Chuối Lũy Thừa Hình Thức Và Hàm Sinh

i Hc Thi Nguyn

Trng i Hc Khoa Hc

Hong Vn Qu

Chui lu tha hnh thc v hm sinh

Chuyn ngnh : Phng Php Ton S Cp

M s: 60.46.40

Lun Vn Thc S Ton Hc

Ngi hng dn khoa hc: PGS.TS. m Vn Nh

Thi Nguyn - 2011

S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

Cng trnh c hon thnh ti

Trng i Hc Khoa Hc - i Hc Thi Nguyn

Phn bin 1: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Phn bin 2: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Lun vn s c bo v trc hi ng chm lun vn hp ti:

Trng i Hc Khoa Hc - i Hc Thi Nguyn

Ngy.... thng.... nm 2011

C th tm hiu ti

Th Vin i Hc Thi Nguyn


Mc lc

1 Kin thc chun b 4

1.1 Khi nim vnh v ng cu . . . . . . . . . . . . . . . . . . . 4

1.1.1 Vnh . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.2 c ca khng. Min nguyn . . . . . . . . . . . . . . 4

1.1.3 ng cu . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.4 Trng . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Vnh a thc v nghim . . . . . . . . . . . . . . . . . . . . . 5

2 Vnh cc chui ly tha hnh thc 11

2.1 Vnh cc chui ly tha hnh thc . . . . . . . . . . . . . . . 11

2.2 Dy hiu ca mt dy . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Hm sinh thng v dy Fibonacci, dy Catalan . . . . . . . . 20

2.4 Hm sinh m v dy s Stirling . . . . . . . . . . . . . . . . . 24

2.5 Hm sinh ca dy cc a thc Bernoulli . . . . . . . . . . . . 27

2.6 Hm sinh Dirichlet v hm Zeta-Riemann . . . . . . . . . . . 34

2.7 Tch v hn . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.8 ng nht thc Newton . . . . . . . . . . . . . . . . . . . . . 41

2.9 Dy truy hi vi hm sinh . . . . . . . . . . . . . . . . . . . . 48

1


M u

Trong ton hc vic s dng cc kin thc ton cao cp gii quyt cc bi

ton ph thng l iu rt quan trng. N khng ch gip ngi lm ton

c nhiu phng php la chn li gii, m rng tm hiu bit ton hc m

cn pht huy c s thng minh v sc sng to, tm bao qut bi ton, m

rng bi ton di nhiu hng khc nhau.

S dng cc kin thc v chui s gii quyt cc bi ton v dy s

l mt vn nh vy. Nh chng ta bit cc vn lin quan n dy

s l mt phn quan trng ca i s v gii tch ton hc. Khi tip cn vn

ny cc em hc sinh gii, sinh vin v kh nhiu thy c gio ph thng

thng rt phi i mt vi rt nhiu bi ton kh lin quan n chuyn

ny.

Trong cc k thi hc sinh gii quc gia, thi Olimpic ton quc t, thi

Olimpic ton sinh vin gia cc trng i hc, cao ng, cc bi ton lin

quan n dy s cng hay c cp v thng loi rt kh, i hi ngi

hc, ngi lm ton phi c mt tm hiu bit rng v rt su sc cc kin

thc v dy s v chui s mi a ra cc phng php gii ton hay v hon

thin c bi ton.

phc v cho vic bi dng hc sinh gii v vic trao i kinh nghim

vi cc thy c gio bi dng hc sinh gii quan tm v tm hiu thm v

phn ny, c s hng dn ca thy m Vn Nh tc gi hc tp thm

v vit ti " Chui lu tha hnh thc v hm sinh".

ti gii quyt cc vn trng tm :

Chng I : Kin thc chun b .Tc gi nhc li cc kin thc c bn nht

v :

1.1 Khi nim vnh v ng cu

1.1.1 Vnh.

1.1.2 c ca khng. Min nguyn.

2


31.1.3 ng cu.

1.1.4 Trng.

1.2 Vnh a thc v nghim.

Chng II : Vnh cc chui lu tha hnh thc. Tc gi gii thiu cc kin

thc.

2.1 Vnh cc chui lu tha hnh thc.

2.2 Dy hiu ca mt dy .

2.3 Hm sinh thng v dy Fibonacci, dy Catalan.

2.4 Hm sinh m v dy s Stirling.

2.5 Hm sinh ca dy cc a thc Bernoulli.

2.6 Hm sinh Dirichlet v hm Zeta-Riemann.

2.7 Tch v hn.

2.8 ng nht thc Newton.

2.9 Dy truy hi vi hm sinh.

Lun vn ny c hon thnh di s hng dn v ch bo tn tnh ca

PGS.TS m Vn Nh - i hc S Phm H Ni. Thy dnh nhiu thi

gian hng dn v gii p cc thc mc ca tc gi trong sut qu trnh lm

lun vn. Tc gi xin by t lng bit n su sc n Thy.

Tc gi xin gi ti cc thy (c) khoa Ton, phng o to Trng i

Hc Khoa Hc - i Hc Thi Nguyn, cng cc thy c tham gia ging

dy kha Cao hc 2009-2011 li cm n su sc v cng lao dy d trong

thi gian qua. ng thi xin gi li cm n tp th lp Cao hc Ton K3B

Trng i Hc Khoa Hc ng vin gip tc gi trong qu trnh hc

tp v lm lun vn ny.

Tc gi xin cm n ti S Ni V, S Gio dc v o to Bc Ninh, Ban

gim hiu v t Ton trng THPT Lng Ti 2 to iu kin gip

tc gi hon thnh kha hc ny.

Tc gi

Hong Vn Qu


Chng 1

Kin thc chun b

1.1 Khi nim vnh v ng cu

1.1.1 Vnh

nh ngha . Ta gi l vnh mt tp hp X cng vi hai php ton hai ngi

cho trong X k hiu theo th t bng cc du + v . (ngi ta thng k

hiu nh vy) v gi l php cng v php nhn sao cho cc iu kin sau

tha mn:

1) X cng vi php cng l mt nhm aben.

2) X cng vi php nhn l mt na nhm.

3) Php nhn phn phi vi php cng: Vi cc phn t ty x, y, z X tac:

x(y + z) = xy + xz(y + z)x = yx+ zx

Phn t trung lp ca php cng th k hiu l 0 v gi l phn t khng.

Phn t i xng (i vi php cng ) ca mt phn t x th k hiu l -x

v gi l i ca x . Nu php nhn l giao hon th ta bo vnh X l giao

hon. Nu php nhn c phn t trung lp th phn t gi l phn t n

v ca x v thng k hiu l e hay 1 .

1.1.2 c ca khng. Min nguyn

nh ngha1 : Ta gi l c ca 0 mi phn t a 6= 0 sao cho c b 6= 0 thamn quan h ab=0.

nh ngha2 : Ta gi min nguyn mt vnh c nhiu hn mt phn t, giao

4


5hon, c n v, khng c c ca 0.

1.1.3 ng cu

nh ngha. Mt ng cu (vnh) l mt nh x t mt vnh X n mt

vnh Y sao cho:

f (a+ b) = f (a) + f (b)f (ab) = f (a) f (b)

vi mi a, b X. Nu X = Y th ng cu f gi l mt t ng cu ca X .Ta cng nh ngha n cu, ton cu, ng cu tng t nh nh ngha

trong nhm.

1.1.4 Trng

nh ngha: Ta gi l trng mt min nguyn X trong mi phn t khc

khng u c mt nghch o trong v nhm nhn X. Vy mt vnh X giao

hon, c n v, c nhiu hn mt phn t l mt trng nu v ch nu

X {0} l mt nhm i vi php nhn ca X.

1.2 Vnh a thc v nghim

Kt qu chnh

Cho vnh giao hon R v mt bin x trn R. Vi cc n N, xt tp hp:

R[x] = {a0 + a1x+ a2x2 + + anxn | ai R} ={ n

i=0

aixi | ai R

}.

Mi phn t f(x) R[x] c gi l mt a thc ca bin x vi cc h sai thuc vnh R. H s an c gi l h s cao nht, cn h s a0 c gil h s t do ca f(x). Khi an 6= 0 th n c gi l bc ca f(x) v c


6k hiu deg f(x). Ring a thc 0 c quy nh c bc l hoc 1.Nu f(x) =

ni=0

aixi, g(x) =

mi=0

bixi R[x] th

f(x) = g(x) khi v ch khi m = n, ai = bi vi mi 0 6 i 6 n

f(x) + g(x) =i=0

(ai + bi)xi, f(x)g(x) =

i=0

(i

j=0

aijbj)xi.

nh l 1.2.1. Ta c R[x] l mt vnh giao hon. Hn na, nu R l mtmin nguyn th R[x] cng l mt min nguyn.

nh l 1.2.2. Gi s k l mt trng. Vi cc a thc f(x), g(x) k[x] vg(x) 6= 0 c hai a thc duy nht q(x), r(x) sao cho f(x) = q(x)g(x)+r(x)vi deg r(x) < deg g(x).

V d 1.2.3. Cho hai s t nhin n v p vi n > p > 1. Tm iu kin cnv xn an chia ht cho xp ap vi a R, a 6= 0.Bi gii: Biu din n = qp+ r trong Z vi 0 6 r < p. Khi c biu din

xn an = (xp ap)(xnp + apxn2p + + a(q1)pxnqp) + aqp(xr ar).Vy, iu kin cn v xn an chia ht cho xp ap l n : p.nh l 1.2.4. Gi s k l mt trng. Khi vnh k[x] l mt vnh chnhv n l vnh nhn t ha.

Gi s R v a thc f(x) =ni=0

aixi R[x]. Biu thc f() =

ni=0

aii R c gi l gi tr ca f(x) ti . Nu f() = 0 th cgi l mt nghim ca f(x) trong R. Gi s s nguyn m > 1 v k.f() = 0 c gi l mt nghim bi cp m ca f(x) trong k nu f(x) chiaht cho (x )m v f(x) khng chia ht cho (x )m+1.nh l 1.2.5. a thc f(x) k[x] bc n > 1. Khi ta c cc kt qu sau:(i) Nu k l nghim ca f(x) th f(x) = (x)g(x) vi g(x) k[x].(ii) f(x) c khng qu n nghim phn bit trong k.


7i khi tm mi lin h gia cc nghim hay mt tnh cht no ca

nghim a thc ta thng s dng kt qu sau y:

nh l 1.2.6. [Vit] Gi s x1, . . . , xn l n nghim ca a thc bc n sauy: f(x) = xn 1xn1 + 2xn2 + (1)nn. Khi c cc h thc

1 = x1 + x2 + + xn2 = x1x2 + x1x3 + + xn1xn...

n = x1x2 . . . xn.

nh l 1.2.7. Gi s f(x1, x2, . . . , xn) k[x1, x2, . . . , xn] l mt a thci xng khc 0. Khi tn ti mt v ch mt a thc s(x1, x2, . . . , xn) k[x1, x2, . . . , xn] sao cho f(x1, x2, . . . , xn) = s(1, 2, . . . , n).

Mt s v d

V d 1.2.8. Gi s f(x) = x4 5x3 + 9x2 10x+ 28. Tnh f(1 + 33).Bi gii: V 1 + 3

3 l nghim ca g(x) = x3 3x2 + 3x 4 = 0 v

f(x) = (x 2)g(x) + 20 nn f(1 + 33) = 20.V d 1.2.9. [VMO 1990] Gi s f(x) = a0x

n+a1xn1+ +an1x+an

R[x] vi a0 6= 0 v tha mn f(x)f(2x2) = f(2x3 + x) vi mi gi tr thcx. Chng minh rng f(x) khng th c nghim thc.

Bi gii: So snh h s ca x3n v x0 hai v, nn t f(x)f(2x2) = f(2x3+x) ta suy ra a20 = a0 v a

2n = an.V a0 6= 0 nn a0 = 1; cn an = 0 hoc an =

1. Nu an = 0 th f(x) = xrg(x) vi g(0) 6= 0. Vy xrg(x)2rx2rg(2x2) =

xr(2x2 + 1)rg(2x3 + x) hay g(x)2rx2rg(2x2) = (2x2 + 1)rg(2x3 + x). Vg(0) 6= 0 nn ta nhn c g(0) = 0 : mu thun. Vy an = 1. Gi sf(x) = 0 c nghim thc x0. Khi x0 6= 0 v an 6= 0. V f(2x30 + x0) =f(x0)f(2x

20) = 0 nn x1 = 2x

30 + x0 cng l nghim thc ca f(x). V hm

y = 2x3 + x l n iu tng nn dy (xr+1 = 2x3r + xr)r>0 v x0 6= 0 lmt dy v hn v mi s hng u l nghim ca f(x) hay f(x) c nhiuv hn nghim: mu thun theo nh l 1.2.5. Vy f(x) khng c nghimthc.


8V d 1.2.10. [IMO 1991] Gi s s hu t a (0; 1) tha mn phngtrnh cos 3pia+ 2 cos 2pia = 0. Chng minh rng a =

2

3.

Bi gii: t x = cospia. Khi 4x3+4x23x2 = 0 hay (2x+1)(2x2+x2) = 0. Nu cos pia = x = 1

2th a =

2

3. Nu x 6= 1

2th 2x2+x2 =

0, v nh vy x l s v t. Do |x| 6 1 nn cospia = x = 1 +

17

4. Bng

quy np, c th ch ra cos 2npia =an + bn

17

4vi s nguyn l an, bn. V

an+1 + bn+1

17

4= cos 2n+1pia = 2 cos2 2npia 1 = 2[an + bn

17

4]2 1

nn an+1 =a2n + 17b

2n 8

2> an. Do dy (an) l mt dy tng nghim

ngt v nh vy tp cc gi tr ca cos 2npia vi n = 0, 1, 2, ... l tp vhn (*) v

17 l s v t. Nhng do a l s hu t nn tp cc gi trca cosmpia vi m = 0, 1, 2, ... phi l hu hn: mu thun vi (*). Do d

a =2

3.

V d 1.2.11. Gi thit a thc f(x) bc n c tt c cc nghim u thc.Khi tt c cc nghim ca af(x) + f (x) cng l nhng s thc.

Bi gii: Gi s f(x) c cc nghim thc x1, x2, . . . , xk vi bi tng ngr1, r2, . . . , rk v ta sp xp x1 < x2 < < xk. Hm s

g(x) =f (x)f(x)

=1

x x1 +1

x x2 + +1

x xkl hm lin tc trong cc khong (;x1), (x1;x2), . . . , (xk1;xk), (xk;).Da vo s bin thin ca cc hm

1

x xj , phng trnh g(x) = a c thmk nghim mi na khc x1, x2, . . . , xk khi a 6= 0. Vy f(x)[g(x) +a] = 0 ctt c (r1 1) + + (rk 1) + k = deg f(x) nghim thc. Vy tt c ccnghim ca af(x)+f (x) u thc. Khi a = 0 th g(x) = 0 c k1 nghimthc mi na. Vy f(x)[g(x)+0] = 0 c tt c (r11)+ +(rk1)+k1 =deg f (x). Tm li tt c cc nghim ca af(x)+f (x) l nhng s thc.


9V d 1.2.12. Gi thit tt c cc nghim ca a thc f(x) v a thc g(x) =a0x

n + a1xn1 + + an u l nhng s thc. Khi tt c cc nghimca F (x) = a0f(x) + a1f

(x) + + anf (n)(x) cng u l nhng s thc.Bi gii: Biu din g(x) = a0(x+ 1)(x+ 2) . . . (x+ n) vi cc j thc.K hiu F0(x) = a0f(x), F1(x) = F0(x) + 1F

0(x) = a0[f(x) + 1f

(x)],F2(x) = F1(x)+2F

1(x) = a0[f(x)+(1+2)f

(x)]+12f (x)],v.v... cuicng Fn(x) = Fn1(x) + nF n1(x) = a0f(x) + a1f

(x) + + anf (n)(x).Theo V d 1.2.11 suy ra tt c cc nghim ca F0, F1, . . . , Fn u thc.

V d 1.2.13. Cho f = cosu + C1n cos(u + )x + + Cnn cos(u + n)xn.Gii phng trnh f(x) = 0.

Bi gii: t g = sinu+ C1n sin(u+)x+ + Cnn sin(u+n)xn. Khi f + ig = z + C1n ztx+ + Cnn ztnxn = z(1 + tx)nf ig = z + C1n ztx+ + Cnn ztnxn = z(1 + tx)n

z = cosu+ i sinu

t = cos + i sin.

Do 2f = z(1 + tx)n + z(1 + tx)n. Phng trnh f(x) = 0 tng

ng vi z(1 + tx)n + z(1 + tx)n = 0 hay(1 + tx

1 + tx

)n= z

z= z2.

Nh vy

(1 + tx1 + tx

)n= cos(2u + pi) + i sin(2u + pi) v c

1 + tx

1 + tx=

cos(2u+ pi + k2pi

n) + i sin(

2u+ pi + k2pi

n) vi k = 0, 1, . . . , n 1. T c x.

V d 1.2.14. Gi s a1, . . . , an, b R \ {0} v 1, . . . , n l nhng s thcphn bit. Khi f(x) = b+

nk=1

a2kx k ch c nghim thc.

Bi gii: Ta c f(c+ id) = b+nk=1

a2kc+ id k = b+

nk=1

a2k(c k id)(c k)2 + d2 .

Phn o Im(f(c + id)) = dnk=1

a2k(a k)2 + b2 6= 0 khi d 6= 0. Vy f(c +

id) 6= 0 khi d 6= 0. Khng hn ch c th coi 1 < 2 < < n1 < n.Hin nhin f(x) = 0 c n 1 nghim thc k tha mn

1 < 1 < 2 < 2 < < n1 < n1 < n


10

v thm ng mt nghim tha mn hoc (, 1) hoc (n,+). T suy ra hm f(x) ch c cc nghim thc.V d 1.2.15. Cho a thc P (x) = 1 + x2 + x9 + xn1 + ...+ xns + x1992

vi n1, ..., ns l cc s t nhin cho trc tha mn 9 < n1 < ... < ns k(k + 1) nn|b0ak1 + b1ak2 + + bk2a1 + bk1| 6 2

(|ak1|+ |ak2|+ + |a1|+ 1)v nh th k(k + 1) < |kak| 6 2

(k + (k 1) + + 2 + 1) = k(k + 1) :mu thun. Nh vy |an| 6 n+ 1 vi mi n.V d 2.1.12. Chng minh rng 2

12 (22)

122 (2n) 12n < 4 vi mi s nguyndng n.

Bi gii: V 212 (22)

122 (2n) 12n = 2

nk=1

k

2knn ch cn chng minh

nk=1

k

2k 1 :an = 2an1 + (2n 1)2an2, bn = 2bn1 + (2n 1)2bn2.

Bi gii: Bng quy np theo n ta nhn c cc cng thc bn =nk=0

(2k+ 1)

v an = (2n + 1)an1 + (1)nn1k=0

(2k + 1). Vyanbn

=an1bn1

+(1)n2n+ 1vi

mi s nguyn n > 0. Nh vy anbn

= 1 13

+1

5+ + (1)

n

2n+ 1. Chuyn

qua gii hn ta c limn+

anbn

= arctan 1 =pi

4.

V d 2.1.14. Cho hai dy s nguyn (an) v (bn) tha mn:{a0 = 1, b0 = 1an = 2n 1, bn = n2, n > 1.Xy dng hai dy cc s nguyn (An) v (Bn) nh sau:{

A0 = 0, B0 = 1, A1 = 1, B1 = a1

An+1 = an+1An + bnAn1, Bn+1 = an+1Bn + bnBn1, n > 1.

(i) Tnh An, Bn theo n.

(ii) Chng minh

AnBn Q \ Z.

(iii) Tm limn

BnAn

.

(iv) Chng minh

nk=1

1

k=

1

1 12

3 22

5 32

.

.

. 2n 3 (n 1)

2

2n 1

.


17

Bi gii: Bng quy np theo n ta nhn c cc cng thc Bn = n! v

An = (nk=1

1

k)n!.

(ii) Ta c

AnBn

=nk=1

1

k Q \ Z.

(iii) V

BnAn

=1nk=1

1

k

nn limn

BnAn

= limx1+

1

ln(1 + x)= 0.

(iii) Do

nk=1

1

k=AnBn

=1

1 12

3 22

5 32

.

.

. 2n 3 (n 1)

2

2n 1

.

2.2 Dy hiu ca mt dy

nh ngha 2.2.1. Cho dy s {an} = {an}nN. Dy {Dan}nN vi Dan =an+1 an, n > 0, c gi l dy hiu ca dy {an}.V dy hiu cng l mt dy s nn ta c th lp dy hiu ca n v k hiu

qua {D2an}. Hin nhinD2an = Dan+1 Dan = an+2 2an+1 + an.Tng qut Dk+1an = D

kan+1 Dkan v Dk(Dhan) = Dk+han.V d 2.2.2. Vi s nguyn dng r, dy (an), trong an =

(nr

), tha mnh thc Dan = an+1 an =

(nr1).

B 2.2.3. Vi hai dy s {an} v {bn} ta cD(ran+sbn) = rDan+sDbnv Dk(ran + sbn) = rD

kan + sDkbn vi mi s r, s v s t nhin k, n.

Chng minh: VD(ran+sbn) = rDan+sDbn = r(an+1an)+s(bn+1bn)nn c ngay kt qu D(ran + sbn) = rDan + sDbn. Tng qut D

k(ran +sbn) = rD

kan + sDkbn c chng minh d dng bng qui np theo k.


18

B 2.2.4. Cho dy s {an}. Nu Dr+1an = 0 vi mi n > 0 th r + 1 shng a0, Da0, . . . , D

ra0 xc nh hon ton tt c cc Dkan vi mi k, n.c bit, nu dy s {bn} tha mn Djb0 = Dja0 v Dr+1bn = 0 vi min > 0, 0 6 j 6 r, th an = bn vi mi n.Chng minh: Hin nhin.

nh l 2.2.5. Cho dy s {an}. Nu c a thc p(x) bc r tha mn an =p(n) vi mi n > 0 thDr+1an = 0 vi mi n > 0. Ngc li, nuDr+1an =0 vi mi n > 0 th

an =

(n

0

)a0 +

(n

1

)Da0 + +

(n

s

)Dsa0 + +

(n

r

)Dra0.

Chng minh: Gi s a thc p(x) bc r tha mn an = p(n) vi mi n > 0.Ta ch ra Dr+1an = 0 bng phng php qui np theo r. Khi r = 0 can = p(n) = a. Vy D

1an = a a = 0. Gi s kt lun ng cho r 1 vp(x) = crx

r+ +c0. V an = p(n) vi mi n > 0 nn Dan = an+1an =p(n+ 1) p(n). t q(x) = p(x+ 1) p(x) tha mn Dan = q(n). V q(x)l a thc bc r 1 nn Dr(Dan) = 0 theo gi thit qui np. Vy ta nhnc Dr+1an = 0.Gi thit dy {an} tha mn Dr+1an = 0 vi mi n > 0. nh ngha dymi {bn} xc nh bi:

bn =

(n

0

)a0 +D

(n

1

)a0 + +Ds

(n

s

)a0 + +Dr

(n

r

)a0, n > 0.

Theo B 2.2.3 ta c ngay

Dbn = D

(n

0

)a0 +D

2

(n

1

)a0 + +Dr+1

(n

r

)a0

=

(n

0

)Da0 +

(n

1

)D2a0 + +Dr+1

(n

r 1)Dra0

v Dr+1a0 = 0. Lp li, vi D2, . . . , Dj v ta nhn c

Djbn =

(n

0

)Dja0 +

(n

1

)Dj+1a0 + +

(n

r j)Dra0

v n Drbn = Dra0(nrr)

= Dra0. Do Dr+1bn = D

r+1a0 = 0 vi min > 0 v Djb0 = Dja0 vi mi 0 6 j 6 r. Vy theo B 2.2.4 c an = bnvi mi n > 0.


19

V d 2.2.6. Cho dy (an) vi a0 = 3 v an+1 = an + 4n+ 1 vi mi n > 0.Chng minh rng vi mi s t nhin dngm u c n an3n1 = 2m2.Bi gii: V Dan = an+1 an = 4n + 1, D2an = Dan+1 Dan = 4(n +1) + 1 4n 1 = 4 v D3an = 0 nn theo nh l 2.2.5 ta c ngayan = 3

(n0

)+Da0

(n1

)+D2a0

(n2

)+ 0 = 3 +n+ 2n(n 1) = 2n2n+ 3 hay

an 3n 1 = 2(n 1)2 vi mi n > 0. Vy vi mi m c am+1 3(m+1) 1 = 2m2.V d 2.2.7. Cho dy (an) vi a0 = 3, a1 = 2 v an+2 = 3an+1 2an 6n2 + 14n 5 vi mi n > 0. Xc nh an theo n.Bi gii: V an+2 2an+1 = an+1 2an 6n2 + 14n 5 nn khi tbn = an+1 2an ta s c bn+1 = bn 6n2 + 14n 5 v dy (bn) vib0 = 4, bn+1 = bn 6n2 + 14n 5 vi mi n > 0. V Dbn = bn+1 bn =6n2 + 14n 5, D2bn = Dbn+1 Dbn = 6(n + 1)2 + 14(n + 1) 5 +6n2 14n + 5 = 12n + 8 v D3bn = 12, D4bn = 0 nn theo nh l2.2.5 ta c ngay bn = 4

(n0

)+Da0

(n1

)+D2a0

(n2

)+D3a0

(n3

)hay

bn = 4 5n+ 8(n

2

) 12

(n

3

)= 2n3 + 10n2 13n 4.

Vy a0 = 3, an+1 = 2an 2n3 + 10n2 13n 4 vi mi n > 0. Vi dykiu ny, ta xt an = u.2

n + an4 + bn3 + cn2 + dn+ e. T y d dng suyra an.

V d 2.2.8. Cho dy (an) vi a0 = 5, a1 = 1 v an+1 = an + 6an1 6n2 + 26n 25 vi mi n > 1. Chng minh rng vi mi t nhin n u can 2.3n + n2(mod 2n).Bi gii: Vi dy kiu ny, trc tin xt a thc c trng x2 x 6 =(x 3)(x+ 2). Tip theo an = u3n + v(2)n + +an3 + bn2 + cn+ d v xt

5 = a0 = u+ v + d

1 = a1 = 3u 2v + a+ b+ c+ d34 = a2 = 9u+ 4v + 8a+ 4b+ 2c+ d

39 = a3 = 27u 8v + 27a+ 9b+ 3c+ d226 = a4 = 81u+ 16v + 64a+ 16b+ 4c+ d

415 = a5 = 243u 32v + 125a+ 25b+ 5c+ d.


20

Gii h c u = 2, v = 3, b = 1 v a = c = d = 0. Nh vy c cng thcan = 2.3

n + 3.(2)n + n2 v an 2.3n + n2(mod 2n) vi mi n > 0.Ch 2.2.9. Nu c a thc g(x) tha mn an = g(n) th an+1 an =g(n+ 1) g(n) l mt a thc ca n vi bc nh i 1.

2.3 Hm sinh thng v dy Fibonacci, dy Catalan

Mt trong nhng ngun gc dn n khi nim hm sinh chnh l nh l

khai trin nh thc Newton (1 + x)n =nk=0

(nk

)xk v khai trin thnh chui

ly tha ca hm phn thc

1

1 x = 1 + x + x2 + + xn + . y lmt k thut gii tch vi nhiu ng dng trong t hp v nghin cu dy s.

Hm sinh c phn ra lm hai loi: Hm sinh thng v Hm sinh m. Ta

bt u vi khi nim hm sinh thng di y:

nh ngha 2.3.1. Cho dy s {an}, hoc tng qut hn l dy hm {an =an(x)}. Chui lu tha hnh thc f(x) =

n=0

anxnc gi l hm sinh

thng ca dy {an}.

Kt qu chnh

nh l sau y c chng minh trong Gii tch.

nh l 2.3.2. Cho s thc dng . Nu chui lu tha hnh thci=0

aixi

hi t ti h(x) th cho mi x (, ) hm h(x) c h(x) =i=1

iaixi1v

h0

f(t)dt =i=0

aixi+1

i+ 1.

nh l 2.3.3. Dy s {an} c gi l dy xc nh kiu tuyn tnh nu dyc dng : a0 = 0, . . . , as1 = s1 v an+s = 1an+s1 + 2an+s2 + +san, n > 0. Khi hm sinh thng f(x) ca dy {an} l mt hm hut

b0 + b1x+ + brxr1 + 1x+ + sxs , r < s, khi v ch khi {an} l dy xc nh kiutuyn tnh.


21

Chng minh: Gi thit a0 = 0, . . . , as1 = s1 v an+s = 1an+s1 +2an+s2 + + san, n 0. t f(x) = a0 + a1x+ a2x2 + a3x3 + . Sdng an+s = 1an+s1 + 2an+s2 + + san, n 0, ta c (1x+ 2x2 + +sxs)f(x) = p(x)+f(x), trong p(x) l a thc vi bc deg p(x) < s.t q(x) = 1x + 2x

2 + + sxs. Ta c (q(x) 1)f(x) = p(x). Vyf(x) =

p(x)

q(x) 1 hay hm sinh thng f(x) ca dy l mt hm hu t.

Ngc li, cho f(x) =b0 + b1x+ + brxr1 + 1x+ + sxs =

p(x)

q(x), r < s.V f(x)q(x) =

p(x) nn khi so snh h s ca cc xn ta c hai ha0 = b0a01 a1 = b1...

a0s1 + a1s2 + as1 = bs1 va0s + a1s1 + as = 0a1s + a2s1 + as+1 = 0a2s + a3s1 + as+2 = 0...

Gii h u c nghim a0 = 0, . . . , as1 = s1. T h sau ta suy raan+s = 1an+s1 + 2an+s2 + + san, n 0. Do {an} l dy xcnh kiu tuyn tnh.

nh l 2.3.4. Nu u(x) = xs + 1xs1 + + s = 0 c cc nghimr1, . . . , rt vi cc bi tng ng 1, . . . , t. Khi tn ti cc a thc p1(n),p2(n), . . . , pt(n) tha mn an = p1(n)r

n1 + p2(n)r

n2 + + pt(n)rnt v

0 6 deg pi(n) 6 i 1 vi i = 1, 2, . . . , t.

Mt vi v d

V d 2.3.5. Dy s a0 = 5, a1 = 13, a2 = 35 v an+3 = 6an+211an+1+6anvi n > 0. Chng minh rng an 2n+1(mod 3n+1) v an 3n+1(mod 2n+1).Bi gii: a thc p(x) = x3 6x2 + 11x 6 = (x 1)(x 2)(x 3). Vyan = a2

n + b3n + c. T iu kin ban u suy ra an = 2n+1 + 3n+1.


22

V d 2.3.6. Xt dy s a0 = 11, a1 = 6, a2 = 18, a3 = 104, a4 = 346 v

an+5 = 6an+4 13an+3 + 14an+2 12an+1 + 8an, n > 0.Tm s nguyn dng ln nht m a2011 chia ht cho 2

m.

Bi gii: a thc p(x) = x5 6x4 + 13x3 14x2 + 12x 8 c vitthnh p(x) = (x 2)3(x i)(x + i). Vy an = p1(n)2n + ain + b(i)n.T iu kin ban u suy ra an = (n

2 + n + 1)2n + 5in + 5(i)n. Vya2011 = (2011

2 + 2011 + 1)22011. S m cn tm bng 2011.

S dng khi nim hm sinh v chui lu tha hnh thc xt mt s

dy s c bit.

V d 2.3.7. Xt dy s Fibonacci a0 = 0, a1 = 1, an+1 = an + an1, n > 1.Cng thc ng cho hm sinh thng ca dy l f(x) =

x

1 x x2 . Tm

an theo n v ch ran=0

an4n+1

=1

11.

Bi gii: t f(x) = a0+a1x+a2x2+a3x

3+ .Khi (1xx2)f(x) =x hay ta c f(x) =

x

1 x x2 . Vi a =1 +

5

2v b =

152

, biu din

f(x) qua chui ly tha f(x) =15

( 11 ax

1

1 bx). Vy c

f(x) =15

((1 + ax+ a2x2 + ) (1 + bx+ b2x2 + )).So snh h s ca xn hai v c an =

an bn5v cng thc ng f(x) =

x

1 x x2 . Vi x =1

4ta nhn c

n=0

an4n+1

=1

11.

T kt qu an =an bn

5ta suy ra cc ng nht thc sau y:

V d 2.3.8. Xt dy s Fibonacci a0 = 0, a1 = 1, an+1 = an + an1, n > 1.Khi ta c

(i) [Phng trnh Biner] an =an bn

5, Ln = a

n + bn [S Lucas]. Do

limn

an+1an

= a.


23

(ii) an =1

2n1[n12 ]k=0

(n

2k+1

)5k =

[n12 ]k=0

(nk1

k

).

(iii) a2n+1 = a2n + a

2n+1, a3n = a

3n + a

3n+1 a3n1.(iv) an = aan + an1, bn = ban + an1.

(v) a3n + (1)nan : a2n.(vi) Nu p > 5 l s nguyn t th ap : p v a2 + 1 = 2 : 2, a3 + 1 =

3 : 3, a5 = 5 : 5.

Bi gii: (i),(ii),(iii),(iv) v (v) u c suy ra t cng thc an =an bn

5.

Vi p > 5 l s nguyn t th t (ii) c ap =1

2p1[p12 ]k=0

(p

2k+1

)5k : p. Vi

V d 2.3.9. Xt dy Catalan a0 = 1, an+1 = a0an+a1an1 + +an1a1 +ana0, n > 0. Cng thc ng cho hm sinh ca dy l f(x) =

11 4x2x

.

Tm cng thc tnh an theo n.

Bi gii: t f(x) = a0 + a1x+ a2x2 + a3x

3 + . Khi f(x)f(x) = a20 + (a0a1 + a1a0)x+ (a0a2 + a1a1 + a2a0)x

2 + = a1 + a2x+ a3x

2 + a4x3 + = f(x) 1

x.

Vy x[f(x)]2 f(x) + 1 = 0. Gii phng trnh ny v do f(0) = 0 nn

f(x) =11 4x

2x=

1

2x 1

2x(1 4x) 12 .

Cng thc ng cho hm sinh f(x) l f(x) =11 4x

2x. Biu din

hm ny qua chui lu tha f(x) =1

2x

[2x +

22

2!x2 +

1.3

3!23x3 + +

1.3.5...(2n 3)n!

2nxx + ]. So snh h s c an = 1n+ 1

(2n

n

)vi mi

s nguyn n > 1.


24

V d 2.3.10. Dy s (an) c xc nh nh sau: a1 = 1 v an =1

n!+

a1(n 1)! +

a2(n 2)! + +

an11!vi n > 1. Xc nh cng thc ng ca

f(x) v ch ra an =F (n)(0)

n!, trong F (x) =

1ex 2 vi mi n.

Bi gii: t a0 = 1. Xt hm sinh f(x) =n=0

anxn. Khi f(x)(ex1) =(

a0+a1x+a2x2+ +anxn+

)( 11!x+

1

2!x2+ + 1

n!xn+ ) = f(x)1.Vy f(x) =

1ex 2 . Da vo Cng thc khai trin Taylor-Maclaurin ta c

an =F (n)(0)

n!, trong F (x) =

1ex 2 vi mi n.

2.4 Hm sinh m v dy s Stirling

Kt qu chnh

Nh mt s tip tc, khi nim hm sinh m s c nh ngha di y.

nh ngha 2.4.1. Cho dy s {an}. Chui lu tha hnh thc biu din trongdng f(x) =

n=0

ann!xn c gi l hm sinh m ca dy {an}.

V d 2.4.2. Vi dy s ((mn

)) hm sinh thng f(x) =

n=0

(mn

)xn c vit

thnh f(x) =n=0

m!

(m n)!n!xn =

n=0

Anmn!xn l hm sinh m ca dy (Anm).

(m l s c nh cho trc )

nh ngha 2.4.3. Cho mt tp hu hn S khc rng. Mt phn hoch caS thnh k phn, vi 1 6 k 6 n, l mt h cc tp con S1, . . . , Sk tha mnba iu kin sau y :

ki=1

Si = S

Si 6= vi mi iSi Sj = vi mi i, j, i 6= j.


25

nh ngha 2.4.4. Cho tp hu hn S khc rng. S cc phn hoch ca tpS thnh k phn c k hiu l S(n, k) v c gi l s Stirling (loi 2).

nh l 2.4.5. Vi hai s nguyn dng n, k tha mn 1 6 k 6 n ta lun c

k!S(n, k) =ki=0

(1)ki(k

i

)in.

Chng minh: Xt tp S = {a1, a2, . . . , an} v R = {1, 2, . . . , k}. TheoBi tp 1.2.22, s cc ton nh t S ln R bng k!S(n, k). Mt khc, biu

din nh x f : S R qua(

a1 a2 . . . anf(a1) f(a2) . . . f(an)

)ta c ngay

{f(a1), f(a2), . . . , f(an)} = {1, 2, . . . , k}. Vit dy s f(a1) . . . f(an) nhmt chnh hp lp chp n ca k s. Nh vy, tng ng mi nh x f

vi ng mt chnh hp lp chp n ca k s. S chnh hp lp ny ngbng kn. K hiu A l tp tt c cc nh x t S vo R v cho mi i k

hiu Ai l tp con ca A gm tt c cc nh x t S vo R \ {i}. Ta c|A| = kn, |Ai| = (k 1)n v

sj=1

Aij = (k s)n. Tp tt c cc ton nh t

S ln R ng bng A \ki=1

Ai. Theo nh l 2.4.5, ta nhn c

k!S(n, k) = |A| |ki=1

Ai| = kn (k

1

)(k 1)n

+

(k

2

)(k 2)n + (1)k

(k

k

)(k k)n.

Vy k!S(n, k) =ki=0

(1)i(ki)(k i)n = ki=0

(1)ki(ki)in.H qu 2.4.6. Vi hai s nguyn dng n, k tha mn 1 6 k 6 n ta lun c

kn =ki=0

(k

i

)i!S(n, i).

Chng minh: t ak = k!S(n, k) v bi = in. Theo nh l 2.4.5, ta c

ak =ki=0

(1)ki(ki)bi. t ck = (1)kak. Khi ck = ki=0

(1)i(ki)bi. K


26

hiu ng thc ny nh sau ck = (1 b)k v hiu l sau khi khai trin thay bibi bi.Vi k hiu hiu hnh thc, ng vi mi gi tr ca x, c th vit ngnht thc nh sau: (c+ x)k = (b+ 1 + x)k. Cho x = 1 ta c (1)kbk =(c 1)k hay bk = (1 c)k =

ki=0

(1)i(ki)ci. Vy kn = ki=0

(ki

)i!S(n, i).

nh l 2.4.7. Hm sinh m ca dy s Stirling l f(x) =n=0

S(n, k)xn

n!c

cng thc ng bng

(ex 1)kk!

.

Chng minh: Theo nh l 2.4.5, f(x) =n=0

1

n!

( 1k!

ki=0

(1)ki(ki)in)xn.Do k!f(x) =

ki=0

(1)ki(ki) n=0

(ix)n

n!=

ki=0

(1)ki(ki)eix = (ex 1)khay f(x) =

(ex 1)kk!

.

Mt vi v d

V d 2.4.8. K hiu D(n) l s cc hon v ca n phn t khng c phn tc nh, chng hn: S cc php hon v pi Sn sao cho pi(k) 6= k vi mi k.D dng ch raD(n) = n!

( 10! 1

1!+

1

2! + (1)

n

n!

). Hm sinh m ca dy

(D(n)) l f(x) =n=0

D(n)

n!xn. Xc nh cng thc ng ca f(x) v chng

minh D(n) = nD(n 1) + (1)n, D(n) = (n 1)(D(n 1) +D(n 2)).Bi gii: Do bi f(x) =

n=0

D(n)

n!xn =

n=0

( 10! 1

1!+

1

2! + (1)

n

n!

)xn

nn f(x) =( k=0

(1)kk!

xk)(

k=0

xk)

= ex.1

1 x =1

ex(1 x) .

T f(x)(1x) = ex suy ra D(n)n!D(n 1)

(n 1)! =(1)nn!v nh vyD(n) =

nD(n 1) + (1)n. T{D(n) = nD(n 1) + (1)nD(n 1) = (n 1)D(n 2) + (1)n1 suyra D(n) + D(n 1) = nD(n 1) + (n 1)D(n 2) hay D(n) = (n 1)(D(n 1) +D(n 2)).


27

V d 2.4.9. K hiu D(n) l s cc hon v ca n phn t khng c phn

t c nh. Chng minh rng

mn=0

(mn

)D(n) = m! vi mi s nguyn m > n.

Bi gii: Do bi f(x) =n=0

D(n)

n!xn =

1

ex(1 x) nn f(x)ex =

1

1 x.

Vy

( n=0

D(n)

n!xn)(

n=0

1

n!xn)

=n=0

xn. So snh h s ca xm hai v ta

c

mn=0

(mn

)D(n) = m!.

V d 2.4.10. Dy s Bell (Bn) c xc nh nh sau: Bn =nk=1

S(n, k)

vi n > 1 v B0 = 1. Xc nh cng thc ng ca f(x) =n=0

Bnn!xn.

Bi gii: D dng ch ra Bn+1 =nk=0

(nk

)Bk. t f(x) =

n=0

Bnn!xn. Vy

f(x)ex =n=0

1

n!Bn+1x

n =n=0

(n + 1)Bn+1

(n+ 1)!xn = f (x). T y suy ra f (x)

f(x)dx =

ex dx hay ln(f(x)) = ex + C. V f(0) = B0 = 1 nn

C = 1 v nh th ln(f(x)) = ex 1. Do f(x) = eex1.

2.5 Hm sinh ca dy cc a thc Bernoulli

nh ngha 2.5.1. Cc a thc Bernoulli {Bn(x)} l nhng a thc tha mnba iu kin sau y:

(i) B0(x) = 1

(ii) Bn(x) = nBn1(x) vi n > 1

(iii)

10

Bn(x)dx = 0 vi n > 1.

nh ngha 2.5.2. S Bernoulli th n l Bn(0) v c k hiu qua Bn vin = 0, 1, 2, . . . .

B 2.5.3. o hm cp s ca Bn(x) l B(s)n (x) = n(n 1) . . . (n s +

1)Bns(x) v Bn(x) l a thc bc n.


28

Chng minh: B(s)n (x) = n(n 1) . . . (n s + 1)Bns(x) c suy ra tiu kin (ii). V B

(n)n (x) = n!B0(x) = n! nn degBn(x) = n.

nh l 2.5.4. Ta c ngay cc h thc sau y: Bn(x) =ns=0

(ns

)Bsx

nsv

B0 = 1,ns=0

(n+1s

)Bs = 0, Bn Q.

Chng minh: Ta lun cBn(x) =ns=0

B(s)n (0)xs

s!. Theo B 2.5.3,Bn(x) =

ns=0

n!Bns(0)xs

s!(n s)! =ns=0

(ns

)Bsx

ns. Cho n 1, t iu kin10

Bn(x)dx = 0

ta suy ra h thc

10

( ns=0

(ns

)Bsx

ns)dx = 0 hay 1n+ 1

ns=0

(n+1s

)Bs = 0. V

B0 = 1,ns=0

(n+1s

)Bs = 0 nn Bn Q vi mi n (bng quy np).


29

T kt qu ny, d dng nhn c mt s ng nht thc v a thc sau:

B0 = 1

2B1 +B0 = 0

3B2 + 3B1 +B0 = 0

4B3 + 6B2 + 4B1 +B0 = 0

5B4 + 10B3 + 10B2 + 5B1 +B0 = 0

6B5 + 15B4 + 20B3 + 15B2 + 6B1 +B0 = 0

. . .ns=0

(n+ 1

s

)Bs = 0.

B0(x) = 1

B1(x) = x 12

B2(x) = x2 x+ 1

6

B3(x) = x3 3

2x2 +

1

2x

B4(x) = x4 2x3 + x2 1

30

B5(x) = x5 5

2x4 +

5

3x3 1

6x.

H qu 2.5.5. Ta c Bn(x) = Bn(x+ 1)Bn(x) = nxn1.Chng minh: Ta lun c h thc di y:

Bn(x+ 1) =ns=0

B(s)n (1)xs

s!=

ns=0

(n

s

)Bs(1)x

ns.

Hin nhin B0(1) = B0(0) = 1. Cho n > 2, t iu kin10

Bm(x)dx = 0

khi m > 1 suy ra 0 =10

nBn1(x)dx =10

Bn(x)dx = Bn(1) Bn(0). Dovy, khi xt b(x) = Bn(x+ 1)Bn(x), t nh l 2.5.4 ta c

b(x) =ns=0

(n

s

)[Bs(1)Bs(0)

]xns =

(n

1

)[B1(1)B1(0)

]xn1.


30

V B1(1)B1(0) = 1 nn Bn(x) = Bn(x+ 1)Bn(x) = nxn1.

V d 2.5.6. Vi s nguyn dng n, s cn1k=0

ks =1

s+ 1

sk=0

(s+1k

)Bkn

s+1k.

Bi gii: Theo H qu 2.5.5 c

n1k=0

ks =1

s+ 1

n1k=0

(Bs+1(k+1)Bs+1(k)

).

Nh vy

n1k=0

ks =1

s+ 1

(Bs+1(n) Bs+1(0)

)=

1

s+ 1

sk=0

(s+1k

)Bkn

s+1k

theo nh l 2.5.4.

V d 2.5.7. Tnh tng T =nk=0

k4 theo n.

Bi gii: Theo v d trn c

nk=0

k4 =1

5

(B5(n + 1) B5

)v nh vy nhn

c T =n5

5+n4

2+n3

3 n

30.

H qu 2.5.8. Ta c Bn =(1)nn!

det

(20

) (21

)0 ... 0(

30

) (31

) (32

)... 0

... ... ... ... ...(n0

) (n1

)... ...

(nn1)(

n+10

) (n+11

)... ...

(n+1n1)

.

Chng minh: T B0 = 1,ni=0

(n+1i

)Bi = 0, ta c h phng trnh tuyn tnh

vi cc n (B0, B1, . . . , Bn) :

(10

)B0 = 1(

20

)B0 +

(21

)B1 = 0(

30

)B0 +

(31

)B1 +

(32

)B2 = 0

...(n+10

)B0 +

(n+11

)B1 + +

(n+1n

)Bn = 0.


31

H ny c nghim (B0, B1, . . . , Bn). Tnh Bn qua nh thc ta c

Bn det

(10

)0 0 ... 0 0(

20

) (21

)0 ... 0 0(

30

) (31

) (32

)... 0 0

... ... ... ... ... ...(n0

) (n1

)... ...

(nn1)

0(n+10

) (n+11

)... ...

(n+1n1) (

n+1n

)

= det

(10

)0 0 ... 0 1(

20

) (21

)0 ... 0 0(

30

) (31

) (32

)... 0 0

... ... ... ... ... ...(n0

) (n1

)... ...

(nn1)

0(n+10

) (n+11

)... ...

(n+1n1)

0

hay Bn =(1)nn!

det

(20

) (21

)0 ... 0(

30

) (31

) (32

)... 0

... ... ... ... ...(n0

) (n1

)... ...

(nn1)(

n+10

) (n+11

)... ...

(n+1n1)

.

Xt dy s Bernoulli {Bn} vi B0 = 1,ni=0

(n+1i

)Bi = 0, n > 1. Hm sinh

m ca dy cc s Bernoulli {Bn} l B(x) =n=0

Bnxn

n!.

nh l 2.5.9. Cng thc ng ca hm sinh m B(x) ca dy cc s

Bernoulli l

x

ex 1 .

Chng minh: V

ns=0

(n+1s

)Bs = 0 nnBn+1 =

n+1s=0

(n+1s

)Bs vi n = 1, 2, . . . .

Th n + 1 qua n c Bn =ns=0

(ns

)Bs vi n = 2, 3, . . . . V B0 =

(00

)B0

v B1 =(10

)B0 +

(11

)B1 1 nn B(x) = x +

n=0

( ns=0

(ns

)Bs)xnn!

=

x+n=0

ns=0

(Bsxss!

)( xns(n s)!

)= x+B(x)ex. Vy B(x) = x

ex 1 .


32

H qu 2.5.10. Cho n 6= 1 v n l c Bn = 0.

Chng minh: V 1 +n=2

Bnxn

n!= B(x) B1x = x

ex 1 +x

2nn 1 +

n=2

Bnxn

n!=x(ex + 1)

2(ex 1) . Vx(ex + 1)

2(ex 1) l hm chn nn Bn = 0 khi n 6= 1 vn l s l.

nh l 2.5.11. [Euler] Ta c

n=0

Bn(x)

n!zn =

zexz

ez 1 .

Chng minh: V Bn(x) =ns=0

(ns

)Bsx

nstheo nh l 2.5.4 nn ta nhn

c biu din

n=0

Bn(x)

n!zn =

n=0

ns=0

(ns

)Bsx

ns

n!zn

=n=0

ns=0

Bsxns

s!(n s)!zn =

s=0

Bszs

s!

r=0

xrzr

r!.

Theo nh l 2.5.9 ta c

n=0

Bn(x)

n!zn =

z

ez 1exz =

zexz

ez 1 .

H qu 2.5.12. Ta lun c

(i)

n1s=0

(ns

)Bs(x) = nx

n1vi mi n > 2.

(ii) Bn(x+ y) =ns=0

(ns

)Bs(x)y

ns.

Chng minh: (i) Theo nh l 2.5.11, ta c biu din

n=0

Bn(x+ 1)

n!zn =

zexz

ez 1ez =

( r=0

Br(x)

r!zr)(

s=0

zs

s!

). So snh h s ca zn hai v, ta c

Bn(x+1) =ns=0

(ns

)Bs(x) hayBn(x+1) = Bn(x)+

n1s=0

(ns

)Bs(x).VBn(x+

1)Bn(x) = nxn1 theo H qu 2.5.5 nnn1s=0

(ns

)Bs(x) = nx

n1, n > 2.


33

(ii) T

ze(x+y)z

ez 1 =zexz

ez 1eyzta suy ra s bng nhau gia hai chui

n=0

Bn(x+ y)

n!zn =

( r=0

Br(x)

r!zr)(

s=0

yszs

s!

)v nh vy

n=0

Bn(x+ y)

n!zn =

r,s=0

Bs(x)yr

s!r!zs+r. So snh h t ca zn c

Bn(x+ y) =ns=0

(ns

)Bs(x)y

ns.

V d 2.5.13.

n=0

Bn(mx)

n!zn =

1

m

m1k=0

n=0

mnzn

n!Bn(x +

k

m) vi mi s

nguyn m > 1.

Bi gii: V

1

ez 1 =1 + ez + e2z + + e(m1)z

emz 1 nn theo nh l 2.5.11,

ta c biu din

n=0

Bn(mx)

n!zn =

zemxz

ez 1 =1

m

mzemxz

ez 1 hayn=0

Bn(mx)

n!zn =

1

m

emxzmz(1 + ez + + e(m1)z)emz 1

=1

m

m1k=0

mzemz(x+

k

m)

emx 1

=1

m

m1k=0

n=0

mnzn

n!Bn(x+

k

m).

Vy

n=0

Bn(mx)

n!zn =

1

m

m1k=0

n=0

mnzn

n!Bn(x+

k

m) vi mi m > 1.

V d 2.5.14. Vi hai s nguyn m,n > 1 hy tm tt c cc a thc f(x)

vi h t cao nht bng 1 tha mn mnf(mx) =1

m

m1k=0

f(x+k

m).

Bi gii: T

n=0

Bn(mx)

n!zn =

1

m

m1k=0

n=0

mnzn

n!Bn(x+

k

m) theo V d 2.5.13

ta suy ra mnBn(mx) =1

m

m1k=0

Bn(x+k

m). Vy Bn(x) tha mn u bi.


34

Tnh duy nht: Gi s c hai a thc phn bit p(x) = xn + axn1 + vq(x) = xn+bxn1+ tha mn u bi. Khi hiu (x) = p(x)q(x) =cxd + vi c 6= 0 v d < n cng tha mn u bi. V mn(mx) =1

m

m1k=0

(x +k

m) nn ta c c = mdnc hay c = 0 : v l, do m > 1 v

d < n.

V d 2.5.15. Nu c

x

ex 1 = 1 +n=1

unn!xn th cc un Q vi mi n.

Bi gii: Gi s

x

ex 1 = 1 +n=1

unn!xn. Khi , qua quy ng v gin c

x, nhn c ng nht thc 1 =(1 +

n=1

1

(n+ 1)!xn)(

1 +n=1

unn!xn). So

snh h s ca xn hai v, cunn!1!

+un1

(n 1)!2! + +u1

1!n!+

1

(n+ 1)!= 0

vi mi n > 1. Biu din dng t hp qua vic nhn hai v vi (n+ 1)! :(n+ 1

1

)un +

(n+ 1

2

)un1 + +

(n+ 1

n

)u1 + 1 = 0, n > 1,

hay vit theo kiu hnh thc (u+ 1)n+1 un+1 = 0 vi ch : Sau khi khaitrin xong phi vit uk thnh uk. T ng nht thc ny ta c h phng

trnh tuyn tnh v hn di y:

2u1 + 1 = 0

3u2 + 3u1 + 1 = 0

4u3 + 6u2 + 4u1 + 1 = 0

5u4 + 10u3 + 10u2 + 5u1 + 1 = 0

6u5 + 15u4 + 20u3 + 15u2 + 6u1 + 1 = 0

. . . .

Nh vy un = Bn Q vi mi n.

2.6 Hm sinh Dirichlet v hm Zeta-Riemann

nh ngha 2.6.1. Vi hm s hc f : N C, v s > 1, chui lu tha hnh


35

thc F (s) =n=1

f(n)

nsc gi l chui Dirichlet tng ng f. Cho dy s

{an}. Chui g(s) =n=1

annscn c gi l hm sinh Dirichlet.

Ta cng nhn hai nh l sau v tnh duy nht, tnh nhn ca chui Dirichlet:

nh l 2.6.2. Cho hai chui Dirichlet F (s) =n=1

f(n)

nsv G(s) =

n=1

g(n)

ns

tng ng cc hm s hc f, g : N C. Nu c s R sao cho F (s) =G(s) vi mi s > th f(n) = g(n) cho mi n.

nh l 2.6.3. Cho ba chui Dirichlet Fi(s) =n=1

fi(n)

nstng ng ba hm

s hc fi : N C, i = 1, 2, 3. Nu f3(n) =

u,v,uv=nf1(u)f2(v) vi mi

n N th F3(s) = F1(s)F2(s).nh l 2.6.4. Cho s s > 1 v hai dy s {an}, {bn}. Xt hai hm sinhDirichlet g(s) =

n=1

annsv h(s) =

n=1

bnns. Gi thit hai chui

n=1

|an|nsv

n=1

|bn|nshi t trong khong (s0,). Khi g(s)h(s) cng l mt hm sinhDirichlet sinh ra bi dy {cn} vi cn =

uv=n

aubv trong khong (s0,).

Chng minh: Theo php nhn cc chui ta c

g(s)h(s) = (u=1

auus

)(v=1

bvvs

) =u=1

v=1

aubv(uv)s

=n=1

uv=n

aubv

ns.

Vy g(s)h(s) l mt hm sinh Dirichlet sinh ra bi dy {cn} vi cn =uv=n

aubv.

nh l 2.6.5. Nu f l mt hm nhn thn=1

f(n)

ns=p

(i=0

f(pi)

pis), trong

tch ly theo tt c cc s nguyn t p. Chuin=1

f(n)

nsc gi l hm

sinh Dirichlet ca hm s hc f.


36

Chng minh: Khai trin tch v phi ca h thc trn ta cp

(i=0

f(pi)

pis) =

j=1

(1 +

i=1

f(pij)

pisj

)= f(pj1j1 )f(pj2j2 ) . . . f(pjrjr )

psj1j1

psj2j2

. . . psjrjr

=

f(p11 p22 . . . p

rr ), v f l hm s nhn,

=n=1

f(n)

ns, n = p

j1j1pj2j2. . . p

jrjr,

trong pt l s nguyn t th t. Nh vy h thc trn l ng.

Ch 2.6.6. Ta s dng tch hu hn Tk =ki=1

( j=1

f(pji )

pjsi

). Sau cho

k .nh ngha 2.6.7. Cho s s > 1 v dy s {an}. Chui lu tha hnh thc(s) =

n=1

1

nsc gi l hm zeta Riemann.

nh l 2.6.8. Vi s > 1 ta c

(i) 2(s) =n=1

d(n)

ns.

(ii) (s) =j=1

1

1 psj.

(iii)

1

(s)=n=1

(n)

ns.

Chng minh: (i) Theo nh l 2.6.4, 2(s) =n=1

u|n

1.1

ns=n=1

d(n)

ns.

(ii) Theo nh l 2.6.5 vi f(n) = 1 c (s) =j=1

(i=0

1

pisj) =

j=1

1

1 psj.

(iii) K hiu G(s) l hm sinh Dirichlet ca hm s hc . Khi

G(s) =n=1

(n)

ns=j=1

(i=0

(pij)

pisj) =

j=1

(1 psj ) =1

(s).


37

Nh vy

1

(s)=n=1

(n)

ns.

nh l 2.6.9. Ta c (2) =pi2

6.

Chng minh: Cng thc khai trin Fourier ca hm f(x) trn [pi, pi] :

f(x) =a02

+n=1

(an cosnx+ bn sinnx),

an =

1

pi

pipif(x) cosnx dx

bn =1

pi

pipif(x) sinnx dx

n > 1.

Vi hm s chn f(x) = x2 c x2 =a02

+n=1

an cosnx, an =2

pi

pi0

x2 cosnx dx

cho mi n 0. Vy x2 = pi2

3+ 4

n=1

(1)n cosnxn2

. Cho x = pi ta c

(2) =pi2

6.

2.7 Tch v hn

nh ngha 2.7.1. Vi dy s (an) ta t tch a1a2 . . . an . . . =n=1

an. Tch

ny c gi l mt tch v hn. K hiuAk =k

n=1an.Nu tn ti lim

n+ An =

A th A c gi l gi tr ca tch v hnn=1

an v vit A =n=1

an.

Vn t ra: Khi no tn ti gi tr ca tch v hn

n=1

an i vi mt dy

s cho trc (an).Ta c kt qu sau y ca G.M. Fichtenholz v tnh hi t ca mt tch qua

hi t ca mt tng tng ng:

nh l 2.7.2. [ Fichtenholz] Cho dy cc s dng (an) v dy cc s m(bn). Khi


38

(i) Tch v hn

i=1

(1 + an) hi t khi v ch khi tng v hni=1

an hi t.

(ii) Tch v hn

i=1

(1 + bn) hi t khi v ch khi tng v hni=1

bn hi t.

Chng minh: (i) Hin nhin, nu tch v hn

i=1

(1 + an) hoc tng v hn

i=1

an hi t th limn+ an = 0. Khi limn+

ln(1 + an)

an= 1. Do bi tch v

hn P =i=1

(1 + an) hi t khi v ch khi tng v hn lnP =i=1

ln(1 + an)

hi t. Vy vic hi t hay phn k ca

i=1

ln(1 + an) vi=1

an l tng

ng. T y suy ra tch v hn

i=1

(1 + an) hi t khi v ch khi tng v

hn

i=1

an hi t.

V d 2.7.3. Gi s dy (an) c xc nh nh sau: a1 = 1 v an+1 =(n+ 1)3 1(n+ 1)3 + 1

an vi mi n > 1. Tm limn+ an v tch v hn

n=2

(n3 1n3 + 1

).

Bi gii: Hin nhin ak =k

n=2

(n3 1n3 + 1

). Do bi n + 1 = (n + 2) 1 v

n2 + n + 1 = (n + 1)2 (n + 1) + 1 nn ak = 2(k2 + k + 1)

3k(k + 1). Do

limn+ an =

2

3v tch v hn

n=2

(n3 1n3 + 1

)=

2

3.

V d 2.7.4. t In =pi/20

sinn x dx . Khi In+1 =n

n+ 1In1 vi mi n > 1.T suy ra

(i)

pi/20

sin2n x dx =(2n 1)(2n 3) . . . 3.1

2n(2n 2) . . . 4.2pi

2pi/20

sin2n+1 x dx =2n(2n 2) . . . 4.2

(2n+ 1)(2n 1) . . . 3.1 .


39

(ii)

[ (2n)!!(2n 1)!!

]2 12n+ 1

2

3.

Bi gii: Bin i c T = 43 = 4xyz. V 1 = xy+yz+zx > 3 3

(xyz)2

nn T 6 4

3

9. Ta cn c P = 21 + 73 > 21. V (x + y + z)2 >

3(x+ yz + zx) = 3 nn P > 2

3.


47

V d 2.8.10. Chng minh rng nu a, b, c l ba s thc phn bit th c

a3(b2 c2) + b3(c2 a2) + c3(a2 b2)a2(b c) + b2(c a) + c2(a b) < a

2 + b2 + c2.

Bi gii: D thy c t v mu u chia ht cho (a b)(b c)(c a). VyV T = ab+ bc+ ca < a2 + b2 + c2 v a, b, c phn bit.

V d 2.8.11. Chng minh rng nu a, b, c l ba s thc phn bit th

a2(a+ b)(a+ c)

(a b)(a c) +b2(b+ c)(b+ a)

(b c)(b a) +c2(c+ a)(c+ b)

(c a)(c b) > 3(ab+ bc+ ca).

Bi gii: Sau khi quy ng, c t v mu u chia ht cho (ab)(bc)(ca).Vy V T = (a+ b+ c)2 > 3(ab+ bc+ ca) v a, b, c phn bit.

V d 2.8.12. Chng minh rng nu a, b, c, d R tha mn ab+ ac+ ad+bc+ bd+ cd = 0 th

a3 + b3 + c3 + d3 3(bcd+ cda+ dab+ abc) = (a+ b+ c+ d)3.

Bi gii: t

1 = a+ b+ c+ d

2 = ab+ ac+ ad+ bc+ bd+ cd

3 = abc+ abd+ acd+ bcd

4 = abcd

Nt = at + bt + ct + dt, t = 1, 2, . . . , N0 = 4.

Theo nh

l 2.8.1 c N3 N21 + N12 33 = 0. Vy N3 33 = N21 N12 =31 312. V 2 = 0 nn a3 + b3 + c3 + d3 3

(bcd+ cda+ dab+ abc

)=

(a+ b+ c+ d)3.

V d 2.8.13. a thc T = 2(x7 + y7 + z7) 7xyz(x4 + y4 + z4) c nhnt l x+ y + z.

Bi gii: t A = x + y + z, B = xy + yz + zx, C = xyz. t an =xn + yn + zn. Khi x, y, z l ba nghim ca t3 At2 + Bt C = 0v


48

an+3 = Aan+2 Ban+1 + Can vi s nguyn n > 0 v a0 = 3. Ch a0 = 3

a1 = A

a2 = A2 2B

a3 = Aa2 Ba1 + Ca0 = A3 3AB + 3C = Ak3 + 3Ca4 = Aa3 Ba2 + Ca1 = Ak4 + 2B2a5 = Ak5 5BCa6 = Ak6 B3 + 3C2a7 = Ak7 + 7B

2C.

Vy T = 2a7 7Ca4 = A(2k7 7k4C) c nhn t A = x+ y + z.

2.9 Dy truy hi vi hm sinh

V d 2.9.1. Dy s (an) xc nh bi:

{a0 = 2, a1 = 4, a2 = 31

an+3 = 4an+2 + 3an+1 18anvi mi n > 0. Chng minh rng a2010 1(mod 2011).Bi gii: t f(x) = a0 + a1x+ a2x

2 + a3x3 + . Khi c quan h

f(x)(4x+ 3x2 18x3) = f(x) 9x2 + 4x 2

hay f(x) =9x2 4x+ 2

18x3 3x2 4x+ 1 =1

1 + 2x+

1

(1 3x)2 . T y suy ra

f(x) =n=0

((2)n + (n + 1)3n)xn v c an = (2)n + (n + 1)3n vi mi

n > 0. Nh vy a2010 1(mod 2011).V d 2.9.2. Dy (an) xc nh qua a1 = 1 v an = 1.2.an1 + 2.3.an2 + + (n 1).n.a1 vi mi s nguyn n > 2. Chng minh ng nht thcan+3 4an+2 an+1 = 2

nk=1

ak khi n > 2.

Bi gii: Xt f(x) = a1x+ a2x2 + + anxn + . Khi ta c h thc

f(x)(1.2.x+ 2.3.x2 + + n.(n+ 1).xn + ) = f(x) x.


49

T

1

1 x = 1 + x + x2 + x3 + ta suy ra 1x2 + 2x3 + = x

2

(x 1)2 .

Ly o hm hai v c 1.2.x + 2.3.x2 + = 2x(x 1)3 . Vy f(x) = x +

2x2x3 3x2 + 5x 1 . T f(x)(x

33x2+5x1) = x43x3+3x2x ta nhn rav so snh h s ca xn, n > 4 hai v, nhn c an+3 = 5an+23an+1+anvi mi s nguyn n > 2. Biu din cc mi quan h trong bng h thc sau

y:

a1 = 1a2 + 5a1 = 3a3 + 5a2 3a1 = 3a4 + 5a3 3a2 + a1 = 1a5 + 5a4 3a3 + a2 = 0a6 + 5a5 3a4 + a3 = 0 = an + 5an1 3an2 + an3 = 0an+1 + 5an 3an1 + an2 = 0an+2 + 5an+1 3an + an1 = 0an+3 + 5an+2 3an+1 + an = 0.

Cng v vi v c an+3 +

4an+2 + an+1 + 2nk=1

ak = 0. Nh vy an+3 4an+2 an+1 = 2nk=1

ak.

B 2.9.3. Gi s dy (an) c hm sinh thng f(x) =n=0

anxntha mn

f(s) c xc nh vi s = 0, 1, . . . , k v l cn nguyn thy bc k ca

n v. Khi c

n=0

ank =f(1) + f(2) + + f(k1)

k.


50

Bi gii: Vi hm sinh thng f(x) = a0 + a1x+ + anxn + c

f(1) =n=0

(ank + ank+1 + + a(n+1)k1

)f() =

n=0

(ank + ank+1 + + a(n+1)k1k1

)f(2) =

n=0

(ank + ank+1

2 + + a(n+1)k12(k1))

= f(k1) =

n=0

(ank + ank+1

k1 + + a(n+1)k1(k1)(k1)).

Bi v 1+s+2s+ +s(k1) = 0 vi s = 1, 2, . . . , k1, nn khi cng kng nht thc trn ta nhn c f(1)+f(2)+ +f(k1) = k

n=0

ank.

V d 2.9.4. Vi s nguyn dng n, hy tnh tng an =nk=0

(1)k( n3k). Xttnh tun hon ca dy (an) v ch ra an : 3

[n/2]1.

Bi gii: Xt hm sinh thng f(x) ca dy (bk = (1)k(nk

)). Khi ta c

h thc f(x) =nk=0

(1)k(nk)xk = (1 x)n. Nh vy, vi = 12 + i

3

2c

an =nk=0

(1)k(n

3k

)=

nk=0

(1)3k(n

3k

)=

1

3

(f(1) + f() + f(2)

)

hay an =(1 )n + (1 2)n

3=

(32 i

3

2

)n+(3

2+ i

3

2

)n3

. Tm li

tng an =2(

3)n

3cos

npi

6=

2.33k1(1)k khi n = 6k33k(1)k khi n = 6k + 133k(1)k khi n = 6k + 20 khi n = 6k + 3

33k+1(1)k+1 khi n = 6k + 433k+2(1)k+1 khi n = 6k + 5.

D dng

suy ra (an) khng tun hon v an : 3[n/2]1.


51

V d 2.9.5. Vi mi s nguyn dng n, hy tnh tng an =nk=0

(1)k(5n5k).Bi gii: Xt hm sinh thng f(x) ca dy (bk = (1)k

(5nk

)). Ta c h thc

f(x) =nk=0

(1)k(5nk )xk = (1x)5n. Nh vy, vi = cos 2pi5 + i sin 2pi5 can =

5nk=0

(1)k(

5n

5k

)=

5nk=0

(1)5k(

5n

5k

)=

1

5

4k=0

f(k)

hay

(1 )5n + (1 2)5n + (1 3)5n + (1 4)5n5

.Vy nhn c tng

a2m =210m+1(1)m

5

[sin10m

pi

5+ sin10m

2pi

5

]v a2m+1 = 0.

V d 2.9.6. Dy (an) tha mn a1 = 12v an+1 =

(n+11

)an +

(n+12

)an1 +

+ (n+1n )a1 + 1 vi mi s nguyn n > 1. Chng minh rng a2011 nguynv chia ht cho 2011.

Bi gii: D dng kim tra

( k=0

xk

(k + 1)!

)(1 +

n=1

anxn

n!

)= 1. Nh vy

1 +n=1

anxn

n!=

x

ex 1 =n=0

Bnxn

n!theo nh l 2.5.9. Do an = Bn

vi mi n. c bit a2011 = B2011 = 0 theo H qu 2.5.10 v suy ra a2011nguyn v chia ht cho 2011.

V d 2.9.7. Dy (an) xc nh qua a1 = 1 v an = 1.2.an1 2.3.an2 + + (1)n(n 1).n.a1 vi mi s nguyn n > 2. Chng minh rng, khin > 2 lun c an+3 + 2an+2 + 5an+1 + 6

nk=1

ak = 8.

Bi gii: Xt f(x) = a1x+ a2x2 + + anxn + . Khi ta c h thc

f(x)(1.2.x 2.3.x2 + + (1)n+1n.(n+ 1).xn + ) = f(x) x.

T

1

1 + x= 1x+x2x3+ ta suy ra1+2x3x2+ = 1

(x+ 1)2.

Ly o hm hai v c 1.2.x2.3.x2+ = 2x(x+ 1)3v nh vy nhn c


52

f(x) =x4 + 3x3 + 3x2 + x

x3 + 3x2 + x+ 1. T f(x)(x3+3x2+x+1) = x4+3x3+3x2+x

ta nhn ra v so snh h s ca xn, n > 4 hai v, c a1 = 1, a2 = 2,v a3 = 2, a4 + a3 + 3a2 = 0, an+3 + an+2 + 3an+1 + an = 0 vi mi snguyn n > 2. Biu din cc mi quan h trong bng h thc sau y:

a1 = 1

a2 + a1 = 3

a3 + a2 + 3a1 = 3

a4 + a3 + 3a2 + a1 = 1

a5 + a4 + 3a3 + a2 = 0

a6 + a5 + 3a4 + a3 = 0

= an + an1 + 3an2 + an3 = 0an+1 + an + 3an1 + an2 = 0an+2 + an+1 + 3an + an1 = 0an+3 + an+2 + 3an+1 + an = 0.

Cng v vi v c an+3 + 2an+2 + 5an+1 + 6nk=1

ak = 8.

V d 2.9.8. Dy (an) xc nh bi an =1.3.5 . . . (2n+ 1)

2011n.n!vi mi s nguyn

n > 0. Tnhn=0

an.

Bi gii: Hin nhin an+1 =2n+ 3

2011(n+ 1)an vi mi n > 1. Khi n +

th

an+1an 2

2011. Nh vy chui ly tha f(x) = a0 + a1x+ a2x

2 + +anx

n + lun lun hi t. T h thc 2011(n + 1)an+1 = 2nan + 3ansuy ra 2011(n + 1)an+1x

n = 2x(nanxn1) + 3anxn. Cho n = 0, 1, 2, . . . v

ly tng tt c c 2011( k=0

ak+1xk+1)

= 2x( k=0

akxk)

+ 3( k=0

akxk).

Khi ta c h thc 2011(f(x) 1) = 2xff (x) + 3f(x) hay (2011

2x)f (x) = 3f(x). T y suy ra

f (x)f(x)

=3

2011 2x. Ly tch phn hai


53

v c ln f(x) = 32

ln(2011 2x) + a hay f(x) = (2011 2x)3/2.ea.V f(0) = 1 nn 1 = 20113/2.ea hay ea = 20113/2. Tm li ta c f(x) =

20113/2.(2011 2x)3/2. Vi x = 1 c

n=0an =

(20112009

)32 .

V d 2.9.9. Dy (an) xc nh qua a1 = 1 v an = 1an1+2an2+ +(n1)a1 vi mi s nguyn n > 2.Chng minh a3 = 3a2 v an+23an+1+an = 0vi mi s nguyn n > 2 v xc nh an theo n. T suy ra a2k+1 chia htcho 3 khi k > 1.Bi gii: Xt f(x) = a1x+ a2x

2 + + anxn + . Khi ta cf(x)(1x+ 2x2 + + nxn + ) = f(x) x.T

1

1 x = 1 + x + x2 + x3 + ta suy ra 1x + 2x2 + = x

(x 1)2 .

Vy f(x) = x +x2

x2 3x+ 1 v c f(x)(x2 3x + 1) = x3 2x2 + x.Nhn ra v so snh h s ca xn, n > 1 hai v, nhn c a1 = 1, a2 = 1,a3 = 3a2, v an+2 3an+1 + an = 0 vi mi s nguyn n > 2. T d ccng thc xc nh an.

V d 2.9.10. Dy (an) xc nh qua a1 = 1 v an+1 =2n+ 3

4(n+ 1)an vi mi

s nguyn n > 0. Tnh tng T =n=0

an.

Bi gii: Do 4(n+1)an+1 = 2nan+3an nn 4(n+1)an+1xn = 2xnanx

n1+3anx

nvi mi s nguyn n > 0. Cng tt c cc h thc ny ta nhn c

4( n=0

an+1xn+1)

= 2x( n=0

anxn)

+ 3n=0

anxn.

t f(x) =n=0

anxn. Khi 4(f 1) = 2xf + 3f hay (4 2x)f = 3f v

suy ra

f

f=

3

2

1

2 x. Nh vy (ln f) = 3

2(ln(2x)). dng c f(x) =

(2 x)3

2ea. V f(0) = a0 = 1 nn ea = 2

3

2 . Tm li f(x) =(

1 x2

)32 .

Vi x = 1 c T = f(1) = 2

2.


54

V d 2.9.11. Dy (an) xc nh qua a1 = 1 v an+1 =2n+ 5

3(n+ 2)an vi mi

s nguyn n > 0. Chng minh rngk

n=0an 0. Cng tt c cc h thcny ta nhn c

3( n=0

an+1xn+2)

= 2x( n=0

anxn+1)

+ 3n=0

anxn+1.

t f(x) =n=0

anxn+1. Khi 3(f x) = 2xf + 3f hay (3 2x)f =

3f + 3. t g(x) = f(x) + 1. Khi g

g=

3

3 2x. Nh vy (ln g) =

32

(ln(3 2x)). dng c g(x) = (3 2x)3

2ea. V g(0) = a0 = 1 nn

ea = 3

3

2 . Tm li g(x) = 1 +n=0

anxn+1 =

(1 2x

3

)32 . Vi x = 1 c

T = g(1) 1 = 33 1. Do vyk

n=0an 0. Xc nh an theo n v chng minh

n=0

an+1n!

= e.

Bi gii: Xt hm sinh m ca dy (an) l f(x) =n=0

ann!xn. Khi ta c

f =n=0

an+1n!

xn =n=0

nan 2n2 + 5n 3n!

xn

=n=0

nann!

xn n=0

2n2 5n+ 3n!

xn = xf (2x2 5x+ 3)ex.

Nh vy f = (2x3)ex v suy ran=0

an+1n!

xn =n=0

2n 3n!

xn. Vy an+1 =


55

2n 3 hay an = 2n 5 vi n > 1. Tn=0

an+1n!

xn = (2x 3)ex ta nhn

c

n=0

an+1n!

= e khi x = 1.

V d 2.9.13. Chng minh rng

nk=0

(2kk

)k + 1

(2(nk)nk

)n k + 1 =

(2(n+1)n+1

)n+ 2

.

Bi gii: Xt f(x) =k=0

(2nn

)n+ 1

xn vi 0 < |x| < 14. Bi v 1+xf(x)2 f(x)

theo V d 2.3.9 nn

nk=0

(2kk

)k + 1

(2(nk)nk

)n k + 1 =

(2(n+1)n+1

)n+ 2qua vic so snh h s

ca xn+1 hai v.

V d 2.9.14. Xt dy s hu t a1 = 1, an = (an1

1!+an2

2!+ + a1

(n 1)!)vi mi s nguyn n > 2. Tm tt c cc s nguyn dng n n!an+1 = 1.

Bi gii: Ta c an = 2an +(an1

1!+an2

2!+ + a1

(n 1)!)vi mi s

nguyn n > 2. t f(x) = a1x+ a2x2 + a3x3 + . Khi

f(x)(

2 +x

1!+x2

2!+x3

3!+

)= 2a1x+ (2a2 +

a11!

)x2 + (2a3 +a21!

+a12!

)x3 + = 2a1x+ a2x

2 + a3x3 + a4x

4 + = f(x) + x.Vy f(x)(1 + ex) = f(x) + x hay f(x) = xex. T y suy ra ng nht

a1x+ a2x2 + a3x

3 + = x(

1 x1!

+x2

2! x

3

3!+

).

Do an+1 =(1)nn!vi mi s nguyn n > 1. n!an+1 = 1 cn v nl s chn.

V d 2.9.15. Xt n > 3 s nguyn dng a1 6 a2 6 a3 6 6 an1 6 anvi tnh cht: Khng c ba s no l di ba cnh mt tam gic khng suy

bin. Xc nh gi tr nh nht ca

ana1m n c th t c.


56

Bi gii: Ta bit ba s 0 < a 6 b 6 c l di ba cnh tam gic khi vch khi c < a + b. Vy khng c ba s hng bt k ca dy ca dy l di ba cnh mt tam gic khng suy bin th ap > aq + ar vi mi p, q, r vp > q, r. V dy l dy khng gim nn ta ch cn xt ai + ai+1 6 ai+2 vii = 1, 2, . . . , 2009. Vi nh ngha

(ab

)>(cd

)khi v ch khi a > c v

b > d ta c bt ng thc(an+1an

)>(

1 11 0

)(anan1

)>(

1 11 0

)n1(a2a1

), n > 2.

Gi s

(1 11 0

)n1=

(a bc d

). Khi ta nhn c an > ca2 + da1 v

suy ra an > (c+ d)a1. Do ana1> c+ d. Do an

a1nh nht l bng c+ d

khi dy l n s hng u ca dy Fibonacci v gi tr nh nht ca t s

bng an =an bn

5vi a =

1 +

5

2v b =

152

.

V d 2.9.16. Xt dy a1 = 1, an = 12an1 + 22an2 + + (n 1)2a1 vimi s nguyn n > 2. Khi

{a2 = 4a1 3, a3 = 4a2 2a1 + 3an+3 = 4an+2 2an+1 + an, n > 2.Bi gii: t f(x) = a1x+ a2x

2 + a3x3 + . Khi tch hai chui

f(x)(

12x+ 22x2 + 32x3 + )

= 12a1x2 + (12a2 + 2

2a1)x3 + (12a3 + 2

2a2 + 32a1)x

4 + = a2x

2 + a3x3 + a4x

4 + a5x5 + = f(x) x.

T

1

1 x = 1 + x+ x2 + x3 + x4 + x5 + ta suy ra chui ly tha sau:

1

(1 x)2 = 1 + 2x+ 3x2 + 4x3 + 5x4 + 6x5 + . Do nhn c

x

(1 x)2 = 1x+ 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + v c biu din

x(1 + x)

(1 x)3 = 12x+ 22x2 + 32x3 + 42x4 + 52x5 + 62x6 + . Nh th


57

f(x)(x(1 + x)

(1 x)3)

= f(x)x hay f(x)[x32x2+4x1

]= x43x3+3x2x.T ng nht

[a1x+a2x

2+a3x3+

][x32x2+4x1

]= x43x3+3x2xsuy ra a3 = 4a2 2a1 + 3, an+3 = 4an+2 2an+1 + an vi mi n > 2.V d 2.9.17. Xt dy a1 = 1, an = 1an1+2an2 +(1)n1(n1)a1vi mi s nguyn n > 2. Khi ta c

(i) a2 = 1, a3 + 3a2 = 0, an+2 + 3an+1 + an = 0, n > 2.(ii) Tm d ca php chia an cho 3.

Bi gii: (i) t f(x) = a1x+ a2x2 + a3x

3 + . Tch hai chui ly tha

F (x) = f(x)( 1x+ 2x2 3x3 +

)= 1a1x2 + (1a2 + 2a1)x3 + (1a3 + 2a2 3a1)x4 + = a2x

2 + a3x3 + a4x

4 + a5x5 + = f(x) x.

T

1

1 + x= 1 x+ x2 x3 + x4 x5 + suy ra chui ly tha sau y:

1(1 + x)2

= 1 + 2x 3x2 + 4x3 5x4 + 6x5 . Do nhn cx

(1 + x)2= 1x+ 2x2 3x3 + 4x4 5x5 + 6x6 . Th vo F (x) c

f(x)( x

(1 + x)2

)= f(x) x hay f(x)

[x2 + 3x + 1

]= x3 + 2x2 + x. T

ng nht

[a1x+ a2x

2 + a3x3 +

][x2 + 3x+ 1

]= x3 + 2x2 + x s suy

ra ngay a1 = 1, a2 + 3a1 = 2, a3 + 3a2 = 0, an+2 + 3an+1 + an = 0, n > 2.(ii) Ta c a3 0(mod 3).V an+2+3an+1+an = 0 nn an+2+an 0(mod 3)

khi n > 2. Do , khi s nguyn k > 1 c

a2k+1 0(mod 3)a4k+2 a2 2(mod 3)a4k 1(mod 3).

V d 2.9.18. Xt dy (an), trong an =1.3.5.7 . . . (2n+ 1)

4n.n!vi mi s

nguyn n > 0. t f(x) =k=0

akxk. Tm cng thc ng v tnh f(1).


58

Bi gii: V an+1 =2n+ 3

4(n+ 1)an vi mi n nn

an+1an 1

2< 1. Vy f(x)

hi t. D dng ch ra

(f(x)a0

)= 2xf (x)+3f(x). Vy

f (x)f(x)

=3

2

1

2 x

hay f(x) =(2 x)32ec. Bi v f(0) = a0 = 1 nn ec = 23/2. Tm li

f(x) =(2 x

2

)32v f(1) = 23/2.

Mt s v d tham kho

V d 2.9.19. [VMO-1997] Cho dy s nguyn (an) ,n N c xc nhnh sau: a0 = 1, a1 = 45 v an+2 = 45an+1 7an vi mi n = 0, 1, 2, . . . .Khi hy

(i) Tnh s c dng ca a2n+1 anan+2 theo n.(ii) Chng minh rng 1997a2n + 7

n+1.4 l s chnh phng vi mi n.

V d 2.9.20. [VMO-1998-A] Cho dy s nguyn (an) ,n N c xcnh nh sau: a0 = 20, a1 = 100 v an+2 = 4an+1 + 5an + 20 vi min = 0, 1, 2, . . . . Khi hy

(i) Tm s nguyn dng h nh nht c tnh cht an+h an chia ht cho1998.

(ii) Tm s hng tng qut ca dy.

V d 2.9.21. [VMO-2011] Cho dy s nguyn (an) ,n N c xc nhnh sau: a0 = 1, a1 = 1 v an = 6an1 + 5an2 vi mi n = 2, 3, 4 . . . .Khi hy

(i) Chng minh rng a2012 2010 chia ht cho 2011.(ii) Tm s hng tng qut ca dy .


Kt lun ca lun vn

Trong lun vn tc gi trnh by c cc ni dung chnh sau y:

(1) Vnh, c ca khng, min nguyn, ng cu, trng, vnh a thc v

nghim.

(2) Vnh cc chui ly tha hnh thc, khi nim hm sinh m v hm sinh

thng cng mt vi dy s lin quan.

(3) Nghin cu mt s dy s Fibonacci, dy Catalan, dy Stirling v dy

cc a thc Bernoulli, Hm sinh Dirichlet v hm Zeta-Riemann, tch

v hn.

(4) Tnh c mt s cng thc ng ca mt s dy v chng minh ng

nht thc Newton.

Do thi gian v dung lng nn lun vn mi ch dng li mc tm hiu v

gii thiu v "Vnh cc chui lu tha hnh thc" v mt s "Hm sinh" c

bn v dy s. Trong thi gian ti, nu iu kin cho php, tc gi s nghin

cu, tm hiu k hn c th a ra mt s kt qu c tnh ng dng thc

tin hn phc v qu trnh hc tp v ging dy.

Trong qu trnh thc hin lun vn chc chn khng trnh khi thiu st.

Tc gi rt mong nhn c nhng kin ng gp ca thy c v bn b

hon thin lun vn tt hn.

Tc gi xin chn thnh cm n.

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Ti liu tham kho

[1] H.X. Snh, i s i cng, NXB Gio dc, 2001.

[2] N.V. Hi, N.K. Minh v H.Q. Vinh, Cc bi thi Olympic Ton THPT

Vit Nam (1990-2006), NXB Gio dc, 2007.

[3] R. Merris, Combinatorics, PWS publishing company 20 Park Plaza,

Boston, MA 02116-4324.

[4] K.H. Wehrahn, Combinatorics-An Introduction, Carslaw Publications

1992.

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