TỔ HỢP - XÁC SUẤT

Chuyn 11

T Hp - Xc Sut

1. Hon V - Chnh Hp - T Hp11.1. (B-05) Mt i thanh nin tnh nguyn c 15 ngi, gm 12 nam v 3 n. Hi c bao nhiu cch phn cngi v gip ba tnh min ni sao cho mi tnh c 4 nam v 1 n.

Li gii. Phn cng i thanh nin tnh nguyn theo th t tnh th nht; tnh th hai v tnh th ba.Phn cng cc thanh nin tnh nguyn v tnh th nht c C412.C

13 cch.

Phn cng cc thanh nin tnh nguyn v tnh th hai c C48 .C12 cch.

Phn cng cc thanh nin tnh nguyn v tnh th ba c C44 .C12 cch.

Vy c C412.C13 .C

48 .C

12 .C

44 .C

12 = 207900 cch phn cng i thanh nin tnh nguyn v gip ba tnh min ni.

11.2. (D-06) i thanh nin xung kch ca trng ph thng c 12 hc sinh, gm 5 hc sinh lp A, 4 hc sinh lpB v 3 hc sinh lp C. Cn chn bn hc sinh i lm nhim v sao cho bn hc sinh ny thuc khng qu hai lp.Hi c bao nhiu cch chn.

Li gii. S cch chn 4 hc sinh thuc khng qu hai lp bng s cch chn 4 hc sinh bt k tr i s cch chn4 hc sinh c mt c hc sinh ba lp.

Chn 4 hc sinh bt k trong tng s 12 hc sinh c C412 = 495 cch.Chn 4 hc sinh c mt c hc sinh ba lp c cc trng hp:TH1: Chn 2 hc sinh lp A, 1 hc sinh lp B, 1 hc sinh lp B c C25 .C

14 .C

13 = 120 cch.

TH2: Chn 1 hc sinh lp A, 2 hc sinh lp B, 1 hc sinh lp B c C15 .C24 .C

13 = 90 cch.

TH3: Chn 1 hc sinh lp A, 1 hc sinh lp B, 2 hc sinh lp B c C15 .C14 .C

23 = 60 cch.

Do chn 4 hc sinh c mt c hc sinh ba lp c 120 + 90 + 60 = 270 cch.Vy chn 4 hc sinh thuc khng qu hai lp c 495 270 = 225 cch.

11.3. (B-04) Trong mt mn hc, thy gio c 30 cu hi khc nhau gm 5 cu hi kh, 10 cu hi trung bnh v15 cu hi d. T 30 cu hi c th lp c bao nhiu kim tra, mi gm 5 cu hi khc nhau, sao chotrong mi phi c 3 loi cu hi (kh, trung bnh v d) v s cu hi d khng t hn 2.

Li gii. Lp mt kim tra tha mn yu cu bi ton c cc trng hp sau:TH1: gm 2 cu d, 2 cu trung bnh v 1 cu kh c C215.C

210.C

15 = 23625 cch.

TH2: gm 2 cu d, 1 cu trung bnh v 2 cu kh c C215.C110.C

25 = 10500 cch.

TH3: gm 3 cu d, 1 cu trung bnh v 1 cu kh c C315.C110.C

15 = 22750 cch.

Vy c 23625 + 10500 + 22750 = 56875 cch lp kim tra.

11.4. Mt hp ng 4 bi , 5 bi trng v 6 bi vng. Ngi ta chn ra 4 vin bi t hp . Hi c bao nhiu cchchn s bi ly ra khng ba mu.

Li gii. S cch chn 4 bi khng ba mu bng s cch chn 4 bi bt k tr s cch chn 4 bi c ba mu.Chn 4 bi bt k trong tng s 15 bi c C415 = 1365 cch.Chn 4 bi c ba mu c cc trng hp sau:TH1: Chn 2 bi , 1 bi trng, 1 bi vng c C24 .C

15 .C

16 = 180 cch.

TH2: Chn 1 bi , 2 bi trng, 1 bi vng c C14 .C25 .C

16 = 240 cch.

TH3: Chn 1 bi , 1 bi trng, 2 bi vng c C14 .C15 .C

26 = 300 cch.

Do chn 4 bi c ba mu c 180 + 240 + 300 = 720 cch.Vy chn 4 bi khng ba mu c 1365 720 = 645 cch.

11.5. Chng minh cc h thc saua) An+2n+k + A

n+1n+k = k

2Ann+k. b) k (k 1)Ckn = n (n 1)Ck2n2.

c) PkA2n+1A2n+3A

2n+5 = nk!A

5n+5. d) (B-08)

n + 1

n + 2

(1

Ckn+1+

1

Ck+1n+1

)=

1

Ckn.

1

GV: Le Ngoc Sn Chuyen e luyen thi H - Co ap an chi tiet Lp 11K2

Li gii.

a) An+2n+k + An+1n+k =

(n + k)!

(k 2)! +(n + k)!

(k 1)! =k(k 1)(n + k)!

k!+k(n + k)!

k!=k2(n + k)!

k!= k2Ann+k (pcm).

b) V T = n (n 1)Ck2n2 =n(n 1)(n 2)!(n k)!(k 2)! =

k(k 1)n!(n k)!k! = k (k 1)C

kn = V P (pcm).

c) V T = PkA2n+1A2n+3A

2n+5 = k!

(n + 1)!

(n 1)! .(n + 3)!

(n + 1)!.(n + 5)!

(n + 3)!= k!

n(n + 5)!

n!= nk!A5n+5 = V P (pcm).

d) V T =n + 1

n + 2

(1

Ckn+1+

1

Ck+1n+1

)=n + 1

n + 2

(k!(n + 1 k)!

(n + 1)!+

(k + 1)!(n k)!(n + 1)!

)=n + 1

n + 2

(k!(n k)!(n + 1 k + k + 1)

(n + 1)n!

)=k!(n k)!

n!=

1

Ckn= V P (pcm).

11.6. Gii phng trnh, bt phng trnh v h phng trnha) PxA2x + 72 = 6

(A2x + 2Px

). b) 12A

22x A2x 6xC3x + 10.

c){

2Ayx + 5Cyx = 90

5Ayx 2Cyx = 80 . d) C22x + C

42x + ... + C

2x2x 22003 1.

e) A3n + 2Cn2n 9n. f) C1x + 6C2x + 6C3x = 9x2 14x.

Li gii.a) iu kin: x Z, x 2. Ta c phng trnh tng ng:

PxA2x + 72 6A2x 12Px = 0 Px

(A2x 12

) 6 (A2x 12) = 0 (A2x 12) (Px 6) = 0[A2x 12 = 0Px 6

[x(x 1) 12 = 0x! = 6

x = 4x = 3 (loi)x = 3

Vy phng trnh c nghim x = 3, x = 4.b) iu kin: x Z, x 3. Ta c bt phng trnh tng ng:

1

2(2x)(2x 1) x(x 1) 6

x

x(x 1)(x 2)3!

+ 10 2x2 x x2 + x x2 3x + 2 + 10 x 4

Kt hp iu kin bt phng trnh c nghim x = 3, x = 4.c) iu kin: x, y Z, x y 0. Ta c h phng trnh tng ng:{

2Ayx + 5Cyx = 90

5Ayx 2Cyx = 80 {

Ayx = 20Cyx = 10

{

y!Cyx = 20Cyx = 10

{

y = 2x = 5

Vy h cho c nghim (x; y) = (5; 2).d) iu kin: x N.Xt khai trin (1 + a)2x = C02x + C

12xa + C

22xa

2 + C32xa3 + ... + C2x12x a

2x1 + C2x2xa2x.

Ln lt chn a = 1 v a = 1 ta c:

22x = C02x + C12x + C

22x + C

32x + ... + C

2x12x + C

2x2x (1)

0 = C02x C12x + C22x C32x + ... + (1)2x1C2x12x + (1)2xC2x2x (2)

Cng theo v (1) v (2) ta c: 22x = 2(C02x + C

22x + C

42x + ... + C

2x2x

) C22x + C42x + ... + C2x2x = 22x1 1.Do bt phng trnh cho tng ng vi 22x1 1 22003 1 x 1002.Vy bt phng trnh c tp nghim S =

{x Nx 1002}.

e) iu kin: n Z, n 3. Ta c bt phng trnh tng ng:

A3n + 2C2n 9n n(n 1(n 2) + 2

n(n 1)2!

9n n(n2 2n 8) 0 2 n 4

Kt hp iu kin bt phng trnh c nghim x = 3, x = 4.f) iu kin: x Z, x 3. Ta c phng trnh tng ng:

x + 6x(x 1)

2!+ 6

x(x 1)(x 2)3!

= 9x2 14x x(x2 9x + 14) = 0 x = 0 (loi)x = 2 (loi)x = 7

Vy phng trnh c nghim x = 7.

2


Chuyn 11. T Hp - Xc Sut

11.7. (D-05) Tnh gi tr ca M =A4n+1 + 3A

3n

(n + 1)!bit C2n+1 + 2C

2n+2 + 2C

2n+3 + C

2n+4 = 149.

Li gii. iu kin: n Z, n 3. iu kin cho tng ng vi(n + 1)n

2!+ 2

(n + 2)(n + 1)

2!+ 2

(n + 3)(n + 2)

2!+

(n + 4)(n + 3)

2!= 149

n2 + 4n 45 = 0[n = 5n = 9 (loi)

Do ta c: M =A46 + 3A

35

6!=

3

4.

11.8. (B-06) Cho tp A gm n phn t (n 4). Bit rng s tp con 4 phn t bng 20 ln s tp con gm 2 phnt. Tm k {1, 2, ..., n} sao cho s tp con gm k phn t ca A l ln nht.Li gii. S tp con k phn t ca A l Ckn. Do theo gi thit ta c:

C4n = 20C2n

n(n 1)(n 2)(n 3)4!

= 20n(n 1)

2! n(n 1) (n2 5n 234) = 0 n = 18

D thy c C018 < C118 < C

218 < ... < C

918 > C

1018 > ... > C

1818 .

Vy s tp con gm 9 phn t l ln nht.

11.9. (B-02) Cho a gic u A1A2...A2n ni tip ng trn (O). Bit rng s tam gic c cc nh l 3 trong 2nnh nhiu gp 20 ln s hnh ch nht c cc nh l 4 trong 2n nh. Tm n.

Li gii. S tam gic c cc nh l 3 trong 2n nh a gic l C32n.Gi ng cho i qua tm O ca a gic l ng cho ln th a gic c n ng cho ln.Mi hnh ch nht c cc nh l 4 trong 2n nh a gic c hai ng cho l hai ng cho ln. Ngc li hai

ng cho ln lun to ra mt hnh ch nht. Do s hnh ch nht bng s cp ng cho ln v bng C2n.

Theo gi thit ta c: C32n = 20C2n

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

= 20n(n 1)

2! n(n 1)(2n 16) = 0 n = 8.

2. Xc Sut11.10. (B-2012) Trong mt lp hc gm c 15 hc sinh nam v 10 hc sinh n. Gio vin gi ngu nhin 4 hc sinhln bng gii bi tp. Tnh xc sut 4 hc sinh c gi c c nam v n.

Li gii. Php th l chn 4 hc sinh bt k ln bng nn s phn t khng gian mu l: || = C425 = 12650.Gi A l bin c: "Chn 4 hc sinh ln bng c c nam v n" ta c |A| = C315.C110 +C215.C210 +C115.C310 = 11075.Vy xc sut 4 hc sinh c gi c c nam v n l: P (A) =

|A||| =

11075

12650=

443

506.

11.11. Mt hp ng 4 vin bi , 5 vin bi trng v 6 vin bi vng. Ngi ta chn ra 4 vin bi t hp . Tnhxc sut trong s bi ly ra khng c ba mu.

Li gii. Php th l chn 4 bi bt k trong tng s 15 bi nn s phn t khng gian mu l: || = C415 = 1365.Gi A l bin c: "Chn 4 bi khng c ba mu, suy ra A l bin c: "Chn 4 bi c ba mu".

Ta c: |A| = C24 .C15 .C16 + C14 .C25 .C16 + C14 .C15 .C26 = 720 P(A)

=|A||| =

720

1365=

48

91.

Vy xc sut trong s bi ly ra khng c ba mu l: P (A) = 1 P (A) = 1 4891

=43

91.

11.12. Mt t c 9 nam v 3 n. Chia t thnh 3 nhm mi nhm gm 4 ngi. Tnh xc sut khi chia ngunhin nhm no cng c n.

Li gii. Php th l chia t thnh ba nhm mi nhm gm 4 ngi nn || = C412.C48 .C44 = 34650.Gi A l bin c: "Nhm no cng c n" ta c: |A| = C39C13 .C36C12 .C33C11 = 10080.Vy xc sut nhm no cng c n l: P (A) =

|A||| =

10080

34650=

16

55.

11.13. Mt t c 13 hc sinh, trong c 4 n. Cn chia t thnh ba nhm, nhm th nht c 4 hc sinh, nhmth hai c 4 hc sinh, nhm th ba c 5 hc sinh. Tnh xc sut mi nhm c t nht mt hc sinh n.

3


Li gii. Php th l chia t thnh ba nhm gm 4 hc sinh, 4 hc sinh v 5 hc sinh nn || = C413.C49 .C55 = 90090.Gi A l bin c: "Mi nhm c t nht mt hc sinh n" ta c:

|A| = C39C14 .C36C13 .C33C22 + C39C14 .C36C23 .C33C11 + C39C24 .C36C12 .C33C11 = 60480

Vy xc sut mi nhm c t nht mt hc sinh n l: P (A) =|A||| =

60480

90090=

96

143.

11.14. C hai hp ng bi. Hp mt c 7 bi xanh v 3 bi , hp hai c 6 bi xanh v 4 bi . Ly ngu nhin mihp mt bi. Tm xc sut c t nht mt bi .

Li gii. Php th l ly mi hp mt bi nn s phn t khng gian mu l: || = C110.C110 = 100.Gi A l bin c: "Ly c t nht mt bi ", suy ra A l bin c: "Ly c hai bi xanh".

Ta c: |A| = C17 .C16 = 42 P(A)

=|A||| =

42

100=

21

50.

Vy xc sut ly c t nht mt bi l: P (A) = 1 P (A) = 1 2150

=29

50.

11.15. C hai hp cha cc vin bi ch khc nhau v mu. Hp th nht cha ba bi xanh, hai bi vng v mt bi. Hp th hai cha hai bi xanh, mt bi vng v ba bi . Ly ngu nhin t mi hp mt vin bi. Tnh xc sut ly c hai bi xanh.

Li gii. Php th l ly mi hp mt bi nn s phn t khng gian mu l: || = C16 .C16 = 36.Gi A l bin c: "Ly c hai bi xanh", ta c: |A| = C13 .C12 = 6.Vy xc sut ly c hai bi xanh l: P (A) =

|A||| =

6

36=

1

6.

11.16. Mt ngi gi in thoi, qun hai ch s cui v ch nh rng hai ch s phn bit. Tnh xc sut ngi gi mt ln ng s cn gi.

Li gii. Gi hai ch s cui ca s in thoi l ab. V a, b phn bit nn c tt c 90 s. Ngi ch gi mt lnnn xc sut l 190 .

11.17. Ngi ta s dng 5 cun sch Ton, 6 cun sch L, 7 cun sch Ho (cc cun sch cng loi ging nhau), lm gii thng cho 9 hc sinh, mi hc sinh c hai cun sch khc loi. Trong s hc sinh c hai bn Ngc vTho. Tm xc sut hai bn Ngc v Tho c gii thng ging nhau.

Li gii. Chia 18 cun sch thnh 9 b sch, mi b hai cun khc loi. Gi x, y, z ln lt l s b Ton-L,Ton-Ha v L-Ha, ta c:

x + y = 5

x + z = 6

y + z = 7

x = 2

y = 3

z = 4

Xt php th l pht 9 b sch cho 9 hc sinh ta c: || = C29 .C37 .C44 = 1260.Gi A l bin c: "Ngc v Tho c gii thng ging nhau" ta c cc trng hp:TH1: Ngc v Tho cng nhn b sch Ton-L c C37C

44 = 35 cch pht b sch cho 7 ngi cn li.

TH2: Ngc v Tho cng nhn b sch Ton-Ha c C27C15C


TH3: Ngc v Tho cng nhn b sch L-Ha c C27C35C


Do s phn t ca bin c A l |A| = 35 + 105 + 210 = 350.Vy xc sut hai bn Ngc v Tho c gii thng ging nhau l: P (A) = |A||| =

3501260 =

518 .

11.18. Mt nhm hc tp gm 7 nam v 5 n, trong c bn nam A v bn n B. Chn ngu nhin 6 bn lpmt i tuyn thi hc sinh gii. Tnh xc sut i tuyn c 3 nam v 3 n, trong phi c hoc bn nam A,hoc bn n B nhng khng c c hai.

Li gii. Php th l chn 6 hc sinh trong tng s 12 hc sinh nn || = C612 = 924.Gi A l bin c: "i tuyn c 3 nam v 3 n, trong phi c hoc bn nam A, hoc bn n B nhng khng

c c hai".Ta c: |A| = C26 .C34 + C36 .C24 = 180.Vy xc sut cn tm l: P (A) =

|A||| =

180

924=

15

77.

11.19. C hai ti. Ti th nht cha 3 tm th nh s 1, 2, 3 v ti th hai cha 4 tm th nh s 4, 5, 6, 8.Rt ngu nhin t mi ti mt tm th ri cng hai s ghi trn hai tm th vi nhau. Gi X l s thu c. Lpbng phn b xc sut ca X v tnh E(X).

4



Li gii. Ta c bng phn b xc sut:

X 5 6 7 8 9 10 11P 112

212

312

212

212

112

112

K vng l E(X) = 7, 75.

3. Nh Thc Newton11.20. (D-04) Tm s hng khng cha x trong khai trin thnh a thc ca biu thc

(3x + 14x

)7, x > 0.

Li gii. Ta c:(

3x + 14x

)7=(x

13 + x

14

)7=

7k=0

Ck7

(x

13

)7k(x

14

)k=

7k=0

Ck7x73 k3 x

k4 =

7k=0

Ck7x73 7k12 .

S hng khng cha x tng ng s hng cha k tha7

3 7k

12= 0 k = 4.

Vy s hng khng cha x l C47 = 35.

11.21. (D-07) Tm h s ca x5 trong khai trin thnh a thc ca biu thc x(1 2x)5 + x2(1 + 3x)10.

Li gii. Ta c: x(1 2x)5 + x2(1 + 3x)10 = x5

k=0

Ck5 (2x)k + x210i=0

Ci10(3x)i

=5

k=0

Ck5 (2)kxk+1 +10i=0

Ci103ixi+2.

S hng cha x5 tng ng s hng cha k v i tha{

k + 1 = 5i + 2 = 5

{

k = 4i = 3

.

Vy h s ca s hng cha x5 l C45 (2)4 + C31033 = 3320.

11.22. (A-04) Tm h s ca x8 trong khai trin thnh a thc ca biu thc(1 + x2 (1 x))8.

Li gii. Ta c:(1 + x2 (1 x))8 = 8

k=0

Ck8[x2(1 x)]k = 8

k=0

Ck8x2k(1 x)k

=8

k=0

Ck8x2k

(ki=0

Cik(x)i)

=8

k=0

ki=0

Ck8x2kCik(1)ixi =

8k=0

ki=0

Ck8Cik(1)ix2k+i

S hng cha x8 tng ng s hng cha k v i tha 2k + i = 8.

V 0 i k 8 nn 2k + i = 8{

k = 3i = 2

hoc{

k = 4i = 0

.

Vy h s ca s hng cha x8 l C38C23 (1)2 + C48C04 (1)0 = 238.

11.23. Tm h s ca x4 trong khai trin a thc P (x) =(1 + 2x + 3x2

)10.Li gii. Ta c: P (x) =

(1 + 2x + 3x2

)10=

10k=0

Ck10(2x + 3x2

)k=

10k=0

Ck10

(ki=0

Cik(2x)ki

(3x2)i)

=10k=0

ki=0

Ck10Cik(2x)

ki(3x2)

i=

10k=0

ki=0

Ck10Cik2

ki3ixk+i

S hng cha x4 tng ng s hng cha k v i tha k + i = 4.

V 0 i k 10 nn k + i = 4{

k = 4i = 0

,

{k = 3i = 1

hoc{

k = 2i = 2

.

Vy h s ca s hng cha x4 l C410C042

430 + C310C132

231 + C210C222

032 = 8085.

11.24. t(1 x + x2 x3)4 = a0 + a1x + a2x2 + ... + a12x12. Tnh h s a7.

Li gii. Ta c:(1 x + x2 x3)4 = (1 x + x2(1 x))4 = (1 x)4(1 + x2)4

=

(4

k=0

Ck4 (x)k)(

4i=0

Ci4(x2)i)

=4

k=0

4i=0

Ck4 (1)kxkCi4x2i =4

k=0

4i=0

Ck4Ci4(1)kxk+2i

S hng cha x7 tng ng s hng cha k v i tha k + 2i = 7.

V 0 i, k 4 nn k + 2i = 7{

k = 3i = 2

hoc{

k = 1i = 3

.

Vy h s a7 ca s hng cha x7 l a7 = C34C24 (1)3 + C14C34 (1)1 = 40.

11.25. (D-02) Tm s nguyn dng n tho mn h thc C0n + 2.C1n + 2

2.C2n + ... + 2n.Cnn = 243.

Li gii. Xt khai trin (1 + x)n =nk=0

Cknxk. Chn x = 2 ta c: 3n =

nk=0

Ckn2k.

Li theo gi thit ta c: 3n = 243 n = 5.

5


11.26. (D-08) Tm s nguyn dng n tho mn h thc C12n + C32n + ... + C

2n12n = 2048.

Li gii. Xt khai trin (1 + x)2n =2nk=0

Ck2nxk.

Chn ln lt x = 1 v x = 1 ta c: 22n =2nk=0

Ck2n (1) v 0 =2nk=0

Ck2n(1)k (2)Tr theo v (1) v (2) ta c: 22n = 2

(C12n + C

32n + ... + C

2n12n

).

Li theo gi thit c 22n = 2.2048 22n = 212 n = 6.11.27. Tm s t nhin n sao cho 1.C1n + 2.C

2n + ... + nC

nn = n.2

2009.


Cknxk. Ly o hm hai v ta c: n(1 + x)n1 =

nk=1

Cknkxk1.

Chn x = 1 ta c: n.2n1 =nk=1

Cknk. Li theo gi thit c: n.2n1 = n.22009 n = 2010.

11.28. (A-2012) Cho n l s nguyn dng tha mn 5Cn1n = C3n. Tm s hng cha x

5 trong khai trin nh thc

Newton ca(nx2

14 1x

)n, x 6= 0.

Li gii. iu kin: n Z, n 3. Khi : 5Cn1n = C3n 5n =n(n 1)(n 2)

3! n = 7.

Do (nx2

14 1x

)n=

(x2

2 1x

)7=

7k=0

Ck7

(x2

2

)7k( 1x

)k=

7k=0

Ck71

27k(1)kx143k.

S hng cha x5 tng ng s hng cha k tha 14 3k = 5 k = 3.Vy s hng cha x5 l C37

1

24(1)3x5 = 35

16x5.

11.29. (B-07) Tm h s ca x10 trong khai trin (2 + x)n, bit 3nC0n 3n1C1n + 3n2C2n + ...+ (1)nCnn = 2048.

Li gii. Xt khai trin (x 1)n =nk=0

Cknxnk(1)k. Chn x = 3 ta c: 2n =

nk=0

Ckn3nk(1)k.

Li theo gi thit ta c: 2n = 2048 n = 11. Khi : (2 + x)n = (2 + x)11 =11k=0

Ck11211kxk.

S hng cha x10 tng ng s hng cha k tha k = 10. Vy h s ca s hng cha x10 l C101121 = 22.

11.30. (A-03) Tm h s ca s hng cha x8 trong khai trin(

1x3 +

x5)n

, bit Cn+1n+4 Cnn+3 = 7 (n + 3).

Li gii. Ta c: Cn+1n+4 Cnn+3 = 7 (n + 3) (n+4)(n+3)(n+2)3! (n+3)(n+2)(n+1)3! = 7 (n + 3) n = 12.Khi :

(1x3 +

x5)n

=(x3 + x

52

)12=

12k=0

Ck12(x3

)12k(x

52

)k=

12k=0

Ck12x112 k36.

S hng cha x8 tng ng s hng cha k tha11

2k 36 k = 8. Vy h s ca s hng cha x8 l C812 = 495.

11.31. (A-06) Tm h s ca x26 trong khai trin(

1x4 + x

7)n, bit C12n+1 + C22n+1 + ... + Cn2n+1 = 220 1.

Li gii. Xt khai trin (1 + x)2n+1 =2n+1k=0

Ck2n+1xk. Chn x = 1 ta c: 22n+1 =

2n+1k=0

Ck2n+1.

Li c Ck2n+1 = C2n+1k2n+1 nn 2

2n+1 =2n+1k=0

Ck2n+1 = 2nk=0

Ck2n+1 22n 1 =nk=1

Ck2n+1.

Li theo gi thit c: 22n 1 = 220 1 k = 10.Khi :

(1x4 + x

7)n

=(x4 + x7

)10=

10k=0

Ck10(x4

)10k(x7)k

=10k=0

Ck10x11k40.

S hng cha x26 tng ng s hng cha k tha 11k 40 = 26 k = 6.Vy h s ca s hng cha x26 l C610 = 210.

11.32. (D-03) Vi n l s nguyn dng, gi a3n3 l h s ca x3n3 trong khai trin thnh a thc ca(x2 + 1

)n(x + 2)

n. Tm n a3n3 = 26n.

Li gii. Ta c:(x2 + 1

)n(x + 2)

n=

(nk=0

Cknx2n2k

)(ni=0

Cin2ixni

)=

nk=0

ni=0

CknCin2

ix3n2ki.

S hng cha x3n3 tng ng s hng cha k v i tha 3n2ki = 3n3 2k+i = 3{k = 0

i = 3hoc

{k = 1

i = 1.

Do h s a3n3 ca s hng cha x3n3 l C0nC3n2

3 + C1nC1n2

1 = 4n(n1)(n2)3 + n2.

Theo gi thit a3n3 = 26n 4n(n1)(n2)3 + n2 = 26n n = 5.

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11.33. (A-02) Cho khai trin biu thc(

2x12 + 2

x3

)n= C0n

(2x12

)n+C1n

(2x12

)n1 (2

x3

)+ ...+Cnn

(2

x3

)n. Bitrng trong khai trin C3n = 5C

1n v s hng th t bng 20n. Tm n v x.

Li gii. iu kin: n Z, n 3. Ta c: C3n = 5C1n n(n1)(n2)3! = 5n n(n2 3n 28) = 0 n = 7.

Khi s hng th t l C37(

2x12

)4(2

x3

)3= 35.22x2.2x = 35.2x2.

Theo gi thit ta c: 35.2x2 = 140 2x2 = 4 x = 4.11.34. (A-05) Tm s nguyn dng n tha C12n+1 2.2C22n+1 + 3.22C32n+1 + ... + (1)n22nC2n+12n+1 = 2005.

Li gii. Xt khai trin (1 + x)2n+1 =2n+1k=0

Ck2n+1xk. Ly o hm hai v c (2n+1)(1 + x)2n =

2n+1k=0

Ck2n+1kxk1.

Thay x = 2 ta c: 2n + 1 =2n+1k=0

Ck2n+1k(1)k12k1.Theo gi thit ta c: 2n + 1 = 2005 n = 1002.

11.35. (A-07) Chng minh rng 12C12n +

14C

32n + ... +

12nC

2n12n =

22n12n+1 .

Li gii. Xt khai trin (1 + x)2n =2nk=0

Ck2nxk (1) v (1 x)2n =

2nk=0

Ck2n(1)kxk (2).

Tr theo v (1) v (2) ta c: (1 + x)2n(1 x)2n = 2 (C12nx + C32nx3 + ... + C2n12n ) 12 [(1 + x)2n (1 x)2n] =C12nx + C

32nx

3 + ... + C2n12n x2n1.

Ly tch phn t 0 n 1 c hai v ta c:

1

2

10

[(1 + x)

2n (1 x)2n]dx =

10

(C12nx + C

32nx

3 + ... + C2n12n x2n1) dx

12

[(1 + x)

2n+1+ (1 x)2n+1

2n + 1

]1

0

=

(C12n

x2

2+ C32n

x4

4+ ... + C2n12n

x2n

2n

)10

12

[22n+1 2

2n + 1

]=

1

2C12n +

1

4C32n + ... +

1

2nC2n12n

22n 1

2n + 1=

1

2C12n +

1

4C32n + ... +

1

2nC2n12n (pcm)

11.36. (B-03) Cho n l s nguyn dng. Tnh tng C0n +22 1

2C1n +

23 13

C2n + ... +2n+1 1n + 1

Cnn .


Cknxk. Ly tch phn t 1 n 2 c hai v ta c:

21

(1 + x)ndx =

21

nk=0

Cknxkdx (1 + x)

n+1

n + 1

2

1

=nk=0

Cknxk+1

k + 1

2

1

3n+1 1n + 1

=nk=0

Ckn2k+1 1k + 1

(pcm)

11.37. Chng minh rng 2.1.C2n + 3.2.C3n + 4.3.C

4n + ... + n (n 1)Cnn = n (n 1) 2n2.


Cknxk.

Ly o hm cp hai hai v ta c: n(n 1)(1 + x)n2 =nk=0

Cknk(k 1)xk2.

Chn x = 1 ta c: n(n 1)2n2 =nk=2

Cknk(k 1) (pcm).

11.38. Tnh tnga) S = C02009 + C

22009 + C

42009 + ... + C

20082009 . b) S = C

02009 + 3

2C22009 + 34C42009 + ... + 3

2008C20082009 .c) S = 2C0n + 5C

1n + 8C

2n + ... + (3n + 2)C

nn . d)

(C02010

)2+(C12010

)2+(C22010

)2+ ... +

(C20102010

)2.Li gii.

a) Xt khai trin (1 + x)2009 =2009k=0

Ck2009xk.

Ln lt chn x = 1 v x = 1 ta c: 22009 =2009k=0

Ck2009 (1) v 0 =2009k=0

Ck2009(1)k (2).

7


Cng theo v (1) v (2) ta c: 22009 = 2(C02009 + C

22009 + ... + C

20082009

) C02009 + C22009 + ... + C20082009 = 22008 S = 22008.

b) Xt khai trin (1 + x)2009 =2009k=0

Ck2009xk.

Ln lt chn x = 3 v x = 3 ta c: 42009 =2009k=0

Ck20093k (1) v (2)2009 =

2009k=0

Ck2009(1)k3k (2).Cng theo v (1) v (2) ta c: 42009 22009 = 2 (C02009 + 32C22009 + ... + 32008C20082009) C02009 + 32C22009 + ... +

32008C20082009 =4200922009

2 S = 4200922009

2 .

c) Xt khai trin x2(1 + x3

)n= x2

nk=0

Cknx3k =

nk=0

Cknx3k+2.

Ly o hm hai v ta c: 2x(1 + x3

)n+ 3nx4

(1 + x3

)n1=

nk=0

Ckn(3k + 2)x3k+1.

Chn x = 1 ta c: 2.2n + 3n.2n1 =nk=0

Ckn(3k + 2) S = 2n+1 + 3n.2n1.

d) Xt khai trin (1 + x)4020 =4020k=0

Ck4020xk c h s ca x2010 l C20104020 (1).

Li c: (1 + x)4020 = (1 + x)2010(1 + x)2010 =2010k=0

Ck2010xk

2010i=0

Ci2010xi =

2010k=0

2010i=0

Ck2010Ci2010x

k+i.

S hng cha x2010 tng ng s hng cha k v i tha k + i = 2010.

Do h s ca s hng cha x2010 l

k+i=2010

Ck2010Ci2010 =

2010k=0

Ck2010C2010k2010 =

2010k=0

(Ck2010

)2 (2).T (1) v (2) ta c

2010k=0

(Ck2010

)2= C20104020 .

11.39. Tnh tng S = 12C1201122010 + 22C220112

2009 + ... + 20112C2011201120.


Cknxk.

Ly o hm v o hm cp hai hai v ta c:

2011(2 + x)2010 =

2011k=1

Ck201122011kkxk1

2011.2010(2 + x)2009 =2011k=1

Ck201122011kk(k 1)xk2

Chn x = 1 ta c:

2011.32010 =

2011k=1

Ck201122011kk (1)

2011.2010.32009 =

2011k=1

Ck201122011kk(k 1) (2)

Cng theo v (1) v (2) ta c: 2011.32010+2011.2010.32009 =2011k=1

[Ck20112

2011k (k + k(k 1))] = 2011k=1

k2Ck201122011k.

Vy S = 2011.32010 + 2011.2010.32009 = 2011.2013.32009.

11.40. Trong khai trin nh thc (a + b)50, tm s hng c gi tr tuyt i ln nht, cho bit |a| = |b|3.Li gii. S hng tng qut ca khai trin (a + b)50 l: Tk+1 = Ck50a

50kbk.Khi : |Tk+1| = Ck50|a|50k|b|k = Ck50

(|b|3)50k|b|k = Ck50(3)50k|b|50.Ta c:

|Tk+2||Tk+1| =

Ck+150(

3)49k|b|50

Ck50(

3)50k|b|50 = 50 k3 (k + 1) .

Suy ra:|Tk+2||Tk+1| > 1

50 k3 (k + 1)

> 1 50 k >

3 (k + 1) k < 50

33 + 1

< 18.

Do T1 < T2 < ... < T18 > T19 > ... > T51.Vy s hng c tr tuyt i ln nht l T18 = C17503

16b50.

11.41. (A-08) Cho khai trin (1 + 2x)n = a0 + a1x+ ...+ anxn, (n N) v cc h s a0, a1, a2, ..., an tho mn hthc a0 + a12 +

a24 + ... +

an2n = 4096. Tm s ln nht trong cc s a0, a1, a2, ..., an.

8



Li gii. Xt khai trin (1 + 2x)n = a0 + a1x + a2x2 + ... + anxn.Chn x = 12 ta c: 2

n = a0 +a12 +

a222 + ... +

an2n = 4096 n = 12.

Khi : (1 + 2x)12 =12k=0

Ck122kxk ak+1ak =

2k+1Ck+1122kCk12

= 2(12k)k+1 =242kk+1 .

Do ak+1ak > 1 242kk+1 > 1 24 2k > k + 1 k < 233 < 8. Suy ra a0 < a1 < ... < a8 > a9 > ... > a12.Vy s ln nht trong cc s a0, a1, ..., a12 l a8 = 28.C812 = 126720.

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