H THC TRUY HIGii44.Tm h thc truy hi v iu kin khi to tnh s chui nh phn di n c 3 bit 0 lin tip t Sn l s chui nh phn di n , c 3 bit 0 lin tip: mt chui di n (n>3) s c mt trong cc dng sau: A1 ( A cha 3 bit 0 lin tc, s cch l S(n-1)) B10 (B cha 3 bt 0 lin tc, s cch l S(n-2)) C100 (C cha 3 bt 0 lin tc, s cch l S(n-3) D000 (D ty di n-3, s cch 2^(n-3) ta c cng thc truy hi: Sn=S(n-1)+S(n-2)+S(n-3)+2^(n-3) khi to: S1=S2=0; S3=1. 45. Tm h thc truy hi v iu kin khi to tnh s chui nh phn di n khng c 3 bit 0 lin tip. t Sn l s chui nh phn di n , ko c 3 bit 0 lin tip: mt chui di n tha mn iu kin s c dng: A1 B10 B100 trong A,B,C c di n-1,n-2,n-3 v tha mn ko c 3 bit ko lin tip. nn ta c h thc: Sn=Sn-1 + Sn-2 + Sn-3 ( khi to: S1=2,S2=4,S3=7) 46. Tm h thc truy hi v iu kin khi to tnh s chui nh phn di n c cha chui con 01 Sn l s chui nh phn c cha chui con 01 (*) th s chui nh phn ko cha chui con 01 l 2^n-Sn. chui tha mn (*) c 1 trong 2 dng: Ax (A tha mn (*) , x = 0 hoc 1 ) B01 (B k tha mn (*)) =>Sn=2 Sn-1 + (2^(n-2)-Sn-2) =>Sn=2 Sn-1 - Sn-2 + 2^(n-2) (khi to: S1=0;S2=1) Thc ra Sn c th tnh d dng ko truy hi nh sau: chui ko tha mn (*) ch c th c dng: 111...111000...000 s chui th ny = n+1 =>Sn=2^n-(n+1) 47.Tm h thc truy hi v iu kin khi to tnh s cch i ln n bc thang nu c th i 1,2 hoc 3 bc mt ln. mun bc ln ti bc n(n>3) c 3 cch: bc ti bc n-1 ri bc 1 bc ln n bc ti n-2 ri bc 2 bc ti n bc ti n-3 ri bc 3 bc ti n vy ta xy dng c: Sn = Sn-1 + Sn-2 + Sn-3 khi to: S1=1; S2=2;

S3=3; 48.Tm h thc truy hi v iu kin khi to tnh s chui nh phn di n c mt s chn bit 0. Sn _ s chui c s chng bit 0 Kn _ s chui c s l bt 0 Sn+Kn=2^k mt chui di n c chng bit ko c thnh lp bng 2 cch * s chn bt 0 di n-1 thm 1 * s l bt 0 di n-1 thm 0 vy Sn=S(n-1)+K(n-1) tng t ta c Kn=S(n-1)+K(n-1) =>Sn=Kn =>Sn=2S(n-1) khi to : S1=1 ( s 1 ) 49.Tm h thc truy hi m Rn tho mn, trong Rn l s min ca mt phng b phn chia bi n ng thng nu khng c hai ng no song song v khng c 3 ng no cng i qua mt im trn mt phng c n (vi n>1) ng thng i mt ct nhau v ko c 3 ng ng quy. ta thy rng c mi ng thng b n-1 ng cn li chia lm n phn. bi vy nu ta b i mt ng bt k th khi mt n phn ny , 2 mt min 2 bn lp tc hp li lm mt, iu ngha l nu b i 1 ng thng th s min b mt i l n: tc l Rn=n+R(n-1), R1=2. => cng thc Rn=n(n+1)/2+1 50.Tm h thc truy hi v iu kin khi to tnh s chui k t gm A,B,C c di n cha hai k t lin tip ging nhau. Gi Sn l s chui tha mn : c cha 2 k t lin tip ging nhau () th s chui ko tha mn () l 3^n-Sn chui tha mn c 1 trong 2 dng: * Mx ( M tha (), x c 3 cch) * Nxx (Nx ko tha ()) => Sn=3 Sn-1 + (3^(n-1) - Sn-1) = 2 Sn-1 +3^(n-1) khi to: S1=0, S2=3 c th gii ko truy hi nh sau: u tin tm s chui ko c 2 k t lin tip bng nhau: chn k t u tin c 3 cch, k t tip theo c 2 cch ( tr k t trc n ) vy tt c c : 3.2^(n-1) cch => Sn=3^n-3.2^(n-1) 51.C bao nhiu byte c 3 bit 0 lin tc. 1 byte co 8 bit vay 3 bit lin tc vy 8 v tr coi cn 6 v tr vy c 6! Cch sp xp 52.Cho dy S={Sn}, Sn l s chui n bit khng cha mu 00. a)Tm h thc truy hi v cc iu kin khi to cho dy {Sn} b)Chng t Sn=f(n+1), n=1,2, vi f l dy Fibonacci.

a)chui di n tha mn ko cha 00 th c mt trong 2 dng * A1 ( A ko cha mu 00) * B10 ( B ko cha mu 00) vy Sn=S(n-1)+S(n-2) S1=2 (0;1) , S2=3 (01;10;11) b)quy np: S1=f(2); S2=f(3) ng. gi s ng n n-1 tc l Sk=f(k+1), mi k=1..n-1 =>Sn=S(n-1)+S(n-2)=f(n)+f(n-1)=f(n+1) vy Sn=f(n+1) mi n=1,2,3,.. 53.Tm h thc truy hi v iu kin khi to cho s cch ng cp ngoc n trong biu thc a1*a2**an,n>=2. # nu ch c mt th mt cch ng cho biu thc di n = a1*a2*...*an gi cng thc tnh s cch ng cp ngoc l Sn n>2: mt cch ng s c 1 trong 2 dng * Aan, vi cp ngoc trong A, A di n-1, S(n-1) cch * Aan), du m ngoc A , n-1 cch. => Sn=S(n-1)+n-1 S2=1; s3=3 s4=3+3=6 vy Sn=(n-1)+(n-2)+..+3+2+1=n(n-1)/2 (hoc mt cch ng ngoc chnh l mt t hp chp 2 ca n phn t, p s l nC2=n(n-1)/2) 54.Gi a(n) l s cc ch s 0 c trong cc s t nhin t 0 n (10^n)-1. a)Chng t rng a(n) tho mn h thc truy hi a(n)=a(n-1)+9*(n-1)*10^(n-2) b)Bit gi tr u a(1)=1, gii h thc truy hi trn. a)Gi Kn (n>0) l s cc ch s 0 c trong cc s t nhin t 10^(n-1) n 10^n-1 ( tc c ng n ch s, ch s u tin bn tri khc 0 ) ta s i chng minh Kn=9(n-1).10^(n-2) mt s c n ch s s c dng Ax, trong A c n-1 ch s v x=0..9 Kn l s cc s 0 trong cc s dng Ax s cc s 0 trong Ax bng 10 ln s cc s 0 trong A cng thm cc s 0 trong x. s cc s 0 trong A th bng 10 ln s cc s K_(n-1) ( v Ax, x c 10 cch chn) s cc s 0 dng A0 th bng s cc s dng A, s u khc 0 nn s cch chn l 9.10^(n-2) vy rt cuc ta c cng thc: Kn = 10.K_(n-1)+9.10^(n-2), t y d dng quy np c Kn = 9(n-1).10^(n-2) ( v 10.9(n-2).10^(n-3) + 9.10^(n-2) = 9(n-1).10^(n-3) ) Mt khc An=A_(n-1)+Kn ( chia t 1 n 10^n-1 thnh 2 on, t 1 n 10^(n-1)-1 v 10^(n-1) n 10^n-1. ) vy ta c h thc truy hi: An=A_(n-1) + 9(n-1).10^(n-2) b)ta ci: An=9(n-1).10^(n-2) + 9(n-2).10^(n-3) +...+ 9.1.10^0+A0 ( A0=1 ) x^n-1=(x-1) (x^(n-1) + x^(n-2) +...+ x^2 + x + 1) (x^n-1)/(x-1)=(x^(n-1) + x^(n-2) +...+ x^2 + x + 1) o hm 2 v (n.x^(n-1).(x-1)-(x^n-1))/(x-1)^2=(n-1 + (n-2).x^(n-3) +...+ 2x + 1

thay x=10 ta c: (n.10^(n-1).9-(10^n-1))/9^2 = (n-1).10^(n-2) + (n-2).10^(n-3) +..+ 2.10 + 1 => An = (n.10^(n-1).9-(10^n-1))/9+1 = n.10^(n-1) - (10^n-1)/9 + 1 55.Gi an l s dy bit di n khng c 2 bit 0 k nhau. a)Tm h thc truy hi cho an b)Bit gi tr u a1=2,a2=3, gii h thc truy hi trn. mt dy di n ko c 2 bit 0 k nhau th c 1 trong 2 dng * A10 ( A c n-2 bit v ko c 2 bit 0 k nhau ) * B1 ( B c n-1 bt v ko c 2 bit 0 k nhau ) vy suy ra an=a(n-1)+a(n-2) a1=2; a2=3 dy ny bt u t a1=2 l dy fibonaci gii truy hi bng cch xt nghim ca phng trnh c trng x^2=x+1 x1=(1+5)/2 ; x2=(1-5)/2 an=c1.x1^n+c2.x2^n thay n=1,2 vo tm c1,c2 56.Gi an l s dy bit di n c 2 bit 0 k nhau. a)Tm h thc truy hi cho an b)Bit gi tr u a0=0 v a1=0, tnh s dy bit di 7 c 2 bit 0 k nhau. a)mt s c 2 bt 0 k nhau th c 1 trong 3 dng * A00 (A bt k di n-2 , s trng hp l 2^(n-2)) * B10 (B cha 2 bt 0 lin nhau di n-2 , s trng hp l a(n-2)) * C1 (C cha 2 bt 0 lin nhau di n-1, s trng hp l a(n-1)) vy an=a(n-1)+a(n-2)+2^(n-2) b)a0=0,a1=0 =>a2=0+0+2^0=1 =>a3=1+0+2^1=3 =>a4=3+1+2^2=8 =>a5=8+3+2^3=19 =>a6=19+8+2^4=43 =>a7=43+19+2^5=94

NGUYN L DIRICHLET1.Trong mt phng Oxy ly ngu nhin 5 im ta nguyn. Chng t rng c t nht mt trung im ca cc on ni chng c ta nguyn. 5 im nn c t nht 3 im c tung cng tnh cht chn l ( chia 2 nhm: tung chn v l, c t nht 1 nhm cha khng t hn 3 im, 3 im tha mn ) trong 3 im c t nht 2 im c honh cng tnh cht chn l. vy ta c 2 im m tung v honh cng tnh cht chn l, trung im 2 im ny tt nhin c ta

nguyn! 2.Trong mt phng cho 6 im phn bit ni nhau tng i mt bi cc on thng mu xanh hoc . Chng t rng c 3 im ni nhau bi cc on thng cng mu. xt 1 im A bt k trong 6 im, t im ny c 5 on thng , nn phi c t nht 3 on thng cng mu AB,AC,AD. gi s 3 on ny c mu chng hn. khi *nu 1 trong 3 on BC,CD,DB c mu th ta c tam gic cng mu ( vi 2 trong 3 on kia) *nu c 3 on BC,CD,DB u xanh th ta c tam gic mu xanh 3.Trong mt phng cho 17 im phn bit ni nhau tng i mt bi cc on thng mu xanh, hoc , hoc vng.Chng t rng c 3 im ni nhau bi cc on thng cng mu. ly 1 im A t 17 im, t A c 16 on thng t bng 3 mu nn c t nht 6 on cng mu. gi s mu . 6 im u kia nu c 2 im ni bng mu na th r rng to thnh tam gic mu . cn khng th 6 im ny c ni vi nhau bng 2 mu cn li ( xanh, vng ) theo bi 2 th tn ti 3 im ni vi nhau cng mu. vy bi 3 c chng minh 4.Mt thng cha 10 qu bng mu xanh v 10 qu bng mu . Phi ly ngu nhin t nht bao nhiu qu bng bo m c 3 qu bng cng mu.p s: 5 qu. chng minh: nu ly 5 qu th tn ti 3 qu cng mu tt nhin nu ly 4 qu th trng hp 2 xanh 2 khng tha mn, vy phi ly t nht 5 qu, ly