Tài liệu Bài tập hoán vị, chỉnh hợp, tổ hợp ôn thi đại học ( www.sites.google.com/site/thuvientailieuvip )

trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 1 www.VNMATH.com ho¸n vÞ. chØnh hîp. tæ hîp 1. Ho¸n vÞ Pn = n! = 1·2·3 · · · n (sè c¸ch x¾p xÕp thø tù n ®èi t−îng kh¸c nhau). VÝ dô 1. Rót gän biÓu thøc A= 6! (m + 1)! · . m(m + 1) 4!(m − 1)! Gi¶i. 4! · 5 · 6 (m − 1)!m(m + 1) · = 30. m(m + 1) 4!(m − 1)! Chó ý. n! = (n − k)!(n − k + 1) · · · n. n P k · k!. VÝ dô 2. Rót gän An = A= k=1 Gi¶i.   k · k! = (k + 1) − 1 ·k! = (k + 1)! − k! =⇒ An = (2! − 1!) + (3! − 2!) + · · · + ((n + 1)! − n!) = (n + 1)! − 1. VÝ dô 3. Chøng minh 1 1 1 1 + + + · · · + < 2. 1! 2! 3! n! Gi¶i. 1 =1 1! 1 1 =1− 2! 2 1 1 1 1 = = − 3! 3 · 2 2 3 1 1 1 1 < = − 4! 3 · 4 3 4 ............ 1 1 1 1 < = − . n! (n − 1)n n − 1 n Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 2 www.VNMATH.com Céng vÕ 1 1 1 1 + + · · · + < 2 − < 2. 1! 2! n! n! VÝ dô 4. Gi¶i c¸c pt a) n! − (n − 1)! 1 = , (n + 1)! 6 b) (n + 1)! = 72 (2) ; n nguyªn d−¬ng. (n − 1)! (1)(n nguyªn d−¬ng). Gi¶i. a) Ta cã n(n − 1)! − (n − 1)! 1 n−1 1 = ⇐⇒ = (n + 1)n(n − 1)! 6 (n + 1)n 6 " n=2 ⇐⇒ n2 − 5n + 6 = 0 ⇐⇒ n = 3. (1) ⇐⇒ b) Ta cã (n + 1) · n · (n − 1)! = 72 n nguyªn d−¬ng (n − 1)! " n = −9 (lo¹i) ⇐⇒ n2 + n − 72 = 0 ⇐⇒ n = 8. (2) ⇐⇒ ChØnh hîp E lµ mét tËp hîp gåm n phÇn tö. r phÇn tö ph©n biÖt, cã kÓ thø tù c¸c phÇn tö cña tËp E (1 ≤ r ≤ n) ®−îc gäi lµ mét chØnh hîp n chËp r . Bé Chó ý. (i) Thø tù. (ii) n = r =⇒ chØnh hîp ≡ ho¸n vÞ. Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 3 www.VNMATH.com C«ng thøc. Arn = n(n − 1) · · · (n − r + 1) = n! . (n − r)! Chó ý. Ann = Pn = n! Ann = Arn · An−r n−r , 1 ≤ r ≤ n. A6n + A5n , 6 < n ∈ R. VÝ dô 5. Rót gän A = A4n Gi¶i. n(n − 1) · · · (n − 5) + n(n − 1) · · · (n − 4) n(n − 1) · · · (n − 3) = (n − 4)(n − 5) − (n − 4) = (n − 4)2. A= VÝ dô 6. Gi¶i. 11 9 A12 A10 49 + A49 17 + A17 M= − A10 A817 49 49! 49! 17! 17! + + 37! 38! 7! 8! . = − 49! 17! 39! 9! n+1 2 n VÝ dô 7. Chøng minh An+2 n+k + An+k = k · An+k . Gi¶i.   (n + k)! 1 k(n + k)! k 2(n + k)! VT = 1+ = = = k 2·Ann+k . (k − 2)! k−1 (k − 1)(k − 2)! k! VÝ dô 8. T×m n nguyªn d−¬ng biÕt r»ng a) A3n = 20n. b) A5n = 18 · A4n−2. Gi¶i. Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 4 www.VNMATH.com a) A3n = 20 ⇐⇒ n! n∈Z = 20n ⇐⇒ n2 − 3n − 18 ⇐⇒ n = 6. (n − 3)! b) A5n = 18A4n−2 ⇐⇒ (n − 2)! n! = 18 · (n − 5)! (n − 6)! " ⇐⇒ n2 − 19n + 90 = 0 ⇐⇒ n=9 n = 10. VÝ dô 9. T×m n nguyªn d−¬ng biÕt Pn+3 = 720A5n · Pn−5. §S: n=7. Tæ hîp E lµ mét tËp gåm n phÇn tö. Mét tËp con cña E gåm r phÇn (1 ≤ r ≤ n) ®−îc gäi lµ mét tæ hîp chËp r cña n. C«ng thøc. Cnr = n! . r!(n − r)! Chó ý. (i) (xÕp thø tù) TËp hîp −−−−−−→  S y Tæ hîp (tËp con) (ii) Quy −íc (iii) ho¸n vÞ  r=n y (xÕp thø tù) −−−−−−→ chØnh hîp 0! = 1 =⇒ Cn0 = 1. r−1 r Cnr = Cnn−r , Cnr = Cn−1 + Cn−1 . VÝ dô 10. Rót gän biÓu thøc A= Cn1 Cn2 Cnn + 2 · 1 + · · · + n n−1 . Cn Cn Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 5 www.VNMATH.com Gi¶i. Cn1 = n n! Cn2 2!(n − 2)! =n−1 2· 1 =2· n! Cn 1!(n − 1)! ··············· Cnn 1 n n−1 = n · = 1. n! Cn (n − 1)!1! =⇒ A = n + (n − 1) + · · · + 1 = n(n + 1) . 2 VÝ dô 11. Chøng minh víi c¸c sè r, n nguyªn, kh«ng ©m sao cho 0 ≤ r ≤ n, ta cã Cnr Gi¶i. r−1 nCn−1 . = r r−1 nCn−1 n (n − 1)! n! = · = = Cnr . r r r!(n − r)! r!(n − r)! VÝ dô 12. Chøng minh víi c¸c sè r, n nguyªn, kh«ng ©m sao cho 0 ≤ r ≤ n, ta cã nCnr = (r + 1)Cnr+1 + rCnr . Gi¶i.   n! n! = (n − r) + r · r!(n − r)! r!(n − r)! n! n! = (n − r) · +r· r!(n − r)! r!(n − r)! n! n! = (r + 1) · +r· (r + 1)!(n − r − 1)! r!(n − r)! = (r + 1)Cnr+1 + rCnr . nCnr = n · Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com 6 trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 www.VNMATH.com a) Chøng minh víi c¸c sè r, n nguyªn, kh«ng ©m sao cho VÝ dô 13. r ≤ n, ta cã 0≤ r−1 r−1 r−1 Cnr = Cn−1 + Cn−2 + · · · + Cr−1 . b) Chøng minh víi k, n ∈ N, 3 ≤ k ≤ n ta cã k Cnk + 3Cnk−1 + 3Cnk−2 + Cnk−3 = Cn+3 . Gi¶i. a) n−1 r−1 r r V P = (Cnr − Cn−1 ) + (Cn−1 − Cn−2 ) + · · · + Cr−1 = Cnr = V T. b)
V T = Cnk + Cnk−1 +2 Cnk−1 + Cnk−2 + Cnk−2 + Cnk−3 | {z } {z } | {z } | k = · · · = Cn+3 . VÝ dô 14. Chøng minh víi 0 ≤ k ≤ n vµ k, n ∈ Z ta luèn cã n n n 2 C2n+k · C2n−k ≤ (C2n ) . Gi¶i. Cè ®Þnh n n n, xÐt d·y uk = C2n+k · C2n−k , 0 ≤ k ∈ Z. BÊt ®¼ng thøc cÇn chøng minh ®−îc viÕt l¹i: uk ≤ u0, ∀k ∈ Z, k ≥ 0. Chøng minh d·y (uk )k ®¬n ®iÖu gi¶m. ThËt vËy (2n + k + 1)! (2n − k − 1)! (2n + k)! (2n − k)! · < · n!(n + k + 1)! n!(n − k − 1)! n!(n + k)! n!(n − k)! 2n + k + 1 2n − k < ⇐⇒ n + 2nk > 0 : ®óng. ⇐⇒ n+k+1 n−k uk+1 < uk ⇐⇒ Suy ra n n n 2 uk ≤ u0 víi 0 ≤ k ∈ Z ⇐⇒ C2n+k · C2n−k ≤ (C2n ) : ®pcm. VÝ dô 15. T×m k ∈ N biÕt k k+2 k+1 C14 + C14 = 2C14 . Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 HD: §iÒu kiÖn k ∈ N, k ≤ 7 www.VNMATH.com 12. Ph−¬ng tr×nh trë thµnh k 2 − 12k + 32 = 0 cã hai nghiÖm k = 4, k = 8. VÝ dô 16. T×m c¸c sè x ∈ Z+ tho¶ m n ph−¬ng tr×nh Cx1 + 6Cx2 + 6Cx3 = 9x2 − 14x. HD: §iÒu kiÖn 3 ≤ x ∈ N (∗) (∗) V T = x3. Ph−¬ng tr×nh trë thµnh x3 − 9x2 + 14x = 0 ⇐⇒ x = 7. VÝ dô 17. T×m k sao cho c¸c sè C7k , C7k+1, C7k+2 theo thø tù ®ã lËp thµnh cÊp sè céng. HD: §iÒu kiÖn 5 ≥ k ∈ N. " C7k + C7k+2 = 2C7k+1 ⇐⇒ k 2 − 5k + 4 = 0 ⇐⇒ k=1 k = 4. Bµi tËp 1. Chøng minh r»ng víi k, n ∈ Z, 2 ≤ k ≤ n, ta cã k−2 . k(k − 1)Cnk = n(n − 1)Cn−2 2. Chøng minh r»ng víi k, n ∈ Z, 4 ≤ k ≤ n, ta cã k Cnk + 4Cnk−1 + 6Cnk−2 + 4Cnk−3 + Cnk−4 = Cn+4 . 3. Gi¶i bÊt ph−¬ng tr×nh §S: 3 ≤ x ≤ 4. x, y ∈ Z+ ®Ó ( x=8 §S: y = 3. 4. T×m 6 1 2 A2x − A2x ≤ · Cx3 + 10. 2 x y Cx+1 Cxy+1 Cxy−1 = = . 6 5 2 Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 8 www.VNMATH.com 5. TÝnh §S: 6. TÝnh §S: A25 A510 A= + . P2 7P5 46. S = P1A12 + P2A23 + P3A34 + P4A45 − P1P2P3P4. 2750.  7. TÝnh C= P5 P4 P3 P1 + + + A45 A35 A25 A15  A25. §S: 42. 2A2x + 50 = a22x. 8. Gi¶i ph−¬ng tr×nh §S: x = ±5. 9. T×m n sao cho §S: 10. T×m §S: Pn+3 = 720 · A5n · Pn−5. n = 7. n sao cho 1 A3n + 3A2n = Pn+1. 2 n = 4. 11. Gi¶i c¸c ph−¬ng tr×nh a) A2x = 2. §S: b) 3Px = Ax3 . §S: (*) c) x = 1, x = 2. Pn+5 = 240 · Ak+3 n+3 . Pn−k §S: d) x = 2. 0 ≤ k ≤ 11, n = 11. A2x · Cxx−1 = 48. §S: x = 4. Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 9 www.VNMATH.com e) f) Px+2 = 210. x−4 Ax−1 · P3 §S: x = 5. §S x = 2. §S: x = 5. §S: x = 4. 1 1 1 − = . C4x C5x C6x 12. Gi¶i c¸c ph−¬ng tr×nh a) b) A4x 24 = . A3x+1 − Cxx−4 23 7 Cx1 + Cx2 + Cx3 = x. 2 Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 10 www.VNMATH.com NhÞ thøc Newton vµ øng dông I. NhÞ thøc Newton 1 c«ng thøc nhÞ thøc newton Víi mäi cÆp sè a, b vµ mäi sè nguyªn n > 0, ta cã: (a + b)n = Cn0an + Cn1an−1b + Cn2an−2b2 + · · · + Cnn−1abn−1 + Cnnbn (1) = n X Cni an−ibi. i=0 2 C¸c nhËn xÐt vÒ c«ng thøc khai triÓn 1. Sè c¸c sè h¹ng ë bªn vÕ ph¶i cña c«ng thøc (1) b»ng n + 1, n lµ sè mò cña nhÞ thøc ë vÕ tr¸i. 2. Tæng c¸c sè mò cña a vµ b trong mçi sè h¹ng b»ng n. 3. C¸c hÖ sè cña khai triÓn lÇn l−ît lµ Cn0, Cn1, Cn2, . . . , Cnn−1, Cnn víi chó ý Cnk = Cnn−k , 0 ≤ k ≤ n. 4. Cnk = n−k+1 · Cnk−1. k Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com 11 trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 www.VNMATH.com 3 Mét sè d¹ng ®Æc biÖt 3.1 D¹ng 1 Thay a = 1 vµ b = x vµo (1), ta ®−îc (1 + x)n = Cn0 + Cn1x + Cn2x2 + · · · + Cnn−1xn−1 + Cnnxn. 3.2 D¹ng 2 Thay a = 1 vµ b = −x vµo (1), ta ®−îc (2) (1−x)n = Cn0 −Cn1x+Cn2x2 −· · ·+(−1)k Cnk xk +· · · (−1)nCnnxn. (3) 3.3 Mét sè hÖ thøc gi÷a c¸c hÖ sè nhÞ thøc Thay x = 1 vµo (2) ta ®−îc: Cn0 + Cn1 + Cn2 + · · · + Cnn = 2n. Thay x = 1 vµo (3) ta ®−îc: Cn0 − Cn1 + Cn2 − · · · + (−1)nCnn = 0. II. C¸c vÝ dô më ®Çu 1. Thùc hiÖn a. Khai triÓn (1 + x)10. b. So s¸nh hai sè (1, 1)10 vµ 2. Gi¶i a. (1 + x)10 = 1 + 10x + 45x2 + 120x3 + 210x4 + 252x5 + 210x6 + 120x7 + 45x8 + 10x9 + x10. b. Víi x > 0, ta cã (1 + x)10 > 1 + 10x. Do ®ã víi x = 0, 1 ta cã (1, 1)10 > 1 + 10 · (0, 1) = 2. Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 12 www.VNMATH.com 2. Thùc hiÖn khai triÓn (3x − 4)5. Gi¶i (3−4x)5 = 5 X C5i (3x)5−i(−4)i = 35C50x5−4·34C51x4+· · ·+(−4)5C55. i=0 Cã 6 sè h¹ng, do tÝnh chÊt cña tæ hîp, chØ cÇn t×m Ta cã C50, C51, C52. C50 = 1, C51 = 5, C52 = 10. VËy (3x − 4)5 = 243x5 − 1620x4 + 4320x3 − 5760x2 + 3840x − 1024. 3. TÝnh gi¸ trÞ cña c¸c biÓu thøc sau: a. S1 = C60 + C61 + C62 + · · · + C66. b. S2 = C50 + 2C51 + 22C52 + · · · + 25C55. Gi¶i a. Ta cã ngay S1 = C60 + C61 + C62 + · · · + C66 = 26 = 64. b. Ta cã (1 + x)5 = 5 X C5i xi. (1) i=0 Thay x = 2 vµo (1) ta ®−îc S2 = C50 + 2C51 + 22C52 + · · · + 25C55 = 35 = 243. 4. TÝnh gi¸ trÞ cña c¸c biÓu thøc sau: a. S1 = 2nCn0 + 2n−2Cn2 + 2n−4Cn4 + · · · + Cnn. b. S2 = 2n−1Cn1 + 2n−3Cn3 + 2n−5Cn5 + · · · + Cnn. Gi¶i Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com 13 trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 www.VNMATH.com Ta cã n (2 + 1) = n X Cni 2n−i ⇐⇒ i=0 n (2 − 1) = n X n X Cni 2n−i = 3n i=0 n X Cni 2n−i(−1)i ⇐⇒ i=0 (1) Cni 2n−i(−1)i = 1. (2) 3n + 1 . = 2 (3) i=0 Suy ra • (1)+(2) ta ®−îc n S1 = 2 Cn0 n−2 +2 Cn2 n−4 +2 Cn4 + ··· + Cnn • (1)-(2) ta ®−îc n−1 S2 = 2 Cn1 n−3 +2 Cn3 n−5 +2 Cn5 + ··· + Cnn 3n − 1 = . (4) 2 5. TÝnh gi¸ trÞ cña biÓu thøc sau: 2001 1 2000 k 2001−k 2001 0 0 C2002 + C2002 C2002 + · · · + C2002 C2002 + · · · + C2002 C1 . S = C2002 Gi¶i Ta xÐt 2002! (2002 − k)! 2002! · = k!(2002 − k)! (2001 − k)! k!(2001 − k)! 2002 · 2001! k . = = 2002C2001 k!(2001 − k)! k 2001−k C2002 C2002 = Tõ ®ã S ®−îc viÕt l¹i d−íi d¹ng 0 1 2001 S = 2002(C2001 +C2001 +· · ·+C2001 ) = 2002(1+1)2001 = 1001·22002. 6. (§HBK 98). Khai triÓn (3x − 1)16. Chøng minh r»ng 0 1 16 316C16 − 315C16 + · · · + C16 = 216. 7. (§H khèi D - 2002). T×m sè nguyªn d−¬ng n sao cho Cn0 + 2Cn1 + 4Cn2 + · · · + 2nCnn = 240. Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 14 www.VNMATH.com 8. (§H §µ L¹t 99). TÝnh hÖ sè cña 9. Víi x25y 10 trong khai triÓn (x3 + xy)15. n lµ sè nguyªn d−¬ng, chøng minh r»ng 1 + 4Cn1 + 42Cn2 + · · · + 4n−1Cnn−1 + 4nCnn = 5n. Gi¶i Thay x = 4 vµo (1 + x)n = Cn0 + Cn1x + Cn2x2 + · · · + Cnn−1xn−1 + Cnnxn. 10. Víi n lµ sè nguyªn d−¬ng, chøng minh r»ng Cn0 + Cn2 + · · · = Cn1 + Cn3 + · · · = 2n−1,  0 nÕu m > n m víi gi¶ thiÕt Cn = n!  nÕu m ≤ n. m!(n − m)! Gi¶i Ta cã 2n = Cn0 +Cn1 +Cn2 +· · ·+Cnn = Cn0 +Cn1 +Cn2 +· · ·+Cnn +· · · (1) 0 = Cn0−Cn1+Cn2−· · ·+(−1)nCnn = Cn0−Cn1+Cn2−· · ·+(−1)nCnn · · · (2) Suy ra • (1) + (2) ta ®−îc 2n = 2(Cn0 + Cn2 + · · · ) ⇐⇒ Cn0 + Cn2 + · · · = 2n−1. (3) • (1) − (2) ta ®−îc 2n = 2(Cn1 + Cn3 + · · · ) ⇐⇒ Cn1 + Cn3 + · · · = 2n−1. 11. Víi a. (4) n lµ sè nguyªn d−¬ng, chøng minh r»ng Cn0 − 2Cn1 + 22Cn2 − · · · + (−1)n · 2nCnn = (−1)n. Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com 15 trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 www.VNMATH.com Cn1 − 2Cn2 + 3Cn3 − · · · + (−1)k−1kCnk + · · · + (−1)n−1nCnn. b. Gi¶i Víi mäi x vµ víi n lµ sè nguyªn d−¬ng, ta cã (1−x)n = Cn0 −Cn1x+Cn2x2 −· · ·+(−1)k Cnk xk +· · ·+(−1)nCnnxn. (1) a. Thay x = 2 vµo (1) ta ®−îc (−1)n = Cn0 − 2Cn1 + 22Cn2 − · · · + (−1)n · 2nCnn, ®pcm. b. LÊy ®¹o hµm theo x hai vÕ cña (1), ta ®−îc −n(1 − x)n−1 = −Cn1 + 2Cn2x + · · · + n(−1)nCnnxn−1. (2) Thay x = 1 vµo (2) ta ®−îc 0 = −Cn1 + 2Cn2 + · · · + n(−1)nCnn ⇐⇒ Cn1 − 2Cn2 + 3Cn3 − · · · + (−1)n−1nCnn = 0, ®pcm. 12. Víi n lµ sè nguyªn d−¬ng, chøng minh r»ng a. (§HTCKT): Cn1 + 2Cn2 + · · · + (n − 1)Cnn−1 + nCnn = n · 2n−1. b. 2 · 1Cn2 + 3 · 2Cn3 + · · · + n(n − 1)Cnn = n(n − 1) · 2n−2. c. r (−1)r Crr Cnr + (−1)r+1Cr+1 Cnr+1 + · · · + (−1)nCnr Cnn = 0 víi r nguyªn d−¬ng vµ r ≤ n. Gi¶i Víi mäi x vµ víi n lµ sè nguyªn d−¬ng, ta cã (1 + x)n = Cn0 + Cn1x + Cn2x2 + · · · + Cnn−1xn−1 + Cnnxn. LÊy ®¹o hµm theo (1) x hai vÕ cña (1), ta ®−îc n(1+x)n−1 = Cn1 +2Cn2x+· · ·+(n−1)Cnn−1xn−2 +nCnnxn−1. (2) Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com 16 trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 a. Thay x=1 www.VNMATH.com vµo (2), ta ®−îc n · 2n−1 = Cn1 + 2Cn2 + · · · + (n − 1)Cnn−1 + nCnn, ®pcm. b. LÊy ®¹o hµm theo x hai vÕ cña (2), ta ®−îc n(n − 1)(1 + x)n−2 = 2 · 1Cn2 + 3 · 2Cn3x + · · · + n(n − 1)Cnnxn−2. (3) Thay x = 1 vµo (3), ta ®−îc n(n − 1) · 2n−2 = 2 · 1Cn2 + 3 · 2Cn3 + · · · + n(n − 1)Cnn, ®pcm. c. LÊy ®¹o hµm cÊp r theo x hai vÕ cña (1), ta ®−îc n(n−1) · · · (n−r+1)(1+x) n−r = n X k(k−1) · · · (k−r+1)Cnk xk−r . k=r Chia hai vÕ cña (4) cho (4) r!, ta ®−îc n(n − 1) · · · (n − r + 1)(1 + x)n−r r! n X k(k − 1) · · · (k − r + 1) Cnk xk−r = r! = = k=r n X k=r n X k! Cnk xk−r r!(k − r)! Ckr Cnk xk−r . (5) k=r Thay x = −1 vµo (5), ta ®−îc 0= n X k=r 13. Víi Ckr Cnk (−1)k−r ⇐⇒ n X Ckr Cnk (−1)k = 0, ®pcm. k=r n lµ sè nguyªn d−¬ng, chøng minh r»ng Cn2 + 2Cn3 + · · · + (n − 1)Cnn > (n − 2)2n−1. Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 17 www.VNMATH.com Gi¶i Víi mäi x vµ víi n lµ sè nguyªn d−¬ng, ta cã (1 + x)n = Cn0 + Cn1x + Cn2x2 + · · · + Cnn−1xn−1 + Cnnxn. Thay (1) x = 1 vµo (1), ta ®−îc 2n = Cn0 + Cn1 + Cn2 + · · · + C n−1n + Cnn. LÊy ®¹o hµm theo (2) x hai vÕ cña (1), ta ®−îc n(1+x)n−1 = Cn1 +2Cn2x+· · ·+(n−1)Cnn−1xn−2 +nCnnxn−1. (3) Thay x = 1 vµo (3), ta ®−îc n · 2n−1 = Cn1 + 2Cn2 + · · · + (n − 1)Cnn−1 + nCnn. (4) (4) − (3), ta ®−îc LÊy n · 2n−1 − 2n = −Cn0 + Cn2 + · · · + (n − 2)Cnn−1 + (n − 1)Cnn ⇐⇒ Cn2 + 2Cn3 + · · · + (n − 1)Cnn = (n − 2)2n−1 + 1 > (n − 2)2n−1. 14. Víi n lµ sè nguyªn d−¬ng, chøng minh r»ng Cn1 + 4Cn2 + · · · + n2n−1 · Cnn = = 4·4n−1Cn0 −(n−1)·4n−2Cn1 +(n−2)·4n−3Cn2 +· · ·+(−1)n−1Cnn. Gi¶i Ta cã n (1 + x) = n X Cni xi. i=0 LÊy ®¹o hµm hai vÕ ta cã n(1 + x)n−1 = Cn1 + 2Cn2x + · · · + (n − 1)Cnn−1xn−2 + nCnnxn−1. Thay x = 2 vµo, ta cã n · 3n−1 = Cn1 + 4Cn2 + · · · + n · 2n−1Cnn. (*) Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 18 www.VNMATH.com Ngoµi ra (x − 1)n = Cn0xn − Cn1xn−1 + · · · + (−1)nCnn. LÊy ®¹o hµm hai vÕ, ta cã n(x − 1)n−1 = n · Cn0xn−1 − (n − 1)Cn1xn−2 + · · · + (−1)n−1Cnn−1. Thay x = 4 vµo, ta ®−îc n·3n−1 = n4n−1Cn0−(n−1)4n−2Cn1+(n−2)4n−3Cn2−· · ·+(−1)n−1Cnn−1 (**) So s¸nh (*) vµ (**) ta cã ®iÒu ph¶i chøng minh. 15. Víi n lµ sè nguyªn d−¬ng, chøng minh r»ng a. (§HGTVT 2000): Cn0 b. Cn0 Cn1 Cn2 Cnk Cnn 2n+1 − 1 + + +···+ +···+ = . 1+1 1+2 1+k 1+n 1+n Cn2 Cnn 1 Cn1 n + − · · · + (−1) · = . − 1+1 1+2 1+n 1+n H−íng dÉn Ta cã (1 + x)n = n X Cnk xk . k=0 LÊy tÝch ph©n hai vÕ, ta ®−îc Zt (1 + x)ndx = 0 (1 + x) 1+n = (1 + t) n+1 n X k+1 Cnk · 0 n+1 ⇐⇒
Cnk xk dx k=0 0  n+1 t ⇐⇒ Zt X n = k=0 n X k+1 k=0 x k+1 #t 0 Cnk t . k+1 Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 19 www.VNMATH.com Thay t = 1 ta ®−îc a. Thay t = −1 ta ®−îc b. 16. Víi n, k lµ c¸c sè nguyªn d−¬ng vµ 1 ≤ k ≤ n, chøng minh r»ng k−1 k−2 0 Cn0Cnk − Cn1Cn−1 + Cn2Cn−2 − · · · + (−1)k Cnk Cn−k = 0. Gi¶i Víi mäi x vµ k lµ sè nguyªn d−¬ng, ta cã (1 + x)k = Ck0 + Ck1x + Ck2x2 + · · · + Ckk xk . ⇐⇒ Cnk (1 + x)k = Ck0Cnk + Ck1Cnk x + Ck2Cnk x2 + · · · + Ckk Cnk xk . (1) Ta cã k! k! · m!(k − m)! n!(n − k)! n! (n − m)! = · m!(n − m)! (k − m)!(n − k)! k−m = Cnm · Cn−m . Ckm · Cnk = Do ®ã (1) cã d¹ng k−1 k−2 2 0 Cnk (1+x)k = Cn0Cnk +Cn1Cn−1 x+Cn2Cn−2 x +· · ·+Cnk Cn−k xk . (1) Thay x = −1 vµo (2), ta ®−îc k−1 k−2 0 0 = Cn0Cnk − Cn1Cn−1 + Cn2Cn−2 − · · · + (−1)k Cnk Cn−k = 0, ®pcm. 17. Chøng minh r»ng víi c¸c sè m, n, p nguyªn d−¬ng sao cho p ≤ n vµ p ≤ m, ta cã p p p−1 1 0 Cn+m = Cn0Cm + Cn1Cm + · · · + Cnp−1Cm + CnpCm . Gi¶i Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com trung t©m luyÖn thi ®¹i häc: trÝ ®øc - 32/2 Nói Thµnh §µ n½ng §T: 0905.100.499 - 0511.624925 Víi mäi x vµ víi m, n 20 www.VNMATH.com lµ c¸c sè nguyªn d−¬ng, ta cã (1 + x)m+n = (1 + x)n · (1 + x)m. (1) MÆt kh¸c (1 + x) n+m = n+m X p Cn+m xp . (2) n+m X p X  p−k p ( Cnk Cm )x . p=0 vµ (1+x)n ·(1+x)m = n X Cnk xk · k=0 m X k k Cm x = p=0 k=0 k=0 (3) Do (1) nªn c¸c hÖ sè cña p x , p = 0, n + m trong c¸c khai triÓn (2) vµ (3) b»ng nhau. VËy p Cn+m = p X p−k Cnk · Cm , ®pcm. k=0 NhËn xÐt quan träng (a) Víi p = n = m, ta ®−îc n . (Cn0)2 + (Cn1)2 + · · · + (Cnn)2 = C2n (b) Víi p = r, N = n + m, ta ®−îc r r−1 1 r 0 CNr = CN0 −mCm + CN1 −mCm + · · · + CNr−1 −m Cm + CN −m Cm . (c) B¹n ®äc h·y lÊy ý t−ëng trong bµi tËp trªn ¸p dông víi khai triÓn (1 − x)n+m. Tõ ®ã chøng minh r»ng 0 2 1 2 2n 2 n (C2n ) − (C2n ) + · · · + (C2n ) = (−1)n · C2n . Biªn so¹n: GVC - Th.S Phan V¨n Danh www.VNMATH.com