bÀi toÁn tÍnh tỔng hỮu hẠn

Môc lôc

1 Tæng h÷u h¹n liªn quan ®Õn hµm luü thõa 4

1.1 Tæng h÷u h¹n liªn quan ®Õn hµm luü thõa cña çc sè tù nhiªn . . . . . . . . . . . . . 4

1.2 Tæng h÷u h¹n liªn quan ®Õn hµm lòy thõa vµ çc sè Bernoulli . . . . . . . . . . . . 8

1.2.1 Kh¸i niÖm vÒ sè Bernoulli vµ ®a thøc Bernoulli . . . . . . . . . . . . . . . . . 8

1.2.2 Mét sè hÖ thøc gi÷a tæng lòy thõa vµ sè Bernoulli . . . . . . . . . . . . . . . 10

1.3 Tæng ®an dÊu vÒ lòy thõa cña sè tù nhiªn . . . . . . . . . . . . . . . . . . . . . . . 12

1.4 Tæng h÷u h¹n liªn quan ®Õn lòy thõa vµ giai thõa . . . . . . . . . . . . . . . . . . . 15

1.5 Tæng h÷u h¹n liªn quan ®Õn hµm luü thõa vµ hµm mò . . . . . . . . . . . . . . . . . 17

2 Tæng h÷u h¹n liªn quan ®Õn tæ hîp 20

2.1 Tæ hîp vµ nhÞ thøc Newton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.1.1 Tæ hîp vµ çc tÝnh chÊt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.1.2 NhÞ thøc Newton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2 Mét sè bµi to¸n th«ng dông . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.3 Mét sè mÖnh ®Ò vµ çc bµi to¸n míi . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Tµi liÖu tham kh¶o . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1

Më ®Çu

Çc bµi to¸n tÝnh tæng thêng xuÊt hiÖn trong çc kú thi häc sinh giái, Olympiad,To¸n quèc tÕ hay kú thi vµo çc trêng phæ th«ng chuyªn díi nhiÒu h×nh thøc kh¸cnhau. Çc bµi to¸n trªn, ®¹i bé phËn lµ nh÷ng bµi to¸n khã mµ häc sinh phæ th«ng,nhÊt lµ phæ th«ng c¬ së kÓ c¶ häc sinh chuyªn to¸n tá ra rÊt lóng tóng khi gÆp çcbµi to¸n d¹ng nµy.

HiÖn nay tµi liÖu tham kh¶o vÒ tÝnh tæng h÷u h¹n b»ng tiÕng ViÖt cha cã nhiÒu,cßn ph©n t¸n vµ çc bµi to¸n khã còng cßn Ýt. CÇn thiÕt ph¶i cã sù tæng hîp, ph©nlo¹i, giíi thiÖu çc ph¬ng ph¸p tÝnh tæng mét çch hÖ thèng vµ çc bµi to¸n khãh¬n.

V× vËy, viÖc t×m hiÓu vµ ph¸t triÓn s©u thªm vÊn ®Ò" TÝnh to¸n vµ ®¸nh gi¸çc tæng h÷u h¹n" lµ cÇn thiÕt cho c«ng viÖc häc tËp vµ gi¶ng d¹y to¸n ë bËc phæth«ng. B¶n s¸ng kiÕn kinh nghiÖm nµy nh»m tr×nh bµy mét sè ph¬ng ph¸p vÒ tÝnhtæng h÷u h¹n vµ giíi thiÖu çc bµi to¸n trªn ë çc møc ®é kh¸c nhau.

S¸ng kiÕn kinh nghiÖm gåm çc phÇn: Më ®Çu, hai ch¬ng néi dung, KÕt luËnvµ Tµi liÖu tham kh¶o.

Ch¬ng mét Tæng h÷u h¹n liªn quan ®Õn hµm lòy thõaCh¬ng nµy tr×nh bµy çc kiÕn thøc vÒ tæng h÷u h¹n cña hµm lòy thõa d¹ng:

Sp(n) =n∑

k=1

kp, Lp(n) =n∑

k=1

(2k − 1)p,

Tp(n) =n∑

k=1

(−1)kkp, Fp(n) =n∑

j=1

j!jp.

VËn dông vµ tÝnh to¸n çc bµi to¸n nµy lµ sö dông çc ph¬ng ph¸p kh¸c nhau nh:Ph¬ng ph¸p sö dông çc cÊp sè, ph¬ng ph¸p ®¹o hµm, nhÞ thøc Newton, ph¬ngph¸p truy håi, ph¬ng ph¸p khö liªn tiÕp vµ ph¬ng ph¸p quy n¹p.

Ch¬ng hai Tæng h÷u h¹n liªn quan ®Õn tæ hîpCh¬ng nµy sö dông vµ kÕt hîp çc ph¬ng ph¸p kh¸c nhau nh ph¬ng ph¸p ®¹ohµm, nhÞ thøc Newton vµ mét sè çc phÐp biÕn ®æi ®Ó gi¶i çc bµi to¸n tÝnh tæng

2

3

liªn quan ®Õn tæ hîp. Qua ®ã giíi thiÖu mét sè mÖnh ®Ò vµ bµi to¸n míi chØ ®îcgiíi thiÖu qua çc nghiªn cøu cña çc nhµ chuyªn m«n.

Trong qu¸ tr×nh nghiªn cøu kh«ng tr¸nh khái nh÷ng thiÕu sãt. T¸c gi¶ kÝnh mong®îc sù chØ d¹y, ®ãng gãp ý kiÕn cña çc thÇy c« gi¸o vµ çc ®éc gi¶.

ViÖt yªn, th¸ng 5 n¨m 2011T¸c gi¶

NguyÔn V¨n Thanh

Ch¬ng 1

Tæng h÷u h¹n liªn quan ®Õn hµm luü thõa

Ch¬ng nµy tr×nh bµy çc kiÕn thøc vÒ tæng h÷u h¹n cña hµm lòy thõa liªn quantíi sè tù nhiªn, sè Bernulli, tæng ®an dÊu, tæng giai thõa vµ hµm mò.

1.1 Tæng h÷u h¹n liªn quan ®Õn hµm luü thõa cña çc sè tù nhiªn

Trong môc nµy sÏ tr×nh bµy çch tÝnh çc tæng d¹ng

Sp(n) =n∑

k=1

kp, (1.1)

trong ®ã ta sö dông çc tÝnh chÊt cña cÊp sè céng, cÊp sè nh©n (®îc tr×nh bµy ëch¬ng 4) ®Ó gi¶i çc bµi to¸n. Chóng ta ®· biÕt

S1(n) =n(n+ 1)

2. (1.2)

Bµi to¸n 1.1. TÝnh S2(n) vµ S3(n).

6Lêi gi¶i. a) TÝnh S2(n). Sö dông c«ng thøc

k3 − (k − 1)3 = 3k2 − 3k + 1 ⇔ 3k2 = k3 − (k − 1)3 + 3k − 1.

Ta cã

3n∑

k=1

k2 =n∑

k=1

[k3 − (k − 1)3 + 3k − 1] =n∑

k=1

[k3 − (k − 1)3] + 3n∑

k=1

k −n∑

k=1

1

= n3 + 3n(n+ 1)

2− n =

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

2.

VËy

S2(n) =n(n+ 1)(2n+ 1)

6. (1.3)

4

5

b) TÝnh S3(n). Tõ c«ng thøc

k4 − (k − 1)4 = 4k3 − 6k2 + 4k − 1⇔ 4k3 = k4 − (k − 1)4 + 6k2 − 4k + 1.

Ta cã

4n∑

k=1

k3 =n∑

k=1

[k4 − (k − 1)4 + 6k2 − 4k + 1]

=n∑

k=1

[k4 − (k − 1)4] + 6n∑

k=1

k2 − 4n∑

k=1

k + n

= n4 + 6S2(n)− 4S1(n) + n

= n4 + n(n+ 1)(2n+ 1)− 2n(n+ 1) + n.

VËy

S3(n) =n2(n+ 1)2

4. (1.4)

Bµi to¸n 1.2. TÝnh c¸c tæng Sp(n), p ∈ N, p ≥ 4.

6Lêi gi¶i. Sö dông nhÞ thøc Newton

(a+ b)p =

p∑i=0

C ipa

p−ibi, Cip =

p!

i!(p− i)!.

Ta cã

(k − 1)p =

p∑i=0

C ipk

p−i(−1)i = kp − pkp−1 +p∑

i=2

C ipk

p−i(−1)i,

suy ra

pkp−1 = kp − (k − 1)p +

p∑i=2

C ipk

p−i(−1)i. (1.5)

Céng tõng vÕ c¸c ®¼ng thøc trong c«ng thøc (1.5) theo k = 1, 2, ..., n, ta ®îc

pSp−1(n) = np +

p∑i=2

C ip(−1)iSp−i(n) (1.6)

6

Theo c«ng thøc truy håi (1.6) vµ çc c«ng thøc (1.2) - (1.4) chóng ta tÝnh ®îc Sp−1.

S4(n) =1

30n(n+ 1)(2n+ 1)(3n2 + 3n− 1), (1.7)

S5(n) =1

12n2(n+ 1)2(2n2 + 2n− 1), (1.8)

S6(n) =1

42n(n+ 1)(2n+ 1)(3n4 + 6n3 − 3n+ 1), (1.9)

S7(n) =1

24n2(n+ 1)2(3n4 + 6n3 − n2 − 4n+ 2), (1.10)

S8(n) =1

90n(n+ 1)(2n+ 1)(5n6 + 15n5 + 5n4 − 15n3 − n2 + 9n− 3), (1.11)

S9(n) =1

20n2(n+ 1)2(n2 + n− 1)(2n4 + 4n3 − n2 − 3n+ 3). (1.12)

Bµi to¸n 1.3. TÝnh çc tæng luü thõa cña çc sè tù nhiªn lÎ

Lp(n) =n∑

k=1

(2k − 1)p, p ∈ N∗. (1.13)

6Lêi gi¶i. Ta cã

L1(n) = 1 + 3 + 5 + ...+ (2n− 1) = n2. (1.14)

Víi p > 1 chóng ta biÕn ®æi c«ng thøc (1.13) nh sau

Lp(n) = 1p + 3p + 5p + ...+ (2n− 1)p

= [1p + 2p + 3p + 4p + ...+ (2n− 1)p + (2n)p]

− [2p + 4p + 6p + ...+ (2n)p]

= Sp(2n)− 2pSp(n).

VËy ta cã c«ng thøcLp(n) = Sp(2n)− 2pSp(n), (1.15)

trong ®ã Sp(n) ®îc x¸c ®Þnh theo c«ng thøc (1.1). Theo c«ng thøc (1.15) ta cã

L2(n) = S2(2n)− 22S2(n) =4

3n3 − 1

3n, (1.16)

L3(n) = S3(2n)− 23S3(n) = 2n4 − n2, (1.17)

L4(n) = S4(2n)− 24S4(n) =16

5n5 − 8

3n4 +

7

15n, (1.18)

7

L5(n) = S5(2n)− 25S5(n) =16

3n6 − 20

3n4 +

7

3n2, (1.19)

L6(n) = S6(2n)− 26S6(n) =64

7n7 − 16n5 +

28

3n3 − 31

21n, (1.20)

L7(n) = S7(2n)− 27S7(n) = 16n8 − 112

3n6 +

98

3n4 − 31

3n2, (1.21)

L8(n) = S8(2n)− 28S8(n) =256

9n9 − 256

3n7 +

1568

15n5 − 496

9n3 +

127

15n,

(1.22)

L9(n) = S9(2n)− 29S9(n) =256

5n10 − 192n8 +

1568

5n6 − 248n4 +

381

5n2.

(1.23)

Bµi to¸n 1.4. Víi p ∈ N∗. TÝnh tæng

Pp(n) =n∑

k=1

kp(k + 1)2.

6Lêi gi¶i. Ta biÕn ®æi tæng Pp(n) nh sau

Pp(n) =n∑

k=1

(kp+2 + 2kp+1 + kp) = Sp+2(n) + 2Sp+1(n) + Sp(n), (1.24)


P1(n) =1

4n4 +

7

6n3 +

7

4n2 +

5

6n, (1.25)

P2(n) =1

5n5 + n4 +

5

3n3 + n2 +

2

15n, (1.26)

P3(n) =1

6n6 +

9

10n5 +

5

3n4 +

7

6n3 +

1

6n2 − 1

15n, (1.27)

P4(n) =1

7n7 +

5

6n6 +

17

10n5 +

4

3n4 +

1

6n3 − 1

6n2 − 1

105n, (1.28)

P5(n) =1

8n8 +

11

14n7 +

7

4n6 +

3

2n5 +

1

8n4 − 1

3n3 +

1

21n, (1.29)

P6(n) =1

9n9 +

3

4n8 +

38

21n7 +

5

3n6 +

1

30n5 − 7

12n4 +

1

18n3 +

1

6n2 − 1

105n,

(1.30)

P7(n) =1

10n10 +

13

18n9 +

15

8n8 +

11

6n7 − 14

15n5 +

5

24n4 +

4

9n3 − 1

15n2 − 1

15n.

(1.31)

8

Bµi to¸n 1.5. Víi p ∈ N∗. TÝnh tæng

Qp(n) =n∑

k=1

k(k + 1)p.

6Lêi gi¶i. Ta biÕn ®æi tæng Qp(n) nh sau. §Æt i = k + 1 ta cã

Qp(n) =n+1∑i=2

(i− 1)ip =n+1∑i=1

(i− 1)ip =n∑

i=1

(i− 1)ip + n(n+ 1)p,

hay

Qp(n) =n∑

i=1

ip+1 −n∑

i=1

ip + n(n+ 1)p = Sp+1(n)− Sp(n) + n(n+ 1)p, (1.32)


Q1(n) =1

3n3 + n2 +

2

3n, (1.33)

Q2(n) =1

4n4 +

7

6n3 +

7

4n2 +

5

6n, (1.34)

Q3(n) =1

5n5 +

5

4n4 +

17

6n3 +

11

4n2 +

29

30n, (1.35)

Q4(n) =1

6n6 +

13

10n5 +

47

12n4 +

17

3n3 +

47

12n2 +

31

30n, (1.36)

Q5(n) =1

7n7 +

4

3n6 + 5n5 +

115

12n4 +

59

6n3 +

61

12n2 +

43

42n, (1.37)

Q6(n) =1

8n8 +

19

14n7 +

73

12n6 +

29

2n5 +

473

24n4 +

91

6n3 +

73

12n2 +

41

42n. (1.38)

1.2 Tæng h÷u h¹n liªn quan ®Õn hµm lòy thõa vµ c¸c sè Bernoulli

1.2.1 Kh¸i niÖm vÒ sè Bernoulli vµ ®a thøc Bernoulli

C¸c sè Bernoulli ®îc ra ®êi, sau khi nhµ To¸n häc Thôy SÜ Bernoulli ( Swissmathematician Jacob Bernoulli, 1654-1705 ) sö dông trong bµi to¸n vÒ tæng lòythõa.Bµi to¸n vÒ tæng lòy thõa lµ bµi to¸n tÝnh tæng d¹ng

Sk(n) = 1k + 2k + ...+ nk =n∑

m=0

mk,

9

trong ®ã Sk(n) lµ nghiÖm cña ph¬ng tr×nh sai ph©n{Sk(0) = 0,

Sk(n+ 1)− Sk(n) = (n+ 1)k, (n = 0, 1, 2...).

§Þnh nghÜa 1.1. Çc sè b0, b1, b2, ..., ®îc x¸c ®Þnh theo c«ng thøc{b0 = 1,

C1k+1bk + C2

k+1bk−1 + ...+ Ckk+1b1 = k,

(1.39)

®îc gäi lµ çc sè Bernoulli.

Theo c«ng thøc (1.39) ta t×m ®îc çc sè Bernoulli chØ sè thÊp sau ®©y

b0 = 1, b1 =1

2, b2 =

1

6, b3 = 0, b4 = −

1

30, b5 = 0,

b6 =1

42, b7 = 0, b8 = −

1

30, b9 = 0, b10 =

5

66.

Çc sè Bernoulli tháa m·n hÖ thøc truy håi

bm = 1−m−1∑k=0

Ckm

bkm− k + 1

, b0 = 1, (m = 1, 2, ...). (1.40)

Chó ý. Çc sè Bernoulli cã ®Æc ®iÓm sau ®©y

b0 = 1, b1 =1

2, hoÆc b1 = −

1

2, b2k+1 = 0, (k = 0, 1, ...).

§Þnh nghÜa 1.2. §a thøc Bernoulli Bk(x) ®îc x¸c ®Þnh nh sauB0(x) = 1,

B′

n(x) = nBn−1(x),∫ 1

0 Bn(x)dx = 0, n ≥ 1.

(1.41)

Çc ®a thøc Bernoulli bËc thÊp

B0(x) = 1,

B1(x) = x− 1

2,

B2(x) = x2 − x+ 1

6,

B3(x) = x3 − 3

2x2 +

1

2x,

B4(x) = x4 − 2x3 + x2 − 1

30,

10

B5(x) = x5 − 5

2x4 +

5

3x3 − 1

6x,

B6(x) = x6 − 3x5 +5

2x4 − 1

2x2 +

1

42,

B7(x) = x7 − 7

2x6 +

7

2x5 − 7

6x3 +

1

6x.

MÖnh ®Ò 1.1. §a thøc Bernoulli cã c¸c biÓu diÔn sau

Bm(x) =m∑k=0

Ckm(−1)kbkxm−k, (1.42)

Bm(x) =m∑n=0

1

n+ 1

n∑k=0

Ckn(−1)k(x+ k)m. (1.43)

MÖnh ®Ò 1.2. Víi mçi sè nguyªn d¬ng n ta cã ®¼ng thøc

Bn(x+ 1)−Bn(x) = nxn−1, Bn(1) = bn,

trong ®ã bn lµ sè Bernoulli.

1.2.2 Mét sè hÖ thøc gi÷a tæng lòy thõa vµ sè Bernoulli

MÖnh ®Ò 1.3.

1k+2k+ ...+nk =1

k + 1(nk+1+C1

k+1b1nk+C2

k+1b2nk−1+ ...+Ck

k+1bkn). (1.44)

Chøng minh. DÔ dµng chøng minh r»ng Sk(n) lµ mét ®a thøc bËc k+1 theo n víisè h¹ng tù do b»ng kh«ng. Do ®ã ta ®Æt

(k + 1)(1k + 2k + ...+ nk) = nk+1 + C1k+1α1n

k + C2k+1α2n

k−1 + ...+ Ckk+1αkn.

§¼ng thøc trªn ®îc viÕt l¹i ë d¹ng biÓu thøc ®Æc trng

(k + 1)(1k + 2k + ...+ nk) = (n+ α)k+1 − αk+1, (1.45)

víi quy íc αk = αk. Trong c«ng thøc (1.45) thay n bëi n+ 1 ta ®îc

(k + 1)[(1k + 2k + ...+ (n+ 1)k] = (n+ 1 + α)k+1 − αk+1. (1.46)

Trõ tõng vÕ c¸c ®¼ng thøc (1.45), (1.46) ta ®îc

(k + 1)(n+ 1)k = (n+ 1 + α)k+1 − (n+ α)k+1. (1.47)

11

Trong ®¼ng thøc (1.47) cho n = 0 ta cã

(α + 1)k+1 − αk+1 = k + 1. (1.48)

Khai triÓn c«ng thøc (1.48) víi quy íc thay αk = αk ta ®îc

C1k+1αk + C2

k+1αk−1 + ...+ Ckk+1α1 + α0 = k + 1. (1.49)

NÕu cho α0 = 1 th× tõ c«ng thøc (1.49) ta t×m ®îc αk = bk lµ c¸c sè Bernoulli.

NhËn xÐt 1.1. C«ng thøc (1.44) cßn viÕt ®îc díi d¹ng sau

Sk(n) =1

k + 1

k∑m=1

Cmk+1bmn

k+1−m. (1.50)

MÖnh ®Ò 1.4. Ta cã hÖ thøc sau ®©y

Sp(m) =Bp+1(m+ 1)−Bp+1(1)

p+ 1. (1.51)

Chøng minh. VËn dông mÖnh ®Ò 1.2 ta cã

1p =1

p+ 1[Bp+1(2)−Bp+1(1)],

2p =1

p+ 1[Bp+1(3)−Bp+1(2)],

................................................,

mp =1

p+ 1[Bp+1(m+ 1)−Bp+1(m)].

Céng c¸c ®¼ng thøc trªn theo tõng vÕ ta cã

Sp(m) =Bp+1(m+ 1)−Bp+1(1)

p+ 1.

Bµi to¸n 1.6. Chøng minh bÊt ®¼ng thøc

nk+1

k + 1< Sk(n) <

(n+ 1)k+1 − 1

k + 1. (1.52)

6Lêi gi¶i. Theo bÊt ®¼ng thøc Bernoulli

(1 + x)k ≥ 1 + kx, (x ≥ −1, k > 0).

12

Ta cã (1 +

1

j

)k+1

> 1 +k + 1

j,

(1− 1

j

)k+1

> 1− k + 1

j.

Nh©n çc bÊt ®¼ng thøc trªn víi jk+1. Sau mét sè biÕn ®æi ta ®îc

jk+1 − (j − 1)k+1

k + 1< jk <

(j + 1)k+1 − jk+1

k + 1, (j = 1, 2..., n). (1.53)

Trong bÊt ®¼ng thøc (1.53) lÇn lît cho j = 1, 2, ..., n råi céng tõng vÕ cña çc bÊt®¼ng thøc ta cã ®iÒu ph¶i chøng minh.

1.3 Tæng ®an dÊu vÒ lòy thõa cña sè tù nhiªn

Trong môc nµy tr×nh bµy çch tÝnh tæng cã d¹ng

Tp(n) =n∑

k=1

(−1)kkp, p ∈ N∗. (1.54)

§Þnh lý 1.1. Cho Tp(n) =n∑

k=1

(−1)kkp lµ tæng cña mét chuçi ®an dÊu

víi p ∈ N∗,m, n ≥ 0. Chøng minh r»ng

Tp(n) = (−1)n(n+ 1)p − 1−p∑

m=0

Cmp Tm(n).

Chøng minh. §Æt k = j + 1, ta cã

Tp(n) =n−1∑j=0

(−1)j+1(j + 1)p = −1−n−1∑j=1

(−1)j(j + 1)p

= −1−n∑

j=1

(−1)j(j + 1)p + (−1)n(n+ 1)p

= (−1)n(n+ 1)p − 1−n∑

j=1

(−1)jp∑

m=0

Cmp j

m

= (−1)n(n+ 1)p − 1−p∑

m=0

Cmp

n∑j=1

(−1)jjm

= (−1)n(n+ 1)p − 1−p∑

m=0

Cmp Tm(n),

13

trong ®ã

Tm(n) =n∑

j=1

(−1)jjm.

Bæ ®Ò 1.1. Víi mçi n ≥ 1 ta cã çc ®¼ng thøc sau ®©y

a)2n∑k=1

(−1)k+1 sin kx =sin x

2 − sin(2n+ 12)x

2 cos x2

. (1.55)

b)2n+1∑k=1

(−1)k+1 sin kx =sin x

2 + sin(2n+ 32)x

2 cos x2

. (1.56)

c)2n∑k=1

(−1)k+1 cos kx =cos x

2 − cos(2n+ 12)x

2 cos x2

. (1.57)

d)2n+1∑k=1

(−1)k+1 cos kx =cos x

2 + cos(2n+ 32)x

2 cos x2

. (1.58)

Chøng minh. Tõ ®¼ng thøc (1.55) ta cã

2 cosx

2(sinx−sin 2x+sin 3x− ...−sinnx+ ...−sin 2nx) = sin

x

2−sin(2n+

1

2)x.

Ta thùc hiÖn çc phÐp biÕn ®æi

2 sinx cosx

2= sin

x

2+ sin

3x

2,

− 2 sin 2x cosx

2= − sin

3x

2− sin

5x

2,

...........................................................,

− 2 sin 2nx cosx

2= − sin(2n− 1

2)x− sin(2n+

1

2)x.

Céng tõng vÕ cña çc ®¼ng thøc trªn ta cã ®iÒu ph¶i chøng minh.

c) Tõ ®¼ng thøc (1.57) ta cã

2 cosx

2(cosx−cos 2x+cos 3x−...−cosnx+...−cos 2nx) = cos

x

2−cos(2n+1

2)x.

14

Ta thùc hiÖn çc phÐp biÕn ®æi

2 cosx cosx

2= cos

x

2+ cos

3x

2,

− 2 cos 2x cosx

2= − cos

3x

2− cos

5x

2,

........................................................,

− 2 cos 2nx cosx

2= − cos(2n− 1

2)x− cos(2n+

1

2)x.

Céng tõng vÕ cña çc ®¼ng thøc trªn ta cã ®iÒu ph¶i chøng minh.Çc ý b, d ta chøng minh t¬ng tù nh ý a, c.

NhËn xÐt 1.2. Theo bæ ®Ò (1.1) ta cã çc ®¼ng thøc sau ®©y

i1)2n∑k=1

(−1)k+1k = −n,

i2)2n∑k=1

(−1)kk2 = n(2n+ 1),

i3)2n∑k=1

(−1)kk3 = n2(4n+ 3),

i4)2n∑k=1

(−1)k+1k4 = n(−8n2 − 8n+ 1),

i5)2n∑k=1

(−1)k+1k5 = −16n5 − 20n4 + 5n2,

i6)2n∑k=1

(−1)kk6 = 32n6 + 48n5 − 20n3 + 3n,

i7)2n∑k=1

(−1)kk7 = 64n7 + 112n6 − 70n4 + 21n2,

i8)2n∑k=1

(−1)k+1k8 = −128n8 − 265n7 + 224n5 − 112n3 + 17n,

i9)2n∑k=1

(−1)k+1k9 = −256n9 − 576n8 + 672n6 − 504n4 + 153n2,

i10)2n∑k=1

(−1)kk10 = 512n10 + 1280n9 − 1920n7 + 2016n5 − 1020n3 − 155n.

15

Bµi to¸n 1.7. TÝnh tæng

a) 1− 3

2+

5

4− 7

8+ ...+ (−1)n−12n− 1

2n−1.

b) 3− 5

2+

7

4− 9

8+ ...+ (−1)n−12n+ 1

2n−1.

6Lêi gi¶i. Ta xÐt

P (x) =n∑

k=1

(2k − 1)xk−1 = 2n∑

k=1

kxk−1 −n∑

k=1

xk−1

=2nxn(x− 1)− (x+ 1)(xn − 1)

(x− 1)2.

Q(x) =n∑

k=1

(2k + 1)xk−1 = 2n∑

k=1

kxk−1 +n∑

k=1

xk−1

=(2n+ 1)xn+1 − (2n+ 3)xn − x+ 1

(x− 1)2.

Thay x = −12, ta ®îc

1− 3

2+

5

4− 7

8+ ...+ (−1)n−12n− 1

2n−1=

2n + (−1)n+1(6n+ 1)

9.2n−1.

3− 5

2+

7

4− 9

8+ ...+ (−1)n−12n+ 1

2n−1=

(−1)n+1(6n+ 7) + 3.2n

9.2n−1.

1.4 Tæng h÷u h¹n liªn quan ®Õn lòy thõa vµ giai thõa

Trong môc nµy tr×nh bµy c¸ch tÝnh tæng cã d¹ng

Fp(n) =n∑

j=1

j!jp, p ∈ N∗. (1.59)

§Þnh lý 1.2. Cho Fp(n) =n∑

j=1

j!jp, víi p ∈ N∗,m, n ≥ 0. Chøng minh r»ng

Fp(n) = 1− n!(n+ 1)p+1 +

p+1∑m=0

Cmp+1Fm(n). (1.60)

16

Chøng minh. §Æt j = k + 1, ta cã

Fp(n) =n−1∑k=0

(k + 1)!(k + 1)p =n−1∑k=0

k!(k + 1)p+1

= 1 +n−1∑k=1

k!(k + 1)p+1 = 1 +n∑

k=1

k!(k + 1)p+1 − n!(n+ 1)p+1

= 1− n!(n+ 1)p+1 +n∑

k=1

k!

p+1∑m=0

Cmp+1k

m

= 1− n!(n+ 1)p+1 +

p+1∑m=0

Cmp+1

n∑k=1

k!km

= 1− n!(n+ 1)p+1 +

p+1∑m=0

Cmp+1Fm(n),

trong ®ã

Fm(n) =n∑

k=1

k!km.

NhËn xÐt 1.3. Tõ ®Þnh lý 1.2 ta cã mét sè kÕt qu¶ sau

1) F1(n) = (n+ 1)!− 1, (1.61)

2) F2(n) + F0(n) = n(n+ 1)!, (1.62)

3) F3(n)− F0(n) = (n+ 1)!(n2 − 2) + 2, (1.63)

4) F4(n)− 2F0(n) = (n+ 1)!(n3 − 3n+ 3)− 3, (1.64)

5) F5(n) + 9F0(n) = (n+ 1)!(n4 − 4n2 + 6n+ 4)− 4. (1.65)

Bµi to¸n 1.8. T×m tæng cña tÊt c¶ 7! sè nhËn ®îc tõ ho¸n vÞ c¸c ch÷ sè cñasè 1234567.

( §Ò thi v« ®Þch níc BØ n¨m 1979 )

6Lêi gi¶i. Víi mäi gi¸ trÞ i, j ∈ {1, 2, ..., 7} sè c¸c ch÷ sè mµ trong ch÷ sè j ®øngë hµng thø i lµ 6! do ®ã

S = (6!.1 + ...+ 6!.7) + (6!.1 + ...+ 6!.7)10 + ...+ (6!.1 + ...+ 6!7)106

= 6!(1 + 2 + ...+ 7)(1 + 10 + ...+ 106)

= 720.28.1111111

= 22399997760.

17

Bµi to¸n 1.9. Chøng minh r»ng víi c¸c sè tïy ý m,n ∈ N, sè

Sm,n = 1 +m∑k=1

(−1)k (n+ k + 1)!

n!(n+ k)

chia hÕt cho m!, nhng víi mét sè gi¸ trÞ m,n ∈ N th× sè Sm,n kh«ng chia hÕt chom!(n+ 1).

( §Ò thi v« ®Þch níc Anh n¨m 1981 )

6Lêi gi¶i. Víi mçi gi¸ trÞ cña m ∈ N, ta chøng minh b»ng quy n¹p theo m ∈ N.

Sù ®óng ®¾n cña ®¼ng thøc Sm,n = (−1)m (n+m)!

n!, víi m = 1 ta cã kh¼ng ®Þnh

S1,n = 1− (n+ 2)!

n!(n+ 1)= 1− (n+ 2) = −(n+ 1)!

n!.

Gi¶ sö ®¼ng thøc ®· ®îc chøng minh víi gi¸ trÞ m ∈ N nµo ®ã. Ta chøng minh®óng víi m+ 1, ta cã

Sm+1,n = Sm,n + (−1)m+1 (n+m+ 2)!

n!(n+m+ 1)

= (−1)m (n+m)!

n!+ (−1)m+1 (n+m)!(n+m+ 2)

n!

= (−1)m+1 (n+m)!

n!(−1 + n+m+ 2) = (−1)m+1 (n+m+ 1)!

n!.

NghÜa lµ kh¼ng ®Þnh ®óng víi m+ 1. VËy sè

Sm,n = (−1)m (n+m)!

n!m!m! = (−1)mCm

n+mm! chia hÕt cho m! v× Cmn+m ∈ N.

Víi n = 2,m = 3 th× sè Sm,n = 60 kh«ng chia hÕt cho m!(n+ 1) = 18.Bµi to¸n ®îc chøng minh.

1.5 Tæng h÷u h¹n liªn quan ®Õn hµm luü thõa vµ hµm mò


Sn =n∑

k=0

(a+ kd)qk.

6Lêi gi¶i. Ta cãn∑

k=0

(a+ kd)qk = an∑

k=0

qk + dn∑

k=1

kqk,

18

trong ®ã

an∑

k=0

qk = a(1 + q + q2 + ...+ qn),

dn∑

k=1

kqk = d[(q + q2 + ...+ qn) + (q2 + q3 + ...+ qn) + ...+ (qn)].

Tõ ®ã suy ra

Sn = a(1 + q + q2 + ...+ qn) + d[(q + q2 + ...+ qn)+

+ (q2 + q3 + ...+ qn) + ...+ (qn)]

= aqn+1 − 1

q − 1+ d.

(q.qn − 1

q − 1+ q2.

qn−1 − 1

q − 1+ ...+ qn

q − 1

q − 1

)= a

qn+1 − 1

q − 1+ d

(qn+1 − q + qn+1 − q2 + ...+ qn+1 − qn

q − 1

)= a

qn+1 − 1

q − 1+ d

(n.qn+1 − (q + q2 + ...+ qn)

q − 1

)=

1

q − 1

(a(qn+1 − 1) + d(nqn+1 − qn+1 − q

q − 1)

).

VËy

Sn =1

q − 1

(a(qn+1 − 1) + d(nqn+1 − qn+1 − q

q − 1)

).


(n+ 1)20 + n.21 + ...+ 2.2n−1 + 1.2n.


k=0

(n+ 1− k)2k = (n+ 1)20 + n.21 + ...+ 2.2n−1 + 1.2n.

¸p dông kÕt qu¶ bµi tËp (1.10) víi a = n+ 1, d = −1, q = 2, ta ®îcn∑

k=0

(n+ 1− k)2k = 1

2− 1

((n+ 1)(2n+1 − 1)− (n2n+1 − 2n+1 − 2

2− 1)

)= (n+ 1)2n+1 − (n+ 1)− n2n+1 + 2n+1 − 2

= 2n+2 − (n+ 3).

19


Gn =n∑

k=1

kpk, p 6= 0, p 6= 1.


k=1

pk =n∑

k=0

pk − 1 =pn+1 − 1

p− 1− 1 =

pn+1 − pp− 1

.

LÊy ®¹o hµm hai vÕ cña ®¼ng thøc trªn theo p ta ®îcn∑

k=1

kpk−1 =npn+1 − (n+ 1)pn + 1

(p− 1)2.

Nh©n hai vÕ víi p ta cã

Gn =n∑

k=1

kpk =npn+2 − (n+ 1)pn+1 + p

(p− 1)2.

Bµi to¸n 1.13. Cho d·y sè x0, x1, ..., xn, víi x0 = 2000,

xn =−2000n

n−1∑k=0

xk, n ≥ 1.

TÝnh

S =2000∑n=0

2nxn.

6Lêi gi¶i. Ta cã

x0 = 2000, x1 = −2000x0,

x2 =−2000

2(x0 + x1) =

−122000x1 −

1

22000x0 = −

1

2(2000− 1)x1.

B»ng ph¬ng ph¸p quy n¹p ta cã

xn =−2000n−1

n

n−1∑k=0

xk = −1

n(2000− n+ 1)xn−1

= (−1)n2000(2000− 1)...(2000− n+ 1)

1.2...nx0 = (−1)nCn

2000x0.

VËy

S =2000∑n=0

2nxn = x0

2000∑n=0

(−1)n2nCn2000 = x0(1− 2)2000 = x0 = 2000.

Ch¬ng 2

Tæng h÷u h¹n liªn quan ®Õn tæ hîp

Ch¬ng nµy tr×nh bµy çc bµi to¸n vµ mÖnh ®Ò liªn quan ®Õn tæ hîp, trong ®ãsö dông ph¬ng ph¸p ®¹o hµm, nhÞ thøc Newton, ph¬ng ph¸p sai ph©n vµ mét sètÝnh chÊt ®Ó gi¶i.

2.1 Tæ hîp vµ nhÞ thøc Newton

2.1.1 Tæ hîp vµ çc tÝnh chÊt

§Þnh nghÜa 2.1. Cho E lµ mét tËp gåm n phÇn tö (n ≥ 1). Víi k ∈ N, 0 6 k 6 n.

Mét tæ chËp k cña n phÇn tö thuéc E lµ mét tËp hîp con gåm k phÇn tö cña E. Ký

hiÖu Ckn hay

(n

k

)lµ sè çc tæ hîp chËp k cña n, tøc lµ

Ckn =

(nk

)=

n!

k!(n− k)!. (2.1)

Çc tÝnh chÊt cña tæ hîp

1) C0n = Cn

n = 1, (2.2)

2) C1n = n, (2.3)

3) Ckn = Cn−k

n , (2.4)

4) Ckn + Ck−1

n = Ckn+1, (2.5)

5) C0n + C1

n + C2n + ...+ Cn

n = 2n. (2.6)

20

21

2.1.2 NhÞ thøc Newton

MÖnh ®Ò 2.1. Ta cã c«ng thøc khai triÓn nhÞ thøc sau ®©y

(a+ b)n =n∑

k=0

Ckna

n−kbk. (2.7)

HÖ qu¶ 2.1. Trong nhÞ thøc Newton cho a = 1, b = x ta cã c«ng thøc

(1 + x)n =n∑

k=0

Cknx

k. (2.8)

HÖ qu¶ 2.2. Trong c«ng thøc (2.8) t¬ng øng cho x = 1, x = −1 ta cã

n∑k=0

Ckn = 2n, (2.9)

n∑k=0

(−1)kCkn = 0. (2.10)

HÖ qu¶ 2.3. LÊy ®¹o hµm theo x hai vÕ cña c«ng thøc (2.8) ta cã

n(1 + x)n−1 =n∑

k=0

kCknx

k−1 =n∑

k=1

kCknx

k−1. (2.11)

2.2 Mét sè bµi to¸n th«ng dông


S = 12C1n + 22C2

n + ...+ P 2Cpn + ...+ n2Cn

n .

6Lêi gi¶i. §Æt

f(x) = (1 + x)n = C0n + C1

nx+ ...+ Cnnx

n.

g(x) = x(1 + x)n = C0nx+ C1

nx2 + ...+ Cn

nxn+1.

LÊy ®¹o hµm hai vÕ theo x ta cã

f ′(x) = n(1 + x)n−1 = C1n + 2xC2

n + ...+ nxn−1Cnn .

g”(x) = 2n(1 + x)n−1 + n(n− 1)x(1 + x)n−2

= 2C1n + 3.2xC2

n + 4.3x2C3n + ...+ (n− 1)nxn−1Cn

n .

22

Thay x = 1 ta ®îc

f ′(1) = n2n−1 = C1n + 2C2

n + ...+ nCnn .

g”(1) = 2n2n−1 + n(n− 1)2n−2 = 2C1n + 3.2C2

n + 4.3C3n + ...+ (n+ 1)nCn

n .

LÊy g′′(1)− f ′

(1) ta cã

S = 12C1n + 22C2

n + ...+ P 2Cpn + ...+ n2Cn

n = n(n− 1)2n−2.

Bµi to¸n 2.2. Chøng minh c¸c ®¼ng thøc

a) Ckn + 4Ck−1

n + 6Ck−2n + 4Ck−3

n + Ck−4n = Ck

n+4, (4 6 k 6 n). (2.12)

b) C0mC

kn + C1

mCk−1n + ...+ Cm

mCk−nn = Ck

m+n, (m 6 k 6 n). (2.13)

Chøng minh. a) ¸p dông c«ng thøc biÕn ®æi Crn = Cr

n−1 +Cr−1n−1, (0 ≤ r ≤ n) ta

cã

V T = Ckn + Ck−1

n + 3(Ck−1n + Ck−2

n ) + 3(Ck−2n + Ck−3

n ) + Ck−3n + Ck−4

n

= Ckn+1 + 3Ck−1

n+1 + 3Ck−2n+1 + Ck−3

n+1

= Ckn+1 + Ck−1

n+1 + 2(Ck−1n+1 + Ck−2

n+1) + Ck−2n+1 + Ck−3

n+1

= Ckn+2 + 2Ck−1

n+2 + Ck−2n+2

= Ckn+2 + Ck−1

n+2 + Ck−1n+2 + Ck−2

n+2

= Ckn+3 + Ck−1

n+3

= Ckn+4 = V P

b) Víi mäi x, n,m lµ sè nguyªn d¬ng ta cã

(1 + x)m+n = (1 + x)m(1 + x)n, (2.14)

(1 + x)m+n =m+n∑k=0

Ckm+nx

k. (2.15)

Tõ ®ã suy ra

(1 + x)m(1 + x)n =m∑p=0

Cpmx

pn∑

p=o

Cpnx

p =m+n∑k=0

k∑p=0

(CpmC

k−pn )xk. (2.16)

Tõ c«ng thøc (2.14) - (2.16) suy ra

Ckm+n =

k∑p=0

CpmC

k−pn = C0

mCkn + C1

mCk−1n + ...+ Cm

mCk−nn .

23

Bµi to¸n 2.3. TÝnh tæng gåm 2n sè h¹ng

S =1

2C1

2n −1

3C2

2n + ...+ (−1)k 1kCk−1

2n + ...+ (−1)2n+1 1

2n+ 1C2

2nn,

trong ®ã Ckn lµ hÖ sè cña khai triÓn nhÞ thøc Newt¬n.

( T¹p trÝ To¸n häc vµ tuæi trÎ sè 336 n¨m 2005)

6Lêi gi¶i. ∀k = 2, 3, ..., 2n+ 1 ta cã

1

kCk−1

2n =1

k

(2n)!

(k − 1)!(2n− k + 1)!=

1

2n+ 1

(2n+ 1)!

k!(2n+ 1− k)!=

Ck2n+1

2n+ 1.

Do ®ã

S =2n+1∑k=2

(−1)kCk

2n+1

2n+ 1

=1

2n+ 1

2n+1∑k=0

(−1)kCk2n+1 − C0

2n+1 + C12n+1

=1

2n+ 1[(1− 1)2n+1 − 1 + 2n+ 1]

=2n

2n+ 1.


A = C0n +

22 − 1

2C1

n +23 − 1

3C2

n + ...+2n+1 − 1

n+ 1Cn

n .

( §Ò thi ®¹i häc khèi B n¨m 2003 )

6Lêi gi¶i. Sö dông c«ng thøc(n+ 1

k + 1

)Ck

n = Ck+1n+1,

trong ®ã n lµ sè nguyªn d¬ng, k lµ sè nguyªn kh«ng ©m, k ≤ n ta cã

A =1

n+ 1

((n+ 1)C0

n + (22 − 1)(n+ 1)C1

n

2+ ...+ (2n+1 − 1)

(n+ 1)Cnn

n+ 1

)=

1

n+ 1

((2− 1)C1

n+1 + (22 − 1)C2n+1 + ...+ (2n+1 − 1)Cn+1

n+1

)

24

=1

n+ 1

((2C1

n+1 + 22C2n+1 + ...+ 2n+1Cn+1

n+1

)−(C1

n+1 + ...+ Cn+1n+1

))=

1

n+ 1

((2 + 1)n+1 − 1− (2n+1 − 1)

)=

3n+1 − 2n+1

n+ 1.

Bµi to¸n 2.5. Víi n lµ sè nguyªn d¬ng. Chøng minh r»ng

3nC0n − 3n−1C1

n + 3n−2C2n + ...+ 3Cn

n = 2n.

( §Ò thi §¹i häc Më n¨m 1997)

6Lêi gi¶i. Víi mäi x vµ víi mäi n lµ sè nguyªn d¬ng ta cã

(1− x)n = C0n − C1

nx+ C2nx

2 + ...+ (−1)nCnnx

n.

Thay x =1

3ta ®îc(

1− 1

3

)n

= C0n −

1

3C1

n +1

32C2

n + ...+ (−1)n 1

3nCn

n

2n

3n= C0

n −1

3C1

n +1

32C2

n + ...+ (−1)n 1

3nCn

n

3nC0n − 3n−1C1

n + 3n−2C2n + ...+ 3Cn

n = 2n.

Bµi to¸n 2.6. Chøng minh r»ng víi n ∈ N tïy ý, ta cã

n∑k=0

(2n)!

(k!)2.(n− k!)2= (Cn

2n)2.

( §Ò thi v« ®Þch níc Mü n¨m 1982)

6Lêi gi¶i. Ta cã(1 + x)2n = (1 + x)n(1 + x)n.

Sö dông nhÞ thøc Newton ta ®îc

C02n + C1

2nx+ ...+ C2n2nx

2n = (C0n + C1

nx+ ...+ Cnnx

n)(C0n + C1

nx+ ...+ Cnnx

n).

So s¸nh c¸c hÖ sè cña xn vµ sö dông ®¼ng thøc Ckn = Cn−k

n , (k = 0, 1, ..., n) ta cã

Cn2n = (C0

n)2 + (C1

n)2 + ...+ (Cn

n)2.

25

VËy

n∑k=0

(2n)!

(k!)2(n− k!)2=

(2n)!

n!n!=

n∑k=0

(n!)2

(k!)2(n− k!)2

= Cn2n[(C

0n)

2 + (C1n)

2 + ...+ (Cnn)

2] = (Cn2n)

2.


S = C02000 + 2C1

2000 + 3C22000 + ...+ 2001C2000

2000 .

( §Ò thi §¹i häc An Ninh n¨m 2000 )

6Lêi gi¶i. Ta cã

(1 + x)2000 = C02000 + C1

2000x+ C22000x

2 + ...+ C20002000x

2000.

Nh©n c¶ hai vÕ víi x ta ®îc

x(1 + x)2000 = C02000x+ C1

2000x2 + C2

2000x3 + ...+ C2000

2000x2001.

LÊy ®¹o hµm hai vÕ theo x ta cã

C02000+2C1

2000x+3C22000x

2+ ...+2001C20002000x

2000 = (1+x)2000+2000x(1+x)1999.

Cho x = 1 ta ®îc

S = 22000 + 2000.21999 = 21999(2 + 2000) = 2002.21999.

Bµi to¸n 2.8. Chøng minh c¸c ®¼ng thøc

a) C0n + C3

n + C6n + ... =

1

3

(2n + 2 cos

nπ

3

).

b) C1n + C4

n + C7n + ... =

1

3

(2n + 2 cos

(n− 2)π

3

).

c) C2n + C5

n + C8n + ... =

1

3

(2n + 2 cos

(n− 4)π

3

).

6Lêi gi¶i. a) XÐt ®ång nhÊt thøc

(1 + x)n = C0n + C1

nx+ C2nx

2 + ...+ Cn−1n xn−1 + Cn

nxn.

26

Trong ®¼ng thøc trªn lÇn lît cho x = 1, ε, ε2, trong ®ã ε2 + ε+ 1 = 0 ta cã

2n = Con + C1

n + C2n + ...

(1 + ε)n = C0n + C1

nε+ C2nε

2 + C3nε

3 + ...

(1 + ε2)n = C0n + C1

nε2 + C2

nε4 + C3

nε6 + ...

NÕu k kh«ng chia hÕt cho 3 th× 1+εk+ε2k = 0, nÕu k chia hÕt cho 3 th× 1+εk+ε2k =3.Tõ ®ã ta cã

2n + (1 + ε)n + (1 + ε2)n = 3(Con + C3

n + C6n + ...).

Do

ε = cos2π

3+ i sin

2π

3,

1 + ε = −ε2 = − cos4π

3− i sin 4π

3= cos

π

3+ i sin

π

3,

1 + ε2 = −ε = − cos2π

3− i sin 2π

3= cos

π

3− i sin π

3.

Tõ ®ã suy ra

(1 + ε)n = cosnπ

3+ i sin

nπ

3,

(1 + ε2)n = cosnπ

3− i sin nπ

3.

Khi ®ã ta ®îc

2n + (1 + ε)n + (1 + ε2)n = 2n +(cos

nπ

3+ i sin

nπ

3+ cos

nπ

3− i sin nπ

3

)= 2n + 2 cos

nπ

3.

VËy

C0n + C3

n + C6n + ... =

1

3

(2n + 2 cos

nπ

3

).

b) T¬ng tù ý a ta cã

ε2(1 + ε)n = C0nε

2 + C1nε

3 + C2nε

4 + C3nε

5 + ...

ε(1 + ε2)n = C0nε+ C1

nε3 + C2

nε5 + C3

nε7 + ...

Tõ ®©y suy ra

2n + ε(1 + ε)n + ε2(1 + ε2)n = 3(C1n + C4

n + C7n + ...).

Khi ®ã ta ®îc

27

2n + ε2(1 + ε)n + ε(1 + ε2)n = 2n − (1 + ε)n+1 − (1 + ε2)n+1

= 2n −[cos

(n+ 1)π

3+ i sin

(n+ 1)π

3+ cos

(n+ 1)π

3− i sin (n+ 1)π

3

]= 2n − 2 cos

(n+ 1)π

3

= 2n + 2 cos(n− 2)π

3.

VËy

C1n + C4

n + C7n + ... =

1

3

(2n + 2 cos

(n− 2)π

3

).

c) T¬ng tù ý a ta cã

ε(1 + ε)n = C0nε+ C1

nε2 + C2

nε3 + C3

nε4 + ...

ε2(1 + ε2)n = C0nε

2 + C1nε

4 + C2nε

6 + C3nε

8 + ...

Tõ ®©y suy ra

2n + ε2(1 + ε)n + ε(1 + ε2)n = 3(C2n + C5

n + C8n + ...).

Khi ®ã ta ®îc

2n + ε(1 + ε)n + ε2(1 + ε2)n = 2n − (1 + ε2)(1 + ε)n − (1 + ε)(1 + ε2)n

= 2n − (1 + ε2)(1 + ε)[(1 + ε)n−1 + (1 + ε2)n−1]

= 2n − (1 + ε+ ε2 + ε3)

[cos

(n− 1)π

3+ i sin

(n− 1)π

3+

+cos(n− 1)π

3− i sin (n− 1)π

3

]= 2n − 2 cos

(n− 1)π

3= 2n + 2 cos

(n− 4)π

3.

VËy

C2n + C5

n + C8n + ... =

1

3

(2n + 2 cos

(n− 4)π

3

).

2.3 Mét sè mÖnh ®Ò vµ c¸c bµi to¸n míi

MÖnh ®Ò 2.2.

n∑j=0

(−1)jCjnj

m =

{0, (0 ≤ m ≤ n− 1),

(−1)nn!, (m = n).(2.17)

28

Chøng minh. Chóng ta chøng minh c«ng thøc (2.17) b»ng quy n¹p theo m.Víi m = 0, n ≥ 1 ta cã

n∑j=0

(−1)jCjn = (1− 1)n = 0.

Víi m = 1, n 6= 1 ta cã

n∑j=0

(−1)jjCjn = n(1− 1)n−1 = 0.

Víi m = n = 1 ta cã

1∑j=0

(−1)jjCjn = −1 = (−1)11!.

Cè ®Þnh n ≥ 3 vµ gi¶ sö r»ng c«ng thøc (2.17) ®óng víi m = k,sao cho 0 ≤ k ≤ n− 2.NhËn xÐt r»ng j(j − 1)...(j −m) = jm+1 + bmj

m + ...+ b1j + b0, víi bj lµ c¸c sènguyªn nµo ®ã ta cã

n∑j=0

(−1)jCjnj

m+1 =n∑

j=1

(−1)jCjnj(j − 1)− ..− (j −m)−

m∑k=0

bk

n∑j=0

(−1)jCjnj

k

=n∑

j=m+1

(−1)jCjnj(j − 1)− ...− (j −m)− 0

=n∑

j=m+1

(−1)j n!j!

j!(n− j)!(j −m− 1)!

=n−m−1∑k=0

(−1)m+1+k n!

(n−m− 1− k)!k!

=n−m−1∑k=0

(−1)m+1(−1)kn(n− 1)...(n−m)[(n−m− 1)!]

(n−m− 1− k)!k!

= (−1)m+1n(n− 1)...(n−m)n−m−1∑k=0

(−1)kCkn−m−1 = 0.

Nh vËyn∑

j=0

(−1)jCjnj

m = 0, (0 ≤ m ≤ n− 1).

29

B©y giê xÐt m = n ta cãn∑

j=0

(−1)jCjnj

n =n∑

j=1

(−1)jCjnj(j−1)...(j−n+1)−

n−1∑k=0

bk

n∑j=0

(−1)jCjnj

k. (2.18)

XÐt vÕ ph¶i cña c«ng thøc (2.18) ta ®îcn−1∑k=0

bk

n∑j=0

(−1)jCjnj

k = 0, v× k < n,

n∑j=1

(−1)jCjnj(j − 1)...(j − n+ 1) = (−1)nn!.

VËy mÖnh ®Ò ®îc chøng minh

MÖnh ®Ò 2.3. Gi¶ sö k(k ≥ 1), n lµ c¸c sè tù nhiªn khi ®ãn∑

j=0

Cjn

(−1)j

j + k=n!(k − 1)!

(n+ k)!. (2.19)

Chøng minh. Chóng ta chøng minh ®¼ng thøc (2.19) b»ng quy n¹p theo n.Víi n = 0 ta cã

V T =0∑

j=0

Cj0

(−1)J

j + k=

1

k,

V P =0!(k − 1)!

(0 + k)!=

(k − 1)!

k!=

1

k, (k ≥ 1).

VËy ®¼ng thøc (2.19) ®óng víi n = 0. Gi¶ sö ®¼ng thøc ®óng víi n, xÐt vÕ tr¸i víin+ 1 ta cã

n+1∑j=0

Cjn+1

(−1)j

j + k= C0

n+1

(−1)0

0 + k+

n∑j=1

Cjn+1

(−1)j

j + k+

(−1)n+1

n+ 1 + k

=

(1

k+

n∑j=1

Cjn

(−1)j

j + k

)+

( n∑j=1

Cj−1n

(−1)j

j + k+

(−1)n+1

n+ 1 + k

)

=n∑

j=0

Cjn

(−1)j

j + k−

n∑q=0

Cqn

(−1)j

q + 1 + k=

1

k− 1

k + 1

=n!(k − 1)!

(n+ k)!− n!k!

(n+ k + 1)!=n!(k − 1)!

(n+ k)!

(1− k

n+ k + 1

)=

n!(k − 1)!(n+ 1)

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

(n+ 1)!(k − 1)!

(n+ k + 1)!.

Theo nguyªn lý quy n¹p ta cã ®iÒu ph¶i chøng minh.

30

MÖnh ®Ò 2.4.m∑j=1

(−1)j−1 Cjm

Cjn+j

=mn

(n+m)(n+m− 1. (2.20)

Chøng minh. [Xem: B. Sury, Sum of the reciprocals of the binomial coeffcients,European J. Combin. 14( 1993), 351-353 ]

MÖnh ®Ò 2.5. ( §¼ng thøc Horrace's). Víi c¸c sè nguyªn m ≥ 1, n ≥ 0 ta cã

m∑j=0

(−1)j Cjm

Cjn+j

=n

n+m. (2.21)

Chøng minh. Chøng minh b»ng quy n¹p theo m. Víi m = 1 ta cã

1∑j=0

(−1)j Cj1

Cjn+j

=C0

1

C0n

− C11

C1n+1

= 1− 1

n+ 1=

n

n+m.

VËy ®¼ng thøc (2.21) ®óng víi m = 1. Gi¶ sö (2.21) ®óng víi m, xÐt víi m+ 1 tacã

m+1∑j=0

(−1)jCj

m+1

Cjn+j

=

= (−1)0C0m

C0n

+m∑j=1

(−1)jCj

m+1

Cjn+j

+ (−1)m+1 Cm+1m+1

Cm+1n+m+1

= (−1)0C0m

C0n

+m∑j=1

(−1)j Cjm

Cjn+j

+m∑j=1

(−1)jCj−1m

Cjn+j

+ (−1)m+1 Cm+1m+1

Cm+1n+m+1

=m∑j=0

(−1)j Cjm

Cjn+j

+m∑j=1

(−1)jCj−1m

Cjn+j

+ (−1)m+1 Cm+1m+1

Cm+1n+m+1

=n

n+m+

m−1∑q=0

(−1)q+1 Cqm

Cq+1n+q+1

+ (−1)m+1 Cm+1m+1

Cm+1n+m+1

=n

n+m−

m∑q=0

(−1)q Cqm

Cq+1n+q+1

=n

n+m−

m∑q=0

(−1)q Cqm

Cqn+q+1

n+ 1

q + 1

=n

n+m− 1

n+ 1

m∑q=0

(−1)q(q + 1)Cq

m

Cqn+q+1

31

=n

n+m− 1

n+ 1

m∑q=0

(−1)qq Cqm

Cqn+q+1

− 1

n+ 1

m∑q=0

(−1)q Cqm

Cqn+q+1

=n

n+m+

1

(n+ 1)

m(n+ 1)

(m+ n+ 1)(m+ n)− 1

(n+ 1)

(n+ 1)

(n+m+ 1)

=n(m+ n+ 1) +m− (m+ n)

(m+ n)(m+ n+ 1)=

n

m+ n+ 1.

Theo nguyªn lý quy n¹p ta cã ®iÒu ph¶i chøng minh.

MÖnh ®Ò 2.6. Gi¶ sö k,m lµ nh÷ng sè nguyªn kh«ng ©m, 0 ≤ m ≤ k khi ®ã

k∑j=m

(−1)jCjkC

mj =

{(−1)k, (m = k),

0, (m < k).(2.22)

MÖnh ®Ò 2.7. Gi¶ sö k, n,m lµ nh÷ng sè nguyªn kh«ng ©m, 0 ≤ m ≤ k khi ®ã

k∑j=m

(−1)jCjkC

mj

1

j + n=

(−1)mk!(m+ n− 1)!

m!(k + n)!. (2.23)

Chøng minh. Khi m = 0 theo mÖnh ®Ò (2.3) ta cã

k∑j=0

(−1)jCjkC

0j

1

j + n=

k∑j=0

(−1)jCjk

1

j + n=n!(k − 1)!

(n+ k)!.

Gi¶ sö mÖnh ®Ò ®óng víi m, xÐt víi m+ 1 ta cãk∑

j=m+1

(−1)jCjkC

m+1j

1

j + n=

=k∑

j=m+1

(−1)jCjk

j −m(m+ 1)(j + n)

=k∑

j=m+1

(−1)jCjkC

mj (

j + n

m+ 1− m+ n

m+ 1)

1

j + n

=1

m+ 1

k∑j=m+1

(−1)jCjkC

mj −

m+ n

m+ 1

k∑j=m+1

(−1)jCjkC

mj

1

j + n

32

=1

m

k∑j=m

(−1)jCjk.C

mj

1

m+ 1(−1)mCm

k Cmm +

m+ n

m+ 1(−1)mCm

k Cmm

1

m+ n

− m+ n

m+ 1

k∑j=m

(−1)jCjk.C

mj

1

j + n.

Theo MÖnh ®Ò 2.6 ta cã

1

m

k∑j=m

(−1)jCjkC

mj

1

m+ 1(−1)mCm

k Cmm +

m+ n

m+ 1(−1)mCm

k Cmm

1

m+ n= 0.

Theo gi¶ thiÕt quy n¹p ta ®îc

m+ n

m+ 1

k∑j=m

(−1)jCjkC

mj

1

j + n=

(−1)mk!m!(m+ n)(m+ n+ 1)

.

VËyk∑

j=m

(−1)jCjkC

mj

1

j + n= −m+ n

j + n

(−1)mk!(m+ n− 1)!

m!(k + n)!=

(−1)m+1k!(m+ n)!

(m+ 1)!(k + n)!.

VËy theo nguyªn lý quy n¹p ta cã ®iÒu ph¶i chøng minh.

Bµi to¸n 2.9. Gi¶ sö n lµ sè nguyªn d¬ng cßn k lµ sè nguyªn kh«ng ©m. Chøngminh r»ng

n! = (n+ k)n − C1n(n+ k − 1)n + C2

n(n+ k − 2)n − ...+ Cnnk

n.

6Lêi gi¶i. Ta cã

(n+ k)n − C1n(n+ k − 1)n + C2

n(n+ k − 2)n − ...+ (−1)NCnn(n+ k − n)n

=n∑

j=0

(−1)jCjn(n− j + k)n =

n∑j=0

(−1)jCjn

n∑q=0

Cqn(n− j)qkn−q

=n∑

j=0

n∑q=0

(−1)jCjnC

qn(n− j)qkn−q =

n∑q=0

Cqnk

n−qn∑

j=0

(−1)jCjn(n− j)q

=n∑

q=0

Cqnk

n−qn∑

j=0

(−1)jCn−jn (n− j)q =

n∑q=0

Cqnk

n−qn∑

j=0

(−1)n(−1)jCjnj

q.

Theo mÖnh ®Ò (2.2) ta ®îcn∑

j=0

(−1)jCn−jn (n− j)q =

n∑j=0

(−1)n−jCjnj

q =

{0, (0 ≤ q ≤ n− 1),

n!, (q = n).(2.24)

Tõ ®ã suy ra ®iÒu ph¶i chøng minh.

KÕt luËn

S¸ng kiÕn kinh nghiÖm ®· tr×nh bµy vµ ®¹t ®îc mét sè kÕt qu¶ sau:

1. Ph©n lo¹i vµ tr×nh bµy çc bµi to¸n vÒ tÝnh to¸n vµ ®¸nh gi¸ çc tæng h÷u h¹n.

2. Sö dông vµ phèi kÕt hîp çc ph¬ng ph¸p kh¸c nhau nh ph¬ng ph¸p ®¹ohµm, sè phøc, nhÞ thøc Newton hay ph¬ng ph¸p sö dông tæng cña çc cÊp sè v.v...®ÓnhËn ®îc çc tæng h÷u h¹n vÒ lòy thõa cña sè tù nhiªn, sè Bernulli, tæng ®an dÊu,tæng giai thõa, tæ hîp, mµ Ýt ®îc giíi thiÖu hay cha nãi nhiÒu trong çc tµi liÖuTiÕng viÖt ë bËc THCS - THPT.

3. S¸ng kiÕn kinh nghiÖm ®· tr×nh bµy ®îc mét sè mÖnh ®Ò vµ bµi tËp míi chØ®îc giíi thiÖu qua çc nghiªn cøu cña çc nhµ chuyªn m«n. Bªn c¹nh ®ã cßn giíithiÖu vµ ®¸nh gi¸ ®îc mét sè bµi to¸n tÝnh tæng trong çc k× thi vµo trêng chuyªn,tuyÓn sinh vµo ®¹i häc, häc sinh giái quèc gia, Olimpyad vµ thi v« ®Þnh quèc tÕ.

33

Tµi liÖu tham kh¶o

[1] Phan Huy Kh¶i (2000), To¸n n©ng cao gi¶i tÝch tÝch ph©n vµ gi¶i tÝch tæ hîp,Nhµ xuÊt b¶n Hµ Néi.

[2] NguyÔn V¨n MËu (2003), Mét sè bµi to¸n chän läc vÒ d·y sè, Nhµ xuÊt b¶nGi¸o Dôc.

[3] NguyÔn V¨n MËu (2006), BÊt ®¼ng thøc, ®Þnh lý vµ ¸p dông, Nhµ xuÊt b¶nGi¸o Dôc.

[4] T¹p trÝ to¸n häc vµ tuæi trÎ sè: 186 (1992), 336 (2005), 346 (2006), 349(2006), 359 (2007), 370 (2008), 375 (2008), Nhµ xuÊt b¶n Gi¸o Dôc.

[5] TuyÓn tËp 30 n¨m T¹p chÝ To¸n häc vµ Tuæi trÎ (1998), Nhµ xuÊt b¶n Gi¸oDôc.

[6] TuyÓn tËp 5 n¨m t¹p trÝ to¸n häc vµ tuæi trÎ (2003), Nhµ xuÊt b¶n Gi¸o Dôc.

[7] TuyÓn chän theo chuyªn ®Ò to¸n häc vµ tuæi trÎ ( 2010), Nhµ xuÊt b¶n Gi¸oDôc.

[8] C¸c ®Ò thi v« ®Þch to¸n 19 níc trong ®ã cã ViÖt Nam (TËp 1, 2) (2002), NhµxuÊt b¶n trÎ.

ViÖt yªn, th¸ng 5 n¨m 2011T¸c gi¶

NguyÔn V¨n Thanh

34

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