Toan-daisotohop-chuong5(2)

# Toan-daisotohop-chuong5(2) - AI SO TO HP Chng V NH THC...

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ÑAÏI SOÁ TOÅ HÔÏP Chöông V NHÒ THÖÙC NEWTON (ph n 2) Daïng 2: ÑAÏO HAØM HAI VEÁ CUÛA KHAI TRIEÅN NEWTON ÑEÅ CHÖÙNG MINH MOÄT ÑAÚNG THÖÙC Vieát khai trieån Newton cuûa (ax + b) n . Ñaïo haøm 2 veá moät soá laàn thích hôïp . Choïn giaù trò x sao cho thay vaøo ta ñöôïc ñaúng thöùc phaûi chöùng minh. Chuù yù : Khi caàn chöùng minh ñaúng thöùc chöùa k k n C ta ñaïo haøm hai veá trong khai trieån (a + x) n. . Khi caàn chöùng minh ñaúng thöùc chöùa k(k – 1) k n C ta ñaïo haøm 2 laàn hai veá cuûa khai trieån (a + x) n . Baøi 136. Chöùng minh : a) 12 nn C2 C3 C 3 1 n n . . . n Cn 2 ++ 123 n 1 n nnn C. −+− n1 1 n1 2 n n 2C 2C 3 . 2C . . .(1 )n C n −− −+ + = 0n 1n1 2n22 n n Ca Ca x Ca x . . . Cx + + 2n2 3n32 nn1 n n a 2C a x 3C a x ... nC x + += b) n . . ( 1 ) n C 0 + = n3 3 n1 n c) . Giaûi Ta coù nhò thöùc ( a + x ) n = . Ñaïo haøm 2 veá ta ñöôïc : n(a + x) n-1 = C + + 1 n . . n 2 a) Vôùi a = 1, x = 1, ta ñöôïc : +++ b) Vôùi a = 1, x = –1, ta ñöôïc :

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123 n 1 n nnn n C2 C3 C. . . ( 1 ) n C 0 −+− + = c) Vôùi a = 2, x = –1, ta ñöôïc : . n1 1 n1 2 n3 3 n1 n 2C 2C 3 . 2C . . .(1 )n C n −− −+ + nn n n = 0 k k 100 100 100 100 100 100 (x ) . . .Cx + + 3 97 (1 ) Baøi 137. Cho (x – 2) 100 = a 0 + a 1 x + a 2 x 2 + … + a 100 x 100 . Tính : a) a 97 b) S = a 0 + a 1 + … + a 100 c) M = a 1 + 2a 2 + 3a 3 + … + 100a 100 Ñaïi hoïc Haøng haûi 1998 Giaûi T a c o ù : (x – 2) 100 = (2 – x) 100 = 100 1 99 k 100 . x. . .C2 + a) ÖÙng vôùi k = 97 ta ñöôïc a 97 . V a ä y a 97 = 97 100 = –8. 100 = ! 3!97! 8 100 99 98 6 ×× × f(x) // f( 1 ) = – 1 293 600 b) Ñaët f(x) = (x – 2) 100 = a 0 + a 1 x + a 2 x 2 + … + a 100 x 100 Choïn x = 1 ta ñöôïc S = a 0 + a 1 + a 2 + … + a 100 = (–1) 100 = 1. c) Ta coù : = a 1 + 2a 2 x + 3a 3 x 2 + … + 100a 100 x 99 Maët khaùc f(x) = (x – 2) 100 = 100(x – 2) 99 Vaäy 100(x – 2) 99 = a 1 + 2a 2 x + 3a 3 x 2 + … + 100a 100 x 99 Choïn x = 1 ta ñöôïc M = a 1 + 2a 2 + … + 100a 100 = 100(–1) 99 = –100. Baøi 138. Cho f(x) = (1 + x) n vôùi n 2. a) Tính
b) Chöùng minh 234 n nnn n 2.1.C 3.2.C 4.3.C ... n(n 1)C n(n 1)2 n 2 +++ + = . Ñaïi hoïc An ninh 1998 Giaûi // (x n – 2 ) thöùc Newt f(x) = n x f(x) 2 2 3 3 4 n 1 n n 3x C 4x C ... nx C + + + + n 2 n n n(n 1)x C + . Chöùng minh n1 1 n1 2 nn 2C 2C 3 −− ++ Ñaïi hoïc Kinh teá Quoác daân 2000 1n 1 2n22 3n 33 n n n n C2 x C2 x C2 x . . . Cx + + + + ha c 1 2n 2 23n 3 n n n C2 2xC2 3xC2 . . . nx C + + n x ôïc 3 n3 n n n C 2 . . .n C + + . Baøi 140. Chöùng minh 1 2 3 n n 1 n C 3 2C 3 3C 3 ... nC n4 −−− + = . Ñaïi hoïc Luaät 2001 a) T ù : f(x + x) n a co ) = (1 = n(1 + x) n – 1 f = n(n – 1)(1 + x) ) Vaäy // f (1) = n(n – 1)2 n – 2 . b Do khai trieån nhò on (1 + x) n = 0 CC + 1 22 44 n n n n n x C xC x. . . C + + + + + = n(1 + x) n - 1 = 1 C 2xC + n n ) n - 2 = 23 2 4 n 2C 6xC 12x C ... + f(x ) ′′ = n(n – 1)(1 + x Choïn x = 1 ta ñöôïc n – 2 = 4 n n n 2C 6C 12C ... n(n 1)C + + . n(n – 1)2 Baøi 139 n3 3 4 n n n n . 2C 4 . 2C . . C + + + = n1 n 3 .

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## This note was uploaded on 11/26/2011 for the course MATH 1002 taught by Professor Chuck during the Spring '11 term at University of Western States.

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Toan-daisotohop-chuong5(2) - AI SO TO HP Chng V NH THC...

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