d t n .

f u, = 2
\i = 9uf, + 3(t„,Vn = 1,2,...

2
Cach 1. Dat u„ = - X u , 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„ ^ ^ {
,,,, + 3,,^^. Chon xux^ sao cho { §| +
Bang quy nap, chiing niinh dUdc:

1

v.^, = v l - 3t;„ = ( x f - ' + x f " ) ' - 3{xr-' + x f " )

= xf + 3xf x\f + xj )+xi - Z{x\ x5 )
= x f + 3(xr-' + x f " ' ) + xf - 3(xr-' + xf") = x f + xf.

fi

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
i„+, = ax^+6x^+cx„+d,V7i € N* a > 0 , c = —

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.
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 + X 2 = 3 vk X1X2 = 1. Ta se chiing
minh quy n?ip rang
Vr, = x\ , V n = l , 2 , . . .
Ta e6 t;i = 3 = xi + X2 = x f + xf. Gia sii t;„ = x f + xf'\i do: »

«n =

. „ - l/ 3= +( v^/ 5j \ ' " " + (/ 3- -2v ^/ 5j\ ' " " ' - l . V n = l , 2

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.
'jl..
,
r
V

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:
Vn = ^ ( x f + x f " ' ) , V n = 1,2,...
Suy ra
(2 + Viy

Bhi

+(2-^5)^

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 , . . .
(x„) cho bdi xi = \/2 + \/3
x„+i = x ^ - 4 x 2 + 2 , V n = 1,2,...
Giai. Dat x„ = 2y„. Khi do
2j/„+, = 162/^-16i/2+2,Vn = l , 2 , . . .
= 8j/^ - 8y2 + 1, Vn = 1,2,...
Tac6
^

v/2 +V5
1 + 2 _ '1 + cos^
yi =
—2 — - = cos—.
12
Do cong thiic Ivr^ng gidc
8
8 cos^ a + 1 nen
= Scos" ^ - 8cos2 iL + 1 = cos ( 4 . ^ ) .
y, = Scos" ( 4 . ^ ) - 8cos2 ( 4 . ^ ^ ) + 1 = cos (42.iL).
Tim so h^ng tong qudt cua day so

COS4Q =

CDs'* Q -

2/2

Gia sii y„ = cos

Gilti.D&tx„ = ^ . K h i d 6 y i = 3 v a

+ 2 ,Vn=l,2,...

u„ =
2y/E
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,...

Bai to4n 22.

Bki todn 23. Tim so hQ.ng tong qudt cua day so (x„) cho bdi
= 3^2^ ^^^^ ^ 1 6 ^ ^ 4 _ 8^2x2 + ^ , Vn = 1,2,...

1 ^

v/2

= 16^24 - 8V/24 + ^ , Vn = 1,2,...
4

«B'yn+i=8y;t-8y2

2

Xet so thvtc a sao cho

i .ii'ii* /

a+ -

a2 - 6a

V?tynludata = 3 + 2 v / 2 t h i 3 = i ( a + i ) , 1 = 3 - 2^2. Ta c6
1\

4

a+1>
a + Do do

Bdi v^y

ri / i\
1/

A

o
4
a^+4a2+6+^ + ^ j .

2

4

-8

1

1
2
4

a+

"

^

-

1

Gia s» !M = i (<.'••'+^) Khi d6

V^y theo nguyen h' quy n^p suy ra y„ = cos ( 4 " " ' • j ^ ) - Vn = 1,2,... Do do
x„ = 2 c o s ( 4 " - ' . ^ ) , V n = l , 2 , . . .
18

a = 3 + 2v^
a = 3 - 2V2.

+ 1= 0

Khi do

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

2

+ l,Vn = l,2,...

19

'

1

u

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
so (u„) nhu sau : ui = ^ vd u„+i = — . — ,
„,Vn e N*. T i m so
5
u * - 8 < -I- 8
long qudt cua day so da cho.
G i 4 i . D^t — =?;„. K h i d6 v i = 7.

hmg

' '

5"-17r

Gia sur x„ = cos — - — , k h i do
- 20 cos"-^ f

= 16 cos^ ^

+ 5 cos"- |

5^.

= cos

gia thiet suy ra
Theo nguyen l i quy n^ip, suy ra x„ = cos — - — , Vn = 1,2,...

o
, f

B a i toan 26. Cho day so (x„) n/ii/ sau:
=>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
5

Tim so hang tong qudt cua day so dd cho.

n

1/

4=2r+a

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

V$,y ngu dat a = 2 t h i J = i J^a + i Y

a =
a=

2

2-1,

2V

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

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

^4n-i ^"^2-4"-''

= 1' 2' • • •

1

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

If
Bdi vay x i = -

^ - t i i ^ si>
X2 = 16.

20

2

Suy ra

ri /

= -[a^

l^" _ 5 (

a4- -

2

5

[1 (
1>

16. - (a + 2 V
a;

B a i toka 25. Cho day so ( x „ ) nhu sau

T i m so hQ.ng long qudt cua day so da cho.

5 / 3 0

20 - a + 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
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<3i vigc neu cong thiic roi chiing minh bang quy nap n h u d mpt so
ihi li?u kh4c.

^ ) = -7.

+

16 •x[a + -]
= - a H 5a3 -MOa + — -I2 \j
2 \
a^
J

iM^

Do do u n =

a = -7-v/48
a = -7 + v/48.

= -7 •«> a^ 4- 14a -H 1 = 0

1 fa +

^ = ^. Tudng t i ; n h u 15i giai bai

toan 23, t a chiing minh diMc
1

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

1 /

-20

0

2

l\

a 4- -

2V

i\

5

a-H

1>

u + -A ;
V

3

+3a+-

Tac6

(3)

-I- - r

1\
+ -^

(4)

l^

(5)

-

3

1

4-5

2 1^

a;

2V

(6)

va

a)
- 20

'l
2 ^

1>
a 4- -

a^

21

3

4- 5

1

1\
a4- -

—2 \/

1

2

5^

^\

a" 4- -?
^
aV

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,...
<^yn+i

5„-l

1 M

1

/

gn

16y^ + 20y^ + 5y„, Vn = 1,2,...

Xet so thyc a sao cho

L2\

, ("1 /

=

fa _

1 \

i^

1
2

y„+i = 180y^ - 48y^ + 5y„, Vn = 1,2,...

20

1/

_1\

L2V

5 [1
[2 V

(1)

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

«/J

1^

v3

:x„+i = 180.-^x^ - 48.-^x^ + — x „ , Vn = 1,2,...
= 20x1 - 16x1 + 5x„, Vn = 1,2,...

16

Khi do |a\/3| < 1. D$,t a = arccos(av/3). Khi d6

ay

2V

1 (
2 V

ayj

^\
a/

T3

+ 5' 1

Dodo
a

V^y ta thu dilpc day so (x„) nhu sau
X , = aV3, x„+i = 20x^ - 16x^ + 5x„, Vn = 1,2,...

n

5/
2 V

Tlit (1), (2), (3) ta CO

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

• Trudug h(?p |a| <

a

aj

ia[l(a-i)f = i ( a - - = . . o . - ^ . A . ^

,Vn=l,2,...

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

^Xn+i

l1 =_ - ( 3 - V Y o ) .

/

Taco

Vay so hang tong quat cua day so da oho la

1

• a = 3 + >/l0
3 - VIO.

La =

V?iy neu d^t a = 3 + \/lO thi

Theo nguyen h' quy n»p suy ra

x„ = -

- 6a - 1 = 0 ^

'

1

0

(

A

N

- a'" J

J^=2r

'a^'J

Gi& sut y,. = i ^a*"-' - - i - ; - ^ . Theo (4) suy ra y„+i = ^

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=v3
26 d trang 21.

Khi do ay/3 > 1. Tien hanh tUdng ti; nhu bai to4n
m

v^y x„ = [(yiO + 3)^""' - (v/TO - 3)
23

,Vn = 1,2,...

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
t i m cac ham so f{x) va h{x) sao cho

/ K ) = /i(/K_i)).

Xn

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

5Xn-i4-4

^ _ X n _ i - 4

^ 1 + 2

-

^glLlLli.i

x„_i + 2 ' ~ " ' -

(/(«0)).

(**)

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
the hicu k h a 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:

7x„ - 1, Vn = 1 , 2 , . . .

X„ +

1

_ .gn-JL+l

p2^n-2

^ - % „ _ l - 4

+ 1

7x„

4.62°13 + 1

L i f t i y . Ta xet
I.

5xn + 4

5x„ + 4 - kxn - 2k
x„ + 2
( 5 - f c ) (x
x„ + 2

x„ + 2
Ta chpn A; sao ehp

D

=

7 ( x , . - , - i ) = 7 > ( x . . , - l ) = ...=

7"-'

^ fin

XI - 4

4 6" + l

(^x„--j,Vn = l,2,...

V&y
Xn--

6(x„_i + l )
x„-i + 2

D o d 6 x „ + l = 6 " x „ - 4 . 6 " <^x„ = ^ : r - ^ , V n € N * . V^y x^oia = 62013 _ 1 •
6" - 1

(5-fc)x„ + 4-2fc
g =

+ 1

^1

X„_2-4

G i a i . Ta c6

a^n+i -

X n _ i + 2 ' -

Suy ra

= h{h (/(u„-2))) = h2 (/(«„_2)) = - - = K

Xn+i

+ 4 = 4x„ + 8 •«> x „ = 4, mau thuan v d i gia

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

(*)

Si'r d y n g (•) lien tiep t a t h u dudc.

xi = 3 ,

X n + i = 4 t h i ^"_^_2 =

2 f c - 4 = k^2k-4=-bk-k'^<^k^-3k-4
b-k

=

0^

1

= 4-.'

Vay nen trpng Icri giai tren t a da xet x„ - 4 va x„ + 1 .
D p d 6 x „ = ^ . 7 " - i + i Vn = l , 2 , . . .
. each xet hi?u x„+i - 1- .
L v f u y . Trpng Idi giai tren, quan trpng nhat la biet
^ 1
6
So - dU0c t i m r a n h u sau: T a 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

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 t a c6 u„ > 0 vdi mpi n va:
9u„+i = u„ + 4 + 4v^l + 2un
<:»18u„+i = 2u„ + 8 + S v ' l + 2u„

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

•«>18u„+i + 9 = 2u„ + 1 + 8\/H-

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)^

diem bat dpng eiia ham so /.

<:»3V'2u„+i + 1 = v'2u„ + 1 + 4.

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.

2iin +

D«Lt Vn = v/2u„ + 1 . K h i do v i = 3 va t t t (*), t a c6:
3t;„+i = v„ + 4

1
4
i;„+i = 3^^" + 3

16

.
(*)

•**t;„+i -

2 = -vn

-

-

v„+i

-

2 = -(v„

-

V§Ly so h9iig tong quat cua day so da cho la:

2).

'

x„ = 7 + { a - 7 ) 2 " - ' , V n 6 N * .

Nhu v$.y:

1 {Vn-l

'^n - 2 =

=

livhi y. Trong bai toan nay thi ham so phy la /(x) = x - 7, c6n hhm l?Lp la

-2)

=

{Vn-2

" 2) = ^ ( f n - 2 " 2 )

-=3;;^(-i-2) = 3 ; ^ ( 3 - 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;

.

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,...

V

1
1/ 1
" " = - 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/ so h(mg tSng qudt cua day so bdng phuang phdp ham l^p.
1. L d p h a m g(x) =
+ ax + b.
Taco
g{x) - c = x'^ + ax+ b - c.

^ax^b-c=

= {x-cf=i^x^

-2cx -H

+ ax + b-c

{ -^l^.