Números Binomiais

5.3. Cc bi ton 137

Tm li ta lun c S 1 + 12

+1

3+ ...+

1

p 2+

1

p 1(mod p)

Theo nh l Wolstenholme ta c

1 +1

2+

1

3+ ...+

1

p 2+

1

p 1 0 (mod p) S 0 (mod p)

S... p

V d 5.14. Cho cc s nguyn khng m i; j;n tho mn : i+ j n.Chng minh rng:

2nij

np=0

(n

p

)(p

i

)(p

j

)4

Li gii.Khng mt tnh tng qut gi s i j.

Ta c:(n

p

)(p

i

)(p

j

)=

(n

i

)(n in p

)(p

j

)t Aj =

np=0

(n

i

)(n in p

)(p

j

)Xt biu thc

F (x) =

nj=0

Ajxj

=

nj=0

np=0

(n

i

)(n in p

)(p

j

)xj

=

(n

i

) np=0

(n in p

) nj=0

(p

j

)xj

=

(n

i

) np=0

(n in p

)(1 + x)p

Chuyn ng Thc T Hp N Din n Ton hc

138 5.3. Cc bi ton

=

(n

i

)(1 + x)n

np=0

(n in p

)1

(1 + x)np

=

(n

i

)(1 + x)n

(1 +

1

1 + x

)ni=

(n

i

)(1 + x)i(2 + x)ni

Vy F (x) =(n

i

)(1 + x)i(2 + x)ni v Aj l h s ca xj trong khai

trin ca F (x).D thy l h s ca cc n thc ca x c bc b hn j trong khaitrin (2 + x)ni u chia ht cho 2nij

Do 2nij |Aj . y l pcm.

V d 5.15 (Australia MO). Tm gi tr k t nhin nh nht sao chos

n m : km+ n+ 1

(2n

n+m

) N

4

Li gii.

Trc ht ta chng minh1

m+ 1

(2m

m

) Z

Ta c:

1

m+ 1

(2m

m

)=

(1 m

m+ 1

)(2m

m

)=

(2m

m

) (2m)!

(m 1)!(m+ 1)!

=

(2m

m

) (2m)!

(m 1)!(m+ 1)!=

(2m

m

)(

2m

m 1

) Z

Gi s cho trc s m N. V vi n = m, s km+ n+ 1

(2n

n+m

)=

k

2m+ 1phi l s t nhin nn gi tr phi tm k N phi chia ht

cho 2m+ 1, v th k 2m+ 1.

Din n Ton hc N Chuyn ng Thc T Hp

Ti liu tham kho

[1] http://diendantoanhoc.net/forum/

[2] http://www.artofproblemsolving.com/

[3] http://www.math.net.vn

[4] http://forum.mathscope.org/

[5] 102 Combinatorial Problems, Titu Andreescu, Zuming Feng

171

170 6.5. Bi tp

Bi 2. Chng minh rng vi mi s t nhin n 1 chng ta c:nr=1

r

(n

r

)2= n

(2n 1n 1

).

Bi 3. (IMO 1989) Cho n, k l cc s nguyn dng v S l tp hpgm n im trn mt phng tha mn:(1) Khng c ba im no thuc tp S thng hng.(2) Vi mi im P thuc tp S th c t nht k im thuc Scch u vi im P.Chng minh rng: k m n bng

2m+ 1

n+m+ 1

(2n

n+m

)=

(1 nm

n+m+ 1

)(2n

n+m

)=

(2n

n+m

) (2n)!

(n+m+ 1)!(nm 1)!

=

(2

n+m

)(

2n

n+m+ 1

) Z

Vy gi tr k nh nht bng 2m+ 1.

V d 5.16 (T8/419-THTT). Tm tt c cc cp s nguyn dng n, ktha mn iu kin (

3n

n

)= 3n.nk

4

Li gii.Ta c: (

3n

n

)= 3n.nk

(3n)!n!(2n)!

= 3n.nk

(3n 2)!(3n 1).3n2n2(n 1)!(2n 1)!

= 3n.nk

(3n 2)!(n 1)!(2n 1)!

=2.3n1nk+1

3n 1

V

(3n 2)!(n 1)!(2n 1)!

=

(3n 2n 1

) Z 2.3n1.nk+1

... 3n 1 (5.4)

Li c (3, 3n 1) = 1 v (n, 3n 1) = 1 nn t (5.4) suy ra

2... 3n 1 3n 1 2 n 1


140 5.3. Cc bi ton

Do n = 1. Ta c(

3n

n

)= 3n.nk

(3

1

)= 3.1k

ng thc ny tha mn vi mi s k nguyn dng.Vy cp s (n, k) cn tm l (1, k) vi k l s nguyn dng bt k.

Nhn xt. C th gii bi ton ny bng cch khc nh sau:Vi n = 1, ta c kt qu nh trn; vi n 2, bng quy np ta chng

minh rng(

3n

n

)6... 3n nn bi ton khng tha mn.

V d 5.17 (IMO 1974). Chng minh rng vi mi s nguyn dngn th

nk=0

(2n+ 1

2k + 1

)23k 6

... 54

Li gii (1).Ta c:

nk=0

(2n+ 1

2k + 1

)23k =

nk=0

(2n+ 1

2k

)23(nk)

Mt khc v 16 chia 5 d 1 nn ta c:

23(ni) 12ni

=2i

2n(mod 5)

Suy ra 2n.nk=0

(2n+ 1

2k + 1

).23k S2n+1 (mod 5)

vi S2n+1 =nk=0

(2n+ 1

2k

)2i

Do vy, gi ta s i tnh S2n+1 =nk=0

(2n+ 1

2k

)2i

Xt hm sinh f(x) = (1 + x

2)2n+1 =2n+1i=0

(i

2n+ 1

)2i,theo nh l

RUF th:

S2n+1 =1

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

2

((1 +

2)2n+1 + (1

2)2n+1

)Din n Ton hc N Chuyn ng Thc T Hp

6.5. Bi tp 169

rng 0 = (1 1)r =ri=0

(1)i(r

i

)= 1

ri=1

(1)i1(r

i

).

Vy chng ta cri=1

(1)i1(r

i

)= 1, tc l x c m 1 ln

v phi.

Vy nguyn l b tr c chng minh.

Nhn xt.Qua mt s bi ton v v d trn cc bn c th thy, phng phpm bng hai cch c din t bng ngn ng hon ton d hiu.Bng cch chng ta c th p dng gii c nhanh gn mt s biton m cc phng php khc t ra km hu hiu v phc tp hn.

Bn cnh u im trn th phng php m bng hai cch cng cnhiu nhc im v tng i yu i vi cc bi ton phc tp (nhtng an du, tng cha phn thc). Vn quan trng trong vic sdng phng php ny l ta phi phn tch c bi di dngmt bi ton m! iu ny cn ti s quan st v kh nng nhybn ca mi ngi. Tuy nhin cn mt s cch nhn bit du hiu vchuyn i h thng cho phng php ny, song do thi gian gp rtnn tc gi cha c iu kin gii thiu n bn c trong chuyn ny.Hn gp li cc bn vo mt dp khc!

Sau cng, mi cc bn cng luyn tp vi mt s bi ton sau:

6.5 Bi tp

Bi 1. Chng minh rng vi mi s t nhin n k 1 chng ta c:

k

(n

k

)= n

(n 1k 1

)= (n k + 1)

(n

k 1

)


168 6.4. ng dng m hai cch gii cc bi ton ri rc

T chng ta c:

b

h=

(c

2

)(k

2

) = c(c 1)k(k 1)

V d 6.17 (Nguyn l b tr). Cho A1, A2, ..., An l cc tp bt k.Khi chng ta c:

ni=1

Ai

=ni=1

|Ai|

1i

142 5.3. Cc bi ton

Ta c:(n

k

)=n

k

(n 1k 1

)nn

(qpm

n

)=

(qpm

kq

)=q

kpm

(qpm1

kq1

)

Do (k, p) = 1 v(qpm1

kp1

)l s nguyn dng suy ra

(qpm

n

)... pm

M n 1 nn pm... pmn+1

T ta c iu phi chng minh. Li gii (2).Ta c: (

qpm

n

)=qpm(qpm 1)(qpm n+ 1)

n!(5.5)

Gi a v b th t l s m cao nht ca p trong phn tch tiu chun

ca t v mu trong (5.5) th(qpm

n

)... pab.

Nhn thy a m.

b =

n

p

+

n

p2

+ ...

n

pk

vi k N, pk n pk+1

n(

1

p+

1

p2+ ...+

1

pk

) 3 v tphp M = {1, 2, ..., p}. Vi mi s nguyn k tha mn 1 k p ta t


144 5.3. Cc bi ton

: Ek = {A M : |A| = k} v xk =AEk

(minA+ maxA).

Chng minh rng:

p1k=1

xk

(p

k

) 0 (mod p3)

4

Li gii.Gi s A = {m1;m2; ...;mk} Ek.Suy ra A = {p+ 1m1; ...; p+ 1mk} Ek. Ta c nhn xt sau:Nu m1 = minA th p+ 1m1 = maxA v mk = maxAth p+ 1mk = minASuy ra:

2xk =AEk

(m1 +p+1+ak+p+1ak) =AEk

2(p+1) =

(p

k

)2(p+1)

hay xk =(p

k

)(p+ 1).

Do ta cn chng minh (p+ 1)p1k=1

(p

k

)2 0 (mod p3) hay

p1k=1

(p

k

)2 0 (mod p3) (5.6)

Tht vy,ta c: (p

k

)... p 1

p

(p

k

)=

(p 1)!k!(p k)!

Do (5.6) tng ng vi:p1k=1

((p 1)!k!(p k)!

)2 0 (mod p)

t ak =(p 1)!k!(p k)!

k!ak = (p 1)(p 2)...(p k + 1) (1)k1(k 1)! (mod p) kak (1)k1 (mod p)


6.3. ng dng phng php m gii cc bi ton th 165

vo tp ny mt phn t o l xi vi qui c nu trong 4 s cchn t Ai trong c phn t xi th ng ngha vi trong 2 cpban u c 1 phn t lp li 2 ln trong 3 phn t cn li.

Sau khi chn ra 4 phn t nu ko c xi th ta c(

3

2

)= 3 cch

to chng thnh 2 cp.Nu c xi th c 3 cch gn gi tr cho xi (l mt trong 3 gi trcn li) vi mi gi tr xi ch cho ng 1 cch phn cp. Tm li

trng hp ny s c 3ni=1

(ki + 1

4

).

Bc 3: Xt biu thc v phi (m theo cch 2)

T n tp ny ta to cni=1

(ki2

)cp phn t thuc cng tp.

Do v phi l s cch chn ra 2 cp nh vy.

T y ta c iu phi chng minh.

6.3 ng dng phng php m gii cc bi ton th

V d 6.14. Trong mt bui hp c n ngi tham gia v c mt s cibt tay (mi ci bt tay to thnh t hai ngi, hai ngi bt ri thkhng bt tay li). Chng minh rng nu s ngi tham gia bt tay lmt s l th s ci bt tay c to ra l mt s chn.Bi ton trn tng ng vi bi ton sauCho th G = (V,E). Khi

2|E| =vV

deg(v).

Trong V l s nh v E l s cnh ca th. 4Li gii.

y chng ta gii bi ton trn theo phng php m bng haicch.


164 6.2. ng dng chng minh ng thc t hp

Bc 3: Xt biu thc v phi (m theo cch 2)Gi an l s tp con ca tp (1; 2; ...;n) m khng cha 2 snguyn lin tip.Xt an+1Nu phn t cui cng l n+ 1 th phn t lin trc n khngth l n, nn c an1 tp con.Nu phn t cui cng khng phi l n+ 1 th c an tp con. an+1 = an + an1D thy a0 = 1; a1 = 2 nn

an =5 + 3

5

10

(1 +

5

2

)n+

5 3

5

10

(1

5

2

)nNh vy:

bn+12 ci=0

(i

n i+ 1

)=

5 + 3

5

10

(1 +

5

2

)n+

5 3

5

10

(1

5

2

)n

V d 6.13. Cho k1; k2; ...; kn l cc s nguyn dng ln hn 1. Chngminh rng:

1i

146 5.3. Cc bi ton

Nn:

S

3m1= (n+ 3

3n 1.+ 3

(3n 1)2.2)3m

+ (n+ 3

3n 1.2 + 3

(3n 1)2.)3m

+ (n+ 3

3n 1 + 3

(3n 1)2)3m

t

(n+ 3

3n 1.+ 3

(3n 1)2.2)3m = am

(n+ 3

3n 1.2 + 3

(3n 1)2.)3m = bm(n+ 3

3n 1 + 3

(3n 1)2)3m = cm

Ch 1 + + 2 = 0; 3 = 1 v an + bn + cn l s nguyn vi mi n

Ta c: a1 + b1 + c1 = 3n; a3 + b3 + c3...3n; a5 + b5 + c5

...3n.

Gi s a2i+1 + b2i+1 + c2i+1... 3n vi mi i < k ta c:

a2k+1 + b2k+1 + c2k+1 = (a2k1 + b2k1 + c2k1)(a2 + b2 + c2)

(a2k3 + b2k3 + c2k3)(a2b2 + c2b2 + a2c2)+ a2b2c2(a2k5 + b2k5 + c2k5)

chia ht cho 3n theo gi thit quy np.

Nn theo nguyn l qui np th a2k+1 + b2k+1 + c2k+1... 3n vi mi k

nguyn dng, tc l S chia ht cho n3m.

Hay1

n3m

mk=0

(3m

3k

)(3n 1)k Z

V d 5.22 (Mongolia TST 2011). Cho p l s nguyn t. Chng minhrng

pk=0

(1)k(p

k

)(p+ k

k

) 1 (mod p3)

4

Li gii.Xt hm sinh:

f(x) =p=0

pk=0

(1)k(p

k

)(p+ k

k

)xp


6.2. ng dng chng minh ng thc t hp 163

Bc 3: Xt biu thc v phi (m theo cch 2)Gi xn l s cch chia tha mn bi ton vi n hc sinh.Xt xn+1Gi s nhm c chia cui cng c k ngi, khi c k cchchia nhm ny ( thc ra l k cch chn nhm trng), v gn+1kcch chia n k ngi trc, nn c k.gn+1k cch chia.Quy c g0 = 1, ta c: g1 = 1

gn+1 =nk=0

kgn+1k

gn+1 = 3gn gn1Kt hp vi g0 = 1, g1 = 1 ta c:

gn =15

(1 +52

)2n+1

(1

5

2

)2n+1Nh vy ta c iu phi chng minh.

V d 6.12. Tnh tng:

bn+12 ci=0

(n i+ 1

i

)4

Li gii.

Bc 2: Xt biu thc v tri (m theo cch 1)Ta m s tp con c i phn t ca tp hp (1; 2; 3; ...;n) mkhng cha hai s nguyn lin tip.Gi A l h tt c cc tp con c tnh cht nu trn v B l ttc cc tp con ca tp hp 1; 2; ...;n (r 1).Xt nh x f : A B nh sau:f : a1; a2; ...; ar b1; b2; ...; br vi bi = ai i+ 1, i = 1; r

D thy f l 1 song nh nn |A| = |B| =(n i+ 1

i

)

bn+12 ci=0

(n i+ 1

i

)l s cc tp con ca 1; 2; ...;n



+ Nu tt c cc gia nh u c con c nhn qu th c(n

n

).2n cch pht qu.

Vy tng cng cnk=0

(n

k

).2k cch pht qu.

Bc 3: Xt biu thc v phi (m theo cch 2)Mi gia nh th c 3 cch (c 2 con u khng c qu, c 2 conu c qu, hoc 1 a c qu 1 a khng c qu).

Nh vy c tt c 3n cch pht qu.T y ta c iu phi chng minh.

V d 6.11.Chng minh rng:

nk=1

(n+ k 1

2k 1

)=

15

(1 +52

)2n+1

(1

5

2

)2n+14

Li gii.

Bc 1: Pht biu li thnh bi ton m quen thuc.Mt lp hc c n hc sinh i d ngoi. C gio chia thnh mts nhm v trong mi nhm chn ra mt nhm trng tinqun l.Hi c tt c bao nhiu cch chia?

Bc 2: Xt biu thc v tri (m theo cch 1)Xt s nhm l k. M hnh ha bi ton nh sau:Ta cho trng nhm cm mt ci ct, gia 2 nhm c 1 ci ctnn c tt c 2k 1 ci ct trong n+ k 1 v tr.

Do vi k nhm th c(n+ k 1

2k 1

)cch chia.

Nh vy c tt cnk=1

(n+ k 1

2k 1

)cch chia.


5.3. Cc bi ton 147

Ta c:

f(x) =p=0

pk=0

(1)k(p

k

)(p+ k

k

)xp

=k=0

(1)k(p

k

)xpk

p=0

(p+ k

k

)xk

=k=0

(1)k(p

k

)xpk 1

(1 x)p+1

=xp

(1 x)p+1k=0

(p

k

)(1x

)k=

xp

(1 x)p+1

(1 1

x

)p= 1

1 x= 1 x x2 ...

Suy rap

k=0

(1)k(p

k

)(p+ k

k

) 1 (mod p3)

V d 5.23. Cho p l s nguyn t l. Chng minh rng

T =

pk=0

(p

k

)(p+ k

k

) (2p + 1)

... p24

Li gii.Ta c:

T =

pk=0

(p

k

)(p+ k

k

) (2p + 1) =

pk=0

(p

k

)(p+ k

k

)

(p

k=0

(p

k

)+ 1

)

T =p1k=1

(p

k

)(p+ k

p

)+ 1 +

(2p

p

)

(p1k=1

(p

k

)+ 3

)

=

p1k=1

(p

k

)((p+ k

k

) 1)

+

((2p

p

) 2)


148 5.4. Bi tp

Ta cn chng minhp1k=1

(p

k

)((p+ k

k

) 1)

... p2 th ta ch cn chng

minh(p+ k

k

) 1

... p vi 1 k p 1 v(p

k

)... p.

Tht vy :(p+ k

k

) 1 = (p+ k)!

p!.k! 1 = (p+ 1)(p+ 2)...(p+ k) k!

k!

V (p+ 1)(p+ 2)...(p+ k) k! (mod p) (p+ 1)(p+ 2)...(p+ k) k! chia ht cho p v k!.

M (p, k!) = 1 (p+ 1)(p+ 2)...(p+ k) k!... k!p

(p+ k

k

) 1

... p

p1k=1

(p

k

)[(p+ k

k

) 1]... p2

M ta li c(

2p

p

) 2

... p2 (nh l Wolstenholme, xem bi 5.1)

Do , ta c T... p2.

5.4 Bi tp

Bi 1. Cho p l s nguyn t v p 5. Chng minh rng(

2p

p

) 2

(mod p3)

Bi 2. (Putnam 1997) Cho p l s nguyn t v a, b l s dng thamn a b > 0. Chng minh(

pa

pb

)(a

b

)(mod p)

Bi 3. Cho p l s nguyn t. Chng minh rng

k = 0, p 1 :(p 1k

) (1)k (mod p)



Bc 2: Xt biu thc v tri (m theo cch 1)Sau chn ra 3 bn cho ba v tr nht, nh, ba,..., bt th thtng cn tnh chnh l s cch chn .

Bc 3: Xt biu thc v phi (m theo cch 2)Chn lun hng nht, nh,...bt ngay t n bn v b sung thmmt s bn trong n 3 bn cn li thi vng chung kt.Nu chn kiu ny ny th c n(n 1)(n 2)2n3 gii php v 1s bn chn theo cch kia chnh l 1 tp con trong n 3 c cnli.

T c kt qu cn tm l n(n 1)(n 2)2n3T y ta c iu phi chng minh.

V d 6.10.Chng minh rng:

kn

(n

k

).2k = 3n

4

Li gii.

Bc 1: Pht biu li thnh bi ton m quen thuc.C n gia nh trong 1 cng ty, mi gia nh c 2 ngi con. Nhnngy trung thu, cng ty t chc pht qu cho cc chu c kt quhc tp cao, nhng trong cng 1 gia nh khng c 2 chu nocng c nhn qu. Hi c bao nhiu cch pht qu?

Bc 2: Xt biu thc v tri (m theo cch 1)

+ Nu khng c gia nh no c con c nhn qu th c(n

0

).20 cch pht qu.

+ Nu c 1 gia inh c con c nhn qu th c(n

1

).21 cch

pht qu.

+ ... ...



Bc 2: Xt biu thc v tri (m theo cch 1)Thy ghp n hc sinh nam v n hc sinh n thnh n i. (viclm ny coi nh thc hin t u v khng nh hng g ncch chia v ca thy)

- Chn ra k i v chia cho mi i 1 v - c 2k(n

k

)cch chn

(v mi i c 1 v nn k i s c 2k kt cc khc nhau), nhvy cn li n k v v n k i cn li. Thy tip tc chn ran k

2

i v chia cho mi i 2 v - c

(n knk2

) cch.- By gi thy cn li S = n k 2

n k

2

v.

S = 0 nu n k l s chn (khi n v c chia ht)S = 1 nu n k l s l (khi chic v cn li dnh cho thy)- D thy rng k c th nhn cc gi tr t 0 n n

Bc 3: Xt biu thc v phi (m theo cch 2)Nu n v c chia ngu nhin cho 2n hc sinh v c mnh th

xy ra(

2n+ 1

n

)trng hp.

Do , theo cch chia ca thy ta c tt c:nk=0

2k(n

k

)(n knk2

)cc cch chia n v cho 2n+ 1 ngi.T y ta c iu phi chng minh.

V d 6.9. Tnhnk=3

(k 2)(k 1)k(n

k

)4

Li gii.

Bc 1: Pht biu li thnh bi ton m quen thuc.Gi s c n bn tham gia thi hi khe Ph ng vng s tuyncn chn ra mt s bn vo vng chung kt.


5.4. Bi tp 149

Bi 4. Cho p l s nguyn t v gi bt k k, a N : 0 a pk 1.Chng minh rng(

pk 1a

) (1)a (mod p)

Bi 5. Chng minh rng nu n = 2m 1 th k = 0, n :(n

k

)l s l.

Bi 6. Tm s d ca(

2009

k

)khi chia cho 2011.

Bi 7. Cho k l s t nhin chn v p l s nguyn t l. Chng minh

rng nu k khng chia ht cho p 1 thp1i=1

(p

i

)k ... pk+1Bi 8. Tm tt c s nguyn n > 1 sao cho k N \ {1} :

(kn

n

)... kn.

Bi 9. Chng minh rng:

2.1

(2000

2

)+ 3.2

(2000

3

)+ ..+ 2000.1999

(2000

1999

)... 3998000

Bi 10. Chng minh rng vi mi s t nhin n k :

CLN[(n

k

);

(n+ 1

k

); . . . ;

(n+ k

k

)]= 1



xk+1 > max {x1, x2, ..., xk}Hi c bao nhiu cch chn?

Bc 2: Xt biu thc v tri (m theo cch 1)ng vi mi xk+1 = i+ 1, (1 i n), ta c i cc chn x1, i cchchn x2, ...; i cch chn xk.Do , s cc cch chn l: S = 1k + 2k + ...+ nk

Bc 3: Xt biu thc v phi (m theo cch 2)Ta s chn ra k+ 1 s t n+ 1 s, s ln nht, ta chn lm xk+1,cc s cn li, ta sp th t l xong.Gi i(0 i k) l s cc phn t bng nhau trong nhmx1, x2, ..., xk.

Chn k i+ 1 s khc nhau t n+ 1 s, ta c(

n+ 1

k i+ 1

)cch.

Xp th t k i s khc nhau vo k ch trng (cc ch trng cnli, hin nhin dnh cho i s bng nhau), ta c Akik cch.Vy s cch chn l:

S =k1i=0

Akik

(n+ 1

k i+ 1

)

T y ta c iu phi chng minh.

V d 6.8.

nk=0

2k(n

k

)(n knk2

) = (2n+ 1n

)4

Li gii.

Bc 1: Pht biu li thnh bi ton m quen thuc.Thy Th, GVCN lp 10A gm n hc sinh nam v n hc sinhn. Ti nay, Rp chiu phim quc gia, chiu mt b phim rthay, thy nh t chc cho c lp i xem... Cui cng thy Thch mua c n v. Thy suy ngh:



Bc 2: Xt biu thc v tri (m theo cch 1)Cho 8n vin bi ny vo 4n hp, mi hp c 2 vin:u tin chn ra ng k hp sao cho mi hp c ng 1 vin bic ly ra.

S cch chn 2n 2k hp trong 4n hp l(

4n

2n 2k

) Trong mi hp trong 2n 2k hp trn ta chn ra ng 1 bi,

trong 2 vin bi c trong hp, nn s cch chn l22n2k = 4nk

Chn 2k vin bi cn li trong 2n+2k hp cn li sao cho mi

hp s c ng 2 bi c chn s l k hp nn c(

2n+ 2k

k

)cch chn.

Bc 3: Xt biu thc v phi (m theo cch 2)Ta s m s cch chn 2n vin bi t 8n vin bi ny khi c(

8n

2n

)cch.

T suy ra s cch chn 2n trong 8n vin bi theo cch m th

2 s lnk=0

4nk(

4n

2n+ 2k

)(2n+ 2k

k

)T y ta c iu phi chng minh.

V d 6.7. Chng minh rng:

1k + 2k + ...+ nk =k1i=0

Akik

(n+ 1

k i+ 1

)vi k = 1, 2, 3, ... 4Li gii.

Bc 1: Pht biu li thnh bi ton m quen thuc:T tp cc s nguyn dng A = {1, 2, ..., n+ 1}, ta chn ra bsp th t (x1, x2, ..., xk+1) tha mn iu kin:


Chng

6K thut m bng haicch chng minhng thc t hp

6.1 Nguyn l m bng hai cch 152

6.2 ng dng chng minh ng thc thp 153

6.3 ng dng phng php m gii ccbi ton th 165

6.4 ng dng m hai cch gii cc biton ri rc 167

6.5 Bi tp 169

Hong Minh Qun (batigoal)Nguyn Hin Trang (tranghieu95)

Tm tt ni dung

K thut m bng hai cch l mt phng php ph bin v c nhiu tc gi vit v n. Tuy nhin bn c hiu ti sao li giic nh th, hoc cch xy dng cc bc gii cho bi ton s dngk thut ny nh th no th nhiu bi vit li cha cp n. Trongkhun kh bi vit nh ny tc gi hy vng cung cp c phn no tng ca phng php ny ti bn c.

151

152 6.1. Nguyn l m bng hai cch

6.1 Nguyn l m bng hai cch

Cng mt s lng th kt qu m c theo hai cch lnh nhau .

Nguyn l tng chng nh rt n gin ny nhng li l khi ngunca nhiu tng gii cc bi ton t hp hay v kh. Bi vit sauy s phn tch mt s tng cho vic s dng nguyn l ny. chng minh mt ng thc t hp c dng A = B. Chng ta cth thc hin cc bc d on sau y s dng phng php mbng hai cch:

6.1.1 Cc bc thc hin

Bc 1: Pht biu li bi ton v m mt s kin quen thuc.

Bc 2: m theo v tri ca ng thc.

Bc 3: m theo v phi ca ng thc.

6.1.2 Ghi nh cn thit

Nu v tri (hoc v phi) l tng cc biu thc th cch mv tri (hoc v phi) ta chia thnh cc trng hp ring m dng quy tc cng.

Nu v tri (hoc v phi) l tch cc biu thc th cch mv tri (hoc v phi) ta chia thnh cc cng on cng honthnh m dng quy tc nhn.

Trong bi vit ny, chng ti minh ha k thut m bng hai cchthng qua cc bi ton ni ting v a phn l cc bi ton, cc nhl c tn nhm minh ha cho tng ca bi vit.Sau y l mt s ng dng ca phng php m bng hai cch.Chng ti phn tch v trnh by chi tit hai v d m u, cc v dsau tng phn tch tng t.



V d 6.5. Vi n nguyn dng cho trc. Chng minh rng:

nk=0

(2k

k

)(2n 2kn k

)= 4n

4

Li gii.

Bc 1: Pht biu li thnh bi ton m quen thuc:Mt on thng c di n c t bng 4 mu, D,X, V, T .

Bc 2: Xt biu thc v tri (m theo cch 1)Ta s chn ra mt tp cch t mu D v X sao cho D +X = k

th s cch chn s lki=0

(i

k

)(k

k i

)=

(2k

k

)khi s cch chn ra cc on mu V, T s lkj=0

(n kj

)(n kk j

)=

(2n 2kn k

)

Bc 3: Xt biu thc v phi (m theo cch 2)R rng ta c 4n cch nh th.

Do vi mi k c nh th ta c s cch t s l(

2k

k

)(2n 2kn k

)Cho k chy t 0 n n ta c s cch t mu l

ni=0

(2k

k

)(2n 2kn k

)T y ta c iu phi chng minh.

V d 6.6.

nk=0

4nk(

4n

2n+ 2k

)(2n+ 2k

k

)=

(8n

2n

); n N

4

Li gii.

Bc 1: Pht biu li thnh bi ton m quen thuc: Xt 8nvin bi.



V d 6.4. Chng minh rng vi n m thk0

(n

k

)(k

m

)=

(n

m

)2nm

4

Li gii.

Bc 1: Pht biu li bi ton m quen thuc:Gi s rng t n hc sinh ca lp hc, gio vin ch nhim cnchn ra mt i vn ngh s lng ngi ty , trong c mhc sinh cm micro. Khi gio vin ch nhim c hai phngn thc hin.


Trc ht gio vin chn ra k ngi t n ngi. Khi c(n

k

)cch chn , sau t k ngi ny s chn ly m ngi cm micro.Cho k chy t 0 n n chng ta c

k0

(n

k

)(k

m

) Bc 3: Xt biu thc v phi (m theo cch 2)

Chn ngay m hc sinh cm Micro t n hc sinh ca lp, sau chn b sung thm mt nhm ty t nm ngi cn li. Trongnm ngi ny i vi mi ngi c th c chn hoc khngc chn nn c 2nm cch chn.

Vy c thy c(n

m

)2nm

Do chng ta ck0

(n

k

)(k

m

)=

(n

m

)2nm

Nhn xt. Vi tng tng t v d trn, bn c c th chng minhng thc sau:Chng minh rng vi n,m N th

mr=0

2nr(n

r

)(m

r

)=

nr=0

(n+m r

m

)(n

r

).



6.2 ng dng chng minh ng thc t hp

V d 6.1. (Chng minh ng thc Pascal)Vi mi s nguyn dng n k 1 chng ta c(

n

k

)=

(n 1k 1

)+

(n 1k

)4

Li gii.

Bc 1: Pht biu li thnh bi ton m quen thuc:Tri h ton hc c n hc sinh tham d ban t chc cn chn rak hc sinh lm bi thi mn t hp . Nh vy ban t chc c haicch m s cch chn.

Bc 2: Xt biu thc v tri (m theo cch 1)Nu chn k hc sinh bt k trong n hc sinh th ban t chc c(n

k

)cch chn.

Bc 3: Xt biu thc v phi (m theo cch 2)Quan st v phi ta thy v phi l tng ca hai biu thc thp nn iu gi cho chng ta nh ti xt cc kh nng dng quy tc cng.Gi s Long l mt trong n hc sinh .

Phng n 1 :Nu Long c chn tham d thi mn t hp th cn chnk 1 ngi trong s n 1 ngi cn li. Khi ban t chc

c(n 1k 1

)cch chn.

Phng n 2 :Nu Long khng c chn thi mn t hp th cn chn racho k ngi trong s n 1 ngi cn li. Khi ban t

chc c(n 1k

)cch chn.



Nh vy theo nguyn l m bng hai cch chng ta c ng thc cchng minh.

V d 6.2. Chng minh rng:(n

0

)+

(n

1

)+ ...+

(n

n

)= 2n

4

Li gii.

Bc 1: Pht biu bi ton di li di dng ton m quenthuc: Tm s cch chn mt s s t n s cho trc.


+ Nu chn 0 vin c(n

0

)cch

+ Nu chn 1 vin c(n

1

)cch

+ ... ...

+ Nu chn n vin c(n

n

)cch

Vy tng cng c(n

0

)+

(n

1

)+

(n

2

)+ ...+

(n

n

)cch.

Bc 3: Xt biu thc v phi (m theo cch 2)Mi s s c 2 trng thi (c chn v khng c chn), mc n s nh vy nn c 2n cch chn.

Nh vy ta c iu cn chng minh.

V d 6.3. (Chng minh ng thc Vandermonde.)(n

0

)(m

k

)+

(n

1

)(m

k 1

)+...+

(n

k

)(m

0

)=

(m+ n

k

); vi k n m.

4Li gii.



Bc 1: Pht biu li thnh bi ton m quen thuc. Cng tyX gm n nhn vin nam v m nhn vin n cn chn ra k ngi lp thnh i tnh nguyn.

Bc 2: Xt biu thc v phi (m theo cch 1)Chn ngu nhin k ngi trong cng ty gm n+m ngi th c(m+ n

k

)cch chn.

Bc 3: Xt biu thc v tri (m theo cch 2)Quan st v tri ta thy v tri cc s hng thnh phn l tchca hai biu thc t hp nn iu gi cho chng ta nh tixt cc kh nng dng quy tc nhn.

Chn ra i nhn vin nam v ki nhn vin n th c(n

i

)(m

k i

)cch. V s ngi c chn l ty trong gii hn cho php kngi nn cho i chy t 0 n k, ta c tng tt c cc cch chnnh vy l:(

n

0

)(m

k

)+

(n

1

)(m

k 1

)+ ...+

(n

k

)(m

0

)Vy ng thc c chng minh.

Nhn xt. ng thc Vandermonde c vit thu gn nh sau:ki=0

(n

i

)(m

k i

)=

(m+ n

k

)Khi :a. Vi m = n th chng ta c ng thc quen thuc

ni=0

(n

i

)2=

(2n

n

)b. Vi (0 ki ni); i = 1, r th

k1+k2+...+kr=k

(n1k1

)(n2k2

)...

(nrkr

)=

(n1 + n2 + ...+ nr

k

)


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