CHUYN DY S (BDHSG)
1. KHI NIM DY S 1) Cho A l mt tp con khc rng ca tp s nguyn , hm s u : A
nn u(n) u= c gi l mt dy s, v k hiu l n(u ) hoc { }nu . Thng thng ta hay chn A sao cho phn t nh nht ca A l 1. Dy (un) gi l dy s hu hn (hoc dy s v hn) nu A l tp hp gm hu hn (v hn) phn t. S un c gi l s hng tng qut ca dy (un). 2) Dy s (un) c gi l dy s tng (tng khng nghim ngt, gim, gim khng nghim ngt) nu n n 1u u +< (tng ng n n 1u u + , n n 1u u +> , n n 1u u + ) vi mi n A. 3) Dy s (un) c gi l tun hon nu tn ti s nguyn dng k sao cho n k nu u , n A.+ = S k nh nht tho mn tnh cht ny c gi l chu k ca dy tun hon (un). Nu k = 1 th ta c mt dy hng (tt c cc s hng bng nhau). 4) Dy s (un) c gi l b chn trn nu tn ti s thc M sao cho nu M vi mi n A. Dy s (un) c gi l b chn di nu tn ti s thc m sao cho nu m vi mi n A. Dy s (un) c gi l b chn (hoc gii ni) nu n va b chn trn va b chn di, tc l tn ti s thc M, m sao cho nm u M vi mi n A, hoc tn ti s thc C sao cho nu C, n A. Dy s hu hn hoc tun hon th lun b chn. 2. CP S 1) Cp s cng - Dy s (un) c gi l cp s cng nu mi s hng u tho mn n 1 nu u d+ = (d: hng s, gi l cng sai). - Cng thc truy hi: n 1 nu u d.+ = + Cng thc s hng tng qut: n 1u u (n 1)d, n A.= + Cng thc tnh
tng n s hng u tin: n 1 n 1 1n n n(n 1)
S (u u ) (2u (n 1)d) nu d.2 2 2
= + = + = + Tnh cht cc s hng:
k 1 k 1 ku u 2u .+ + = 2) Cp s nhn - Dy s (un) c gi l cp s nhn nu mi s hng u tho mn n 1 nu u .q+ = (q: hng s, gi l cng bi).
- Cng thc truy hi: n 1 nu u .q.+ = Cng thc s hng tng qut: n 1
n 1u u .q .= Cng thc tnh tng n s hng
u tin: n 1S nu= nu q = 1, n 1
n 11 q
S u1 q
+=
nu q 1. Tnh cht cc s hng: 2k 1 k 1 ku .u u .+ =
3) Cp s nhn cng - Dy s (un) c gi l cp s nhn cng nu mi s hng u tho mn n 1 nu u .q d+ = + (q, d l hng s). 4) Cp s iu ho
- Dy s (un) c gi l cp s iu ho nu mi s hng ca n u khc 0 v tho mn n 1 n 1nn 1 n 1
2u uu ,
u u +
+=
+
hay n n 1 n 1
1 1 1 1( ).
u 2 u u += + (Hc sinh t n tp cc dng ton v cp s)
3. XC NH S HNG TNG QUT CA DY S 3.1. D ON S HNG TNG QUT V CHNG MINH BNG QUI NP BI TP 1) Xc nh s hng tng qut ca dy s cho bi:
n1 n 1 1 n 1 n
n
2 n1 n 1 n n 1 n 1
n
ua)u 1, u , n 1, 2,3,... b)u 2, u 2 u , n 1, 2,3,...
1 u
3 1 ( 3 1)u3 5 3 1c)u 1, u u u 1, n 1, 2,3,... d)u , u , n 1, 2,3,.
2 2 3 1 3 1 ( 3 1)u
+ +
+ +
= = = = = + =+
+ += = + + = = = =+ +
..
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2 21 n 1 n 1 n 1 n n
1 1e)u ,u 2u 1, n 1,2,3,... f )u , u 2u 1 u , n 1, 2,3,...
2 2+ += = = = = =
3.2. MT S DNG TRUY HI C BIT Vi dy s cho bi cng thc truy hi dng n 1 nu u f (n)+ = + th n 1u u f (1) f (2) ... f (n 1).= + + + + Vi dy s cho bi cng thc truy hi dng n 1 nu u .g(n)+ = th n 1u u .g(1).g(2)...g(n 1).= BI TP 2) Xc nh s hng tng qut ca dy s cho bi:
2n
1 n 1 n 1 n 1
3 2 n1 n 1 n 1 n 1 n
1 n 1 n
(n 1) ua)u 1, u u n!.n, n 1, 2,3,... b)u 1, u , n 1, 2,3,...
n(n 2)
c)u 1, u u n 3n 3n 1, n 1, 2,3,... d)u 3, u u 3 , n 1, 2,3,...
e)u 1, u u (n 1).
+ +
+ +
+
+= = + = = = =+
= = + + + = = = + =
= = + + n 1 n 1n
2
1 n 1 n 1 n 1 n
12 , n 1, 2,3,... f )u 2, u 2 , n 1, 2,3,...
u
n 1 ng)u 1, u u , n 1, 2,3,... h)u 0, u (1 u ), n 1, 2,3,...
n n 1
+
+ +
= = = =
= = = = = + =+
3.3. PHNG TRNH C TRNG Ta ch xt hai trng hp n gin sau y:
a) Xt dy s (un) cho bi 1 2 n 2 n 1 nu ,u ,u a.u b.u , n * (a,b const). + += + = Khi phng trnh 2x ax b 0 =
c gi l phng trnh c trng ca dy s cho.
Nu phng trnh trn c hai nghim thc phn bit 1 2x , x th n n
n 1 2u A.x B.x .= +
Nu phng trnh trn c hai nghim thc trng nhau 1 2x x= th n
n 1u (A nB).x .= + Nu phng trnh trn c 0 < , gi hai nghim phc ca n l 1 2x , x , v biu din hai s phc ny dng lng gic 1 2x r(cos i.sin ), x r(cos i.sin ),= + = vi r, l cc s thc, r l mun ca 1x v 2x ,
[ )0;2 , i l n v o, th nnu r (A.cos n B.sinn ).= + ( cc hng s A, B c xc nh nh 1 2u , u ) b) Xt dy s (un) cho bi 1 2 3 n 3 n 2 n 1 nu ,u ,u ,u a.u b.u c.u , n * (a,b,c const) + + += + + = c phng trnh c
trng 3 2x ax bx c 0. = Nu phng trnh trn c ba nghim thc phn bit 1 2 3x , x , x th
n n nn 1 2 3u A.x B.x C.x .= + +
Nu phng trnh trn c ba nghim thc 1 2 3x , x , x m 1 2 3x x x = th n n
n 1 2u A.x (B nC).x .= + +
Nu phng trnh trn c ba nghim thc 1 2 3x , x , x v 1 2 3x x x= = th 2 n
n 1u (A nB n C).x .= + + Nu phng trnh trn c ba nghim 1 2 3x , x , x trong 1x l nghim thc, cn hai nghim
2x r(cos i.sin ),= + 3x r(cos i.sin )= l hai nghim phc (khng phi l s thc) th n n
n 1u A.x r (B.cos n C.sinn ).= + + ( cc hng s A, B, C c xc nh nh 1 2 3u , u ,u )
VD1. Cho dy s n(u ) xc nh bi 1 2 n 2 n 1 nu 1,u 0, u u u , n *.+ += = = Chng minh n(u ) b chn.
HD. Phng trnh c trng ca dy s cho l 2x x 1 0 + = c hai nghim phc 1x cos i.sin ,3 3 = +
2x cos i.sin ,3 3
= nn nnn n
u 1 (A.cos B.sin ), n *.3 3
= + Do 1 2u 1,u 0,= = nn ta c
1
2
A B 3 A 11 A.cos B.sin ( u ) 13 3 2 2 .32 2 BA B 30 A.cos B.sin ( u ) 0 33 3 2 2
== + = + = = = + = + =
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Suy ra nn 3 n
u cos .sin , n *.3 3 3
= + Vy 2 2nn 3 n 3 2
u cos .sin 1 ( ) , n *,3 3 3 3 3
= + + = hay
n(u ) l dy b chn.
VD2. Cho n(u ) c 1 2 3 n 3 n 2 n 1 nu 0,u 16,u 47,u 7u 11u 5u , n *.+ + += = = = + Tm d khi chia 2011u cho 2011.
HD. Phng trnh c trng 3 2x 7x 11x 5 0 + = c 3 nghim thc 1x 5= (nghim n), 2 3x x 1= = (nghim
kp) do n nnu A.5 (B nC).1 , n *.= + + V 1 2 3u 0,u 16,u 47= = = nn 1
A ,B 13,C 12.5
= = = Suy ra
n 1nu 5 12n 13, n *.
= + T 20102011u 5 12.2011 13.= + Theo nh l Phcma th 20105 1 2011 (nh
l Phcma: Nu p l s nguyn t, a l s nguyn v (a, p) = 1, th p 1a 1 (mod p)). Vy 2011u chia cho 2011 d 12 (hay d 1999).
VD3. Cho hai dy s n n(x ), (y ) tho mn 1 1 n 1 n n n 1 n nx y 1, x 4x 2y , y x y , n *.+ += = = = + Xc nh cng thc ca n nx , y .
HD. Ta c n 2 n 1 n 1 n 1 n n n 1 n n n 1 n n 1 nx 4x 2y 4x 2(x y ) 4x 2x 2y 4x 2x x 4x+ + + + + + += = + = = + hay
n 2 n 1 nx 5x 6x ( n *). + += Dy n(u ) c phng trnh c trng 2x 5x 6 0 x 2 + = = hoc x = 3. Suy ra
n nnx A.2 B.3 , n *.= + M 1 2 1 1x 1, x 4x 2y 2,= = = nn
1A , B 0,
2= = v ta c n 1nx 2 , n *.
= T
v n 1n 1 n n nx 4x 2y y 2 .
+ = = Vy n 1
n nx y 2 , n *.= =
3.4. PHNG PHP DY S PH VD4. Cho dy s n 1 2 n 2 n 1 n(u ) : u 1,u 2,u 2.u u 1, n *.+ += = = + t n n 1 nv u u .+= Chng minh n(v ) l cp s cng v tm nu .
HD. a) Ta c 1 2 1v u u 1.= = V n 2 n 1 n n 2 n 1 n 1 nu 2.u u 1, n * u u u u 1, n *+ + + + += + = +
n 1 nv v 1, n *.+ = + Vy n(v ) l cp s cng vi s hng u tin 1v 1,= cng sai d = 1. b) T cu a ta c n 1v v (n 1).d 1 (n 1).1 n,= + = + = hay n 1 nu u n, n *.+ = Suy ra n n n 1 n 1 n 2u (u u ) (u u ) = + +
2
2 1 1(n 1).n n n
... (u u ) u [(n 1) (n 2) ... 2 1] 1 1 1, n *.2 2 2
+ + + = + + + + + = + = +
VD5. Cho dy s n 1 2 n 2 n 1 n2 1
(u ) : u 0, u 1,u .u u , n *.3 3+ +
= = = + t n n 1 nv u u .+= Chng minh n(v ) l
cp s nhn v tm nu .
HD. Ta c 1 2 1v u u 1.= = V n 2 n 1 n n 2 n 1 n2 1
u .u u , n * 3u 2u u , n *3 3+ + + +
= + = +
n 2 n 1 n 1 n3(u u ) (u u ), n *+ + + = n 1 n1
v v , n *.3+ = Vy n(v ) l cp s nhn vi s hng u tin
1v 1,= cng bi 1
q .3 =
Ta c n 1 n 1n 11
v v .q ( ) ,3
= = n *. Suy ra n n n 1 n 1 n 2 2 1 1 n 1 n 2u (u u ) (u u ) ... (u u ) u v v = + ++ + + = + +
n 1
2 n 22 1 1 n
11 ( )1 1 1 3 93... v v u 0 1 ( ) ( ) ... ( ) 1. , n *.
13 3 3 4 4.( 3)1 ( )3
+ + + + = + + + + + = = +
VD6. Tm s hng tng qut ca dy s ( nu ) c 1 1, . , *,+= = + + nnu c u q u an d n a, c, d, q l hng s.
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HD. Vi q = 1 th 1 , *nnu u an d n+ = + + . Ta nhn thy 2 1 3 2u u a d,u u 2a d,...,= + + = + +
n 1 n 2 n n 1 2 3 n 1 n 1 2 n 2 n 1
n 1 n
... ...
n(n 1) n(n 1)a(1 2 ... (n 2) (n 1)) (n 1)d a (n 1)d c a (n 1)d, (n *).
2 2
u u (n 2)a d,u u (n 1)a d u u u u u u u u
u u u
+ + + + = + + + + + + + + + + + = + + = + +
= + + = + +
Khi q 1, ta s xt mt dy ph ( nv ) tha mn , *,n nu v bn e n= + + b, e l nhng hng s, v ta c gng chn b, e thch hp ( nv ) l cp s nhn. T ng thc truy hi ban u ta c
1 ( ) ( ), *,+ = + + + + n nv qv qb a b n qe d b e n v d thy ( nv ) l cp s nhn th cn c
qb a b+ = qe d b e+ = 0 2 ,(q 1),1 ( 1)a d a qd
b eq q
= =
. Lc ny (vi b, e nh trn) do 1n nv qv+ =
nn ( nv ) l cp s nhn c cng bi q. Suy ra nv = 11
.n
qv = 11 2( ).( 1)naq dq du q
q+ +
, t ta tnh c s hng
tng qut ca ( nu ) l 1
1 2 2.( ). 1 ( 1)( 1)
nn
aq dq d d a qda nu u q q qq= + + + +
(n 2).
Vy, s hng tng qut ca ( nu ) cho l : 1
2 2
( 1)( 1) , 1
2
, 1.( ). 1 ( 1)( 1)
+ + ==
+ + + +
nn
n nc a n d khi q
khi qu
aq dq d d a qda nc qq qq
.
BI TP
3) Cho n(u ) : n
1 n 1n
3 2uu 0,u , n
