Đại số tổ hợp - Chương V: Nhị thức Newton (phần 2)

ÑAÏI SOÁ TOÅ HÔÏP 
 Chöông V 
NHÒ THÖÙC NEWTON (phần 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 knC ta ñaïo haøm hai veá trong khai trieån (a 
+ x)n.. 
• Khi caàn chöùng minh ñaúng thöùc chöùa k(k – 1) knC ta ñaïo haøm 2 laàn hai veá cuûa 
khai trieån (a + x)n. 
Baøi 136. Chöùng minh : 
 a) 1 2n nC 2C 3C
3 n n 1
n n... nC n2
−+ +
1 2 3 n 1 n
n n nC 2C 3C .
−− + −
n 1 1 n 1 2
n n n n2 C 2 C 3.2 C ... ( 1) nC n
− −− + − + − =
0 n 1 n 1 2 n 2 2 n n
n n n nC a C a x C a x ... C x
− −+ + + +
1 n 1 2 n 2 3 n 3 2 n n 1
n n n na 2C a x 3C a x ... nC x
+ + = 
 b) n.. ( 1) nC 0+ − =
n 3 3 n 1 n− − c) . 
Giaûi 
 Ta coù nhò thöùc 
 (a + x)n = . 
 Ñaïo haøm 2 veá ta ñöôïc : 
 n(a + x)n-1 = C − − − −+ + + +
1 2 3 n n 1
n n n nC 2C 3C ... nC n2
a) Vôùi a = 1, x = 1, ta ñöôïc : 
 −+ + + + = 
b) Vôùi a = 1, x = –1, ta ñöôïc : 
 1 2 3 n 1 nn n n nC 2C 3C ... ( 1) nC 0
−− + − + − =
c) Vôùi a = 2, x = –1, ta ñöôïc : 
 . n 1 1 n 1 2 n 3 3 n 1 n2 C 2 C 3.2 C ... ( 1) nC n− − − −− + − + −n n n n =
0 k k 100 100
100 100 100 100( x) ... C x− + +
3 97( 1)−
Baøi 137. Cho (x – 2)100 = a0 + a1x + a2x2 +  + a100x100 . Tính : 
 a) a97 
 b) S = a0 + a1 +  + a100 
 c) M = a1 + 2a2 + 3a3 +  + 100a100 
Ñaïi hoïc Haøng haûi 1998 
Giaûi 
 Ta coù : 
 (x – 2)100 = (2 – x)100 
 = C 2 100 1 99 k 100C 2 .x ... C 2 −− + +
 a) ÖÙng vôùi k = 97 ta ñöôïc a97. 
 Vaäy a97 = 97100C 2
 = –8. 100 = !
3!97!
8 100 99 98
6
− × × ×
f (x)′
f (x)′
≥
//f (1)
 = – 1 293 600 
 b) Ñaët f(x) = (x – 2)100 = a0 + a1x + a2x2 +  + a100x100 
 Choïn x = 1 ta ñöôïc 
 S = a0 + a1 + a2 +  + a100 = (–1)100 = 1. 
 c) Ta coù : = a1 + 2a2x + 3a3x2 +  + 100a100x99 
 Maët khaùc f(x) = (x – 2)100 
 ⇒ = 100(x – 2)99 
 Vaäy 100(x – 2)99 = a1 + 2a2x + 3a3x2 +  + 100a100x99 
 Choïn x = 1 ta ñöôïc 
 M = a1 + 2a2 +  + 100a100 = 100(–1)99 = –100. 
Baøi 138. Cho f(x) = (1 + x)n vôùi n 2. 
 a) Tính 
 b) Chöùng minh 
 2 3 4 nn n n n2.1.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) = nx 
⇒ f (x)′ 2 2 3 3 4 n 1 nn3x C 4x C ... nx C−+ + + + 
n 2 n
nn(n 1)x C
−+ − 
. Chöùng minh 
 n 1 1 n 1 2n n2 C 2 C 3
− −+ +
Ñaïi hoïc Kinh teá Quoác daân 2000 
1 n 1 2 n 2 2 3 n 3 3 n n
n n n nC 2 x C 2 x C 2 x ... C x
− − −+ + + + 
 ha c 
1 n 1 2 n 2 2 3 n 3 n 1 n
n n n nC 2 2xC 2 3x C 2 ... nx C
− − − −+ + + + 
n x ôïc 
 n 1 1 n 1 2 3 n 3 nn n n n2 C 2 C 3C 2 ... nC
− − −+ + + + . 
Baøi 140. Chöùng minh 1 n 1 2 n 2 3 n 3 n n 1n n n nC 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 (x)′ 
 ⇒ f = n(n – 1)(1 + x)) 
 Vaäy //f (1) = n(n – 1)2n – 2 . 
 b Do khai trieån nhò on 
 (1 + x)n = 0C C+ 1 2 2 3 3 4 4 nn n n n n nx C x C x C x ... C+ + + + +
 = n(1 + x)n - 1 = 1n nC 2xC+ n n
)n - 2 = 2 3 2 4n n n2C 6xC 12x C ...+ + +⇒ f (x)′′ = n(n – 1)(1 + x
 Choïn x = 1 ta ñöôïc 
n – 2 = 2 3 4 nn n n n2C 6C 12C ... n(n 1)C+ + + + − . n(n – 1)2
Baøi 139
n 3 3 n 4 4 n
n n n.2 C 4.2 C ... nC
− −+ + + = n 1n3 − . 
Giaûi 
 Ta coù : 
 (2 + x)n = 0 nnC 2 +
 Ñaïo øm 2 veá ta ñöôï
 n(2 + x)n – 1 = 
 Choï = 1 ta ñö
 n3n – 1 =
Giaûi 
n n n n
 ha
x) n n 1n... nC x
 Ta coù : 
 (3 + x)n = 0 nnC 3 + 1 n 1 2 n 2 2 3 n 3 3 n nC 3 x C 3 x C 3 x ... C x− − −+ + + + 
 Ñaïo øm 2 veá ta ñöôïc 
 n(3 + n – 1 = 1 n 1 2 n 2 2 3 n 3n n nC 3 2xC 3 3x C 3
− − −+ + −+ + 
h
1 = 1 n 1 2 n 2 3 n 3 nn n n nC 3 2C 3 3C 3 ... n
− − −+ + + + . 
Baøi 141. Tính A = 1 2 3 4 n 1C 2C 3C 4C ... ( 1) nC−− + − + + −
Ñaïi hoïc Baùch khoa Haø Noäi 1999 
n n n
n1) C x− 
ña ñöôïc 
n n n 1
n... ( 1) nC x
 C oïn x = 1 
 ⇒ n4n – C
n
n n n n n
Giaûi 
 Ta coù : 
 (1 – x )n = 0 1n nC C x C− + 2 2 3 3x C x−n n ... (+ +
 Laáy ïo haøm hai veá ta
 –n(1 – x)n – 1 = 1 2 2 3n n nC 2xC 3x C
−− + − + + − 
n x ta coù : 
C 2+
öùn nh vôùi 
 Choï = 1 
 0 = − 1 2 3 n nn n n nC 3C ... ( 1) nC− + + − 
 ⇒ A = 1 2 3n n nC 2C 3C ... ( 1− + + + − n 1 nn) nC 0− = 
Baøi 142. Ch g mi n ∈ N vaø n > 2 
1 2 3 n
n n n
1 (C 2C 3C ... n!
n
+ + + + (*) 
Giaûi 
n n
n... x C+ 
ña eá ta ñöôïc : 
 1 = 1 2 n 1 nn n nC 2xC ... nx C
−+ + + 
n x 
2n – 1 = 1 2 nn nC 2C nC+ + 
nnC ) <
 Ta coù : (1 + x)n = 0 1 2 2n n nC xC x C+ + +
 Laáy ïo haøm theo x hai v
 n(1 + x)n –
 Choï = 1 ta ñöôïc 
 n n ...+
 Vaäy (*) ⇔ n 11 (n.2 )− < n! ⇔ 2n – 1 < n! 
n
(**) 
u = 22 < 3! = 6 
û ! > 2k – 1 
k – 1 
 k – 1 k do k > 3 neân k + 1 > 4 ) 
 Keát quaû (**) seõ ñöôïc chöùng minh baèng qui naïp 
 (**) ñ ùng khi n = 3. Thaät vaäy 4
 G ö (**) ñuùng khi n = k vôùi k > 3 nghóa laø ta ñaõ coù : kiaû s
 Vaäy (k + 1)k! > (k + 1)2
⇔ (k + 1)! > 2 . 2 = 2 (
 Do ñoù (**) ñuùng khi n = k + 1. 
n – 1 Keát luaän : 2 2. 
Baøi 143. 
 a) 
Chöùng minh 
2 3 n 2
n n n1.2C 2.3C ... (n 1)nC n(n 1)2
n−+ + + − = − 
 b) 2 3 n 2 nn n n1.2C 2.3C ... ( 1) (n 1)nC 0
−− + + − − = 
 c) n 1 2 n 2 n 2n n2 C 3.2 C 1)3
− − −− 
 d) n 1 2 n 2 3 nn2 C 3.2 C 3.4.2
− −− +
 Ta coù nhò thöùc 
 n nnC x+ . 
 2n
3 n 4 4 n
n n3.4.2 C ... (n 1)nC n(n
−+ + + + − =
4 4 n 2 n
n nC ... ( 1) (n 1)nC n(n 1)
− −− + − − = − . n
Giaûi 
(a + x)n = 0 n 1 n 1 2 n 2 2nC a C a x C a x ...
− −+ + +n n
Ñaïo haøm 2 veá 2 laàn , ta ñöôïc : 
2 n 2 3 n 3 n(n – 1)(a + x) = n nn n1.2C a 2.3C a x ... (n 1)nC x
n – 2 − − −+ + + − 
 Vôùi a = 1, x = 1, ta ñöôïc : 
n n 2
n n n1.2C 2.3C ... (n 1)nC n(n 1)2
a) 
2 3 −+ + + − = − 
 Vôùi a = 1, x = – 1, ta ñöôïc : 
 n 2 nn n n1.2C 2.3C ... ( 1) (n 1)nC 0
−− + + − − = 
c) Vôùi a = 2, x = 1, ta ñöôïc : 
 n 2 2 n 3 3 n n 2n n1.2.2 C 2.3.2 C ... (n 1)nC n(n 1)3
− − −+ + + − = − 
n 4 4 n n 2
n n n n2 C 3.2 C 3.4.2 C ... (n 1)nC n(n 1)3
− −+ + + + − = − 
 d) Vôùi a = 2, x = –1, ta ñöôïc : 
b) 
2 3
n
n 1 2 n 2 3− − ⇔ 
 nn 2 2 n 3 3 n 4 4 n 2 nn n n n1.2.2 C 2.3.2 C 3.4.2 C ... ( 1) (n 1)nC n( 1)
− − − −− + − + − − = −
− . 
aø
+ . 
 b) 0 1 nn n3C 4C ... ( 1) (n− + + −
Giaûi 
n
n 
 ñöôïc : 
1 n 1 4 2 n 2 5 n n 3
nC a x C a x ... C x
 ⇔ n 1 2n2 C 3− − n 2 3 n 4 4 n 2 nn n n.2 C 3.4.2 C ... ( 1) (n 1)nC n(n 1)− − −+ − + − − =
B i 144. Chöùng minh : 
 a) n)0 1 n n 1n n n3C 4C ... (n 3)C 2 (6
−+ + + + =
n
n3)C 0+ = . 
 Ta coù nhò thöùc (a + x)n = 0 nC 1 n 1 2 n 2 2 nn n na C a x C a x ... C x
− −+ + + +
 Nhaân 2 veá vôùi x3, ta
 x3(a + x)n = 0 n 3nC a x n n
− − + . 
1 n 1 3 n n 2
na x ... (n 3)C x
+ + + +
 Ñaïo haøm 2 veá, ta ñöôïc : 
 3x2(a + x)n + nx3(a + x)n – 1 = 0 n 2n n3C a x 4C
− ++ + + + . 
a = 1, x = 1, ta ñöôïc : 
n n n 1 n 1
n3)C 3.2 n2 2 (6 n)
− −= + = + . 
a = , x = –1, ta ñöôïc : 
n n
n) (n 3)C 0+ = . 
-- ------------- 
Daïng 
TÍCH PH ON ÑEÅ 
ÄT ÑAÚNG THÖÙC 
 + Laáy tích phaân xaùc ñònh hai veá thöôøng laø treân caùc ñoaïn : [0, 1], [0, 2] hay [1, 2] 
c ñaúng thöùc caàn chöùng minh. 
öùa 
 a) Vôùi 
 0 1n n3C 4C ... (n+ + + +
 b) Vôùi 1
 0 1n n3C 4C ... ( 1− + + −
--------------------------
3: 
AÂN HAI VEÁ CUÛA NHÒ THÖÙC NEWT
CHÖÙNG MINH MO
 + Vieát khai trieån Newton cuûa (ax + b)n. 
ta seõ ñöôï
 Chuù yù : 
• Caàn chöùng minh ñaúng thöùc ch
k
nC
k 1
 ta laáy tích phaân vôùi caän thích hôïp hai veá +
trong khai trieån cuûa (a + x)n. 
• Caàn chöùn minh ña g thg ún öùc chöùa 1
k m 1+ +
k
nC ta laáy tích phaân vôùi caän thích hôïp 
g khai trieån cu xm(a + x)n. 
Baøi 145. Cho n N vaø n 2. 
 a) Tính I = 
 b) Chöùng minh : 
hai veá tron ûa 
∈ ≥
1 2 3 n
0
x (1 x ) dx+∫ 
n 1
0 1 2 n
n n n n
1 1 1 1 2C C C 1C
3 6 9 n 1) 3(n 1)
+ −...
3(
+ + + =+ + . 
Ñaïi hoïc Môû 1999 
+
Giaûi 
 a) Ta coù : I = 
1
x ( = 2 3 n1 x ) dx+
0∫ 13
1 3 n 3
0
(1 x ) d(x 1)+ + ∫
 I = 
13 n 11 (1 x+
3
.
0
⎥⎦
 = )
+ ⎤
n 1+
n 12 1
3(n 1)
+1 ⎡ ⎤−⎣ ⎦+ . 
 b Ta coù : (1 + x3)n = 0 1 3 2 6n n nC C x C x ... C+ + + +) 
n 2 n
nC 
 = 
n 3n
nx 
 + x3)n = 2 0 5 1 8 2 3n n nx C x C x C ... x+ + + + ⇒ x2(1 +
Laáy tích phaân töø 0 ñeán 1 hai veá ta ñöôïc : 
 I
13 6 9 3n 3
0 1 2
n n n
0
x x x xC C C ...
3 6 9 3n 3
+⎡ ⎤+ + + +⎢ ⎥+⎣ ⎦
 Vaäy : 
+ −n 1 22 1 1= + + + ++ +
0 1 n
n n n n
1 1 1C C C ... C
3(n 1) 3 6 9 3n 3
Baøi 146. 
n 12 1
n 1
+ −
+Chöùng minh 
k
n
k 1
 = 
n C
+∑ k 0=
Ñaïi hoïc Giao thoâng Vaän taûi 2000 
Giaûi 
 Ta coù : (1 + x)n = 0 1 2 2 n nn n n nC C x C x ... C x+ + + + 
 Vaäy 
1 n
0
(1 x) dx+∫ = ( )1 0 1 2 2 n nn n n n0 C C x C x ... C x dx+ + + +∫ 
⇔ 
1
0
x)
n 1
+⎢ ⎥+⎣ ⎦
 = 
n 1(1 +⎡ ⎤ 12 3 n 10 1 2 n
n n n n
0
x x xC x C C ... C
2 3 n 1
+⎡ ⎤+ + + +⎢ ⎥+⎣ ⎦
⇔ 
n 12 1
n 1
+ −
+
0 1 2
n n n
1 1 1C C C ... C
2 3 n 1
+ + + + + = 
n
n
⇔ 
n 12 1
n 1
+ −
+ = 
kn
n
k 0
C
k 1= +∑ 
2 3 n 1
n1... C− . 0 1 2n n n2 1 2 1 2C C C2 3 n 1
+− −+ + + + +
Tuyeån sinh Ñaïi hoïc khoái B 2003 
Baøi 147. Tính : n
 Giaûi 
 Ta coù : (1 + x)n = 0 1 2 2 3 3 n nn n n n nC C x C x C x ... C x+ + + + + 
 Vaäy 
2
1
(1∫ n dx = x)+ ( )2 0 1 2 2 3 3 n nn n n n n1 C C x C x C x ... C x+ + + + +∫ dx 
⇔ 
2n 1(1 x)
1
+⎡ ⎤+
n 1⎢ ⎥+⎣ ⎦
 = 
22 3 4 n 1
0 1 2 3 n
n n n n n
x x x xC x C C C ... C
1n 1
+⎡ ⎤⎢ ⎥+⎣ ⎦
⇔
2 3 4
+ + + + +
n 1 n 13 2+ +
n 1 n 1+ +− =
2 21 1 20 2 1 2 2 3 n n 1
n 1
C [x] C x C x ... C x
n 1
+1
n 1 n n1 12 3
⎡ ⎤ ⎡ ⎤ ⎡ ⎤+ + + + ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ +
⇔ 
n 1 n 13 2+ +− = 
n 1+
2 3 n 1
1 2 n
n n n n
2 1 2 1 2 1C C ... C
2 3 n 1
+−0C − −+ ++ + + 
Chöùng minh : 
Baøi 148. 
n n( 1)0 2 1 3 2 n 1 n
n n n
1 1 ( 1) 12C 2 .C 2 .C ... 2 C
2 3 n 1 n 1
+−
n
+− + + + =+ +
d
− 
Ñaïi hoïc Giao thoâng Vaän taûi 1996 
Giaûi 
 Ta coù : (1 – x)n C = C0 1 2 2 n n nn n n nx C x ... ( 1) C x− + + + − 
 Vaäy 
2 n
0
(1 x)−∫ x = ( )nx dx 
⇔
2 0 1 2 2 n n
n n n n0
C C x C x ... ( 1) C− + + + −∫
2n 1
0
(1 x)
n 1
+⎡ ⎤−−⎢ ⎥+⎣ ⎦
 =
23 n n 1
0 2 1 2 n
n n n n
0
1 x ( 1) xC x x C C ... C
2 3 n 1
+⎡ ⎤−− + + +⎢ ⎥+⎣ ⎦
⇔ 
n 1( 1)
n 1
+− −− + = 
1 2 3 n n 10 1 2 n
n n n n
2 2 ( 1) 22C C C ...
2 3 n 1
+−− + + + + C
 ⇔ 
n1 ( 1)+ − = 
2 3 n n 1
0 1 2 n
n n n n
2 2 ( 1) 22C C C ... C
2 3 n 1
+−− + + + + n 1+
Baøi 149. Chöùng minh : 
a) 
n
n 0 n 1 1 n
n n n
1 1 (( 1) C ( 1) C ... C
2 n 1 n
− 1)
1
−− + − + + =+ + 
0 1 n n
n n n
1 1C C ... ( 1) C− + + − = . b) 1
2 n 1 n 1+ +
 = 
Giaûi 
höùc Ta coù nhò t
(a + x) 1 nan = 0 nn nC a C
1 2 n 2 2 n n
n nx C a x ... C x
− −+ + + + . 
Vaäy : ∫1 n(a x) dx+0 ( )1 0 n 1 n 1 n nn n n0 C a C a x ... C x dx−+ + +∫ 
⇔ 
1n 1(a x) ++ = 
0n 1+
1
0 n 1 n 1 2 n n 1
n n n
0
1 1C a x C a x ... C x
2 n 1
− +⎛ ⎞+ + +⎜ ⎟+ ⎝ ⎠
 ⇔ 
n 1 n 1(a 1) a+ ++ − = 
n 1+
0 n 1 n 1 n
n n n
1 1C a C a ... C
2 n 1
−+ + + + . 
a) ôùi a = –1 , ta ñV öôïc : 
+
− − − − − + − + + = =+ + +n n2 n 1 n 1 n
n 1 n
n 1 1 n1 1 ( 1) ( 1)) C ... C
1
) 
 Vaäy (a x) dx+∫ = 
n 0 ( 1n( 1) C
b Ta coù nhò thöùc 
 (a + x) +n = 0 nC a
1 n−
1 n 1 2 n 2 2 n n
n n n nC a x C a x ... C x
− −+ + + . 
(
0
)nC a C a x ... C x dx+ + +∫ 
1 0 n 1 n 1 n n− −
n n0
1n 1
⇔ 
0n 1+
 = (a x)
−++ 1n n 1
n
0
1 1... C x
2 n 1
−
− +⎛ ⎞+⎜ ⎟+ 
0 n 1 n 1 2
n nC a x C a x+ +⎝ ⎠
 ⇔ 
n 1 n 1(a 1) a+ +− − = 
n 1+
0 n 1 n 1 n 1 n
n n n
1 1C a C a ... ( 1) C
2 n 1
− +− + − + − + . 
a = 1, ta ñöôïc : 
 Vôùi 
0 1 n 1 n
n n n
1 1 1C C ... ( 1) C
2 n 1 n
+ −− + − + − =
1+ + . 
 ⇔ 0 1 n nn n n1 1 1C C ... ( 1) C2 n 1 n 1− + + − = + . +
Baøi 150. Tính 
 Ruùt goïn S = 
1 19
0
x(1 x) dx−∫ 
0 1 2 18 19
19 19
1 1C C− 19 19 191 1 1C C C ...2 3 4 20 21− + + +
Ñaïi hoïc Noâng nghieäp Haø Noäi 1999 
iaûi 
• ⇒ dt = –dx 
 Ñoåi caän 
G
Ñaët t = 1 – x 
x 0 1 
t 1 
 I = 
0
 = Vaäy
1 19x(1 x) dx−∫
0 
0 19
1
(1 t)t ( dt)− −∫ 
20t )dt = 
1
20 21 ⇔ I = 1 19
0
(t −∫
0
t t
20 21
−1 1 ⎤⎥⎦ = 
1 1
21
 = 
20
− 1
420
 Ta coù : 1 2 2 18 18 19 1919x C x ... C x C x− + + + − 
x 18 19 19 2019C x− 
Vaä I = 19x) dx− = 
• (1 – x)19 = C019 C19 19 19
)19 = 0 1 2 2 319 19 19 19xC x C x ... C x− + + + ⇒ x(1 C– 
 y 
0
x( 
1
1∫
12 3 20 21
0 1 18 19
19 19 19 19
0
x x x xC C ... C C
2 3 20 21
⎡ ⎤− + −+⎢ ⎥⎦
⎣
1 ⇔ 
420
 = 0 1 18 191 1 1 1C ... C C
2 3 20 21
− + + − 
y S = 
19 19 19 19C
 Vaä 1
420
. 
1 2 nx ) dx 
 b) Chöùng minh 
Baøi 151. 
 a) Tính 
0
x(1−∫
n
0 1 2 3
n n n n
1 1 1 1 (C C C C ...
2 4 6 8 2n
− + − + + nn1) 1C2 2(n 1)
− =+ + 
Ñaïi hoïc Baùch khoa Haø Noäi 1997 
Giaûi 
 a) T co I = a ù : = 
1 2 n
0
x(1 x ) dx−∫ 1 2 n 201 (1 x ) d(1 x )2− − −∫ 
12 n 1
0
1 (1 x )
2 n 1
+⎡ ⎤− ⇔ I = − +⎢ ⎥⎣ ⎦
 = n 11 0 1
2(n 1)
+⎡ ⎤− −⎣ ⎦+ 
 ⇔ I = 1 . 
2(n 1)+
 b) Ta coù : 
 (
0 1
n n
1 – x ) = n n
– x2)n = xC
2 n 0 1 2 2 4 3 6 n n 2nC C x C x C x ... ( 1) C x− + − + + − n n n
3 2 5 3 7 n n 2n 1
n n nC x C x ... ( 1) C x⇒ x(1 C x +− + − + + − 
Vaäy I = 
1
0
x(1∫ 2 n) dx = x−
1n
3 2n 2 n
n n
0
( 1)C ... x C
2n 2
+⎡ ⎤−+⎢ ⎥+⎣ ⎦
2 4 6 8
0 1 2
n n n
x x x xC C C
2 4 6 8
− + − +
⇔ 1 = 
2(n 1)+
n
1 2 3 n
n n n
1 1 ( 1)C C C ... C
2 4 6 8 2n 2
−+ − + + + 
* .Chöùng minh : 
0
n n
1 1C −
Baøi 152
n 1 2 n0 1 n
n n
1 1 1 2 (n 2) 2C C ... C
n
+ + +
n3 (n 1)(n 2)(n 3)3 4
−+ + + = . 
 (a + x) = n n nC a
2(a + x)n = 2 1 n 1 3 n n 2n n nC x C a x ... C x
+ + + +
Giaûi 
 a) Ta coù nhò thöùc 
n 0 n 1 n 1 n nC a x ... C x−+ + + 
 Suy ra : x 0 na − ++ + + 
0
( Vaäy 1 2 nx (a x) dx+∫ = )1 0 n 2 1 n 1 3 n 2n n n0 C a x C a x ... C x dx− ++ + + n∫
 = 0 n 1 n 11 1C a C a ...−+ + nn n n1 C4 n 3+ + 
 Ñeå tính tích phaân ôû veá traùi, ñaët t = a + x ⇒ dt = dx 
 Ñoåi caän : 
3
x 0 1 
t a a + 1 
 Suy ra : 
 = 
 = = 
1 2 n
0
x (a x) dx+∫ a 1 2 na (t a) t dt+ −∫ 
a 1 n 2 n 1 2 n
a
(t 2at a t )dt
+ + +− +∫
a 1n 3 n 2 2 n 1t 2at a t
+
a
n 3 n 2 n 1
+ + +⎛ ⎞− +⎜ ⎟ + + +⎝ ⎠
n 2 n 2 2 n 1 n 1n 3 n 3 2a (a 1) a a (a 1) a(a 1) a
+ + + ++ + ⎡ ⎤ ⎡+ − + −+ − = 
n 3 n 2 n 1
⎤⎣ ⎦ ⎣ ⎦− + + 
+ +
 Vôùi a = 1, ta ñöôïc : 
1 2 n
0
x (a x) dx+∫ = n 3 n 2 n 12 1 2(2 1) 2 1n 3 n 2 n 1
+ + +− − −− ++ + + 
 = n 1 4 4 1 2 12
n 3 n 2 n 1 n 2 n 3 n
+ ⎛ ⎞ ⎛− + + − −⎜ ⎟ ⎜+ + + + + +⎝ ⎠ ⎝
1
1
⎞⎟⎠ 
 = 
2
n 1 n n 2 22
(n 1)(n
+ + +
+ +
 = 
2)(n 3) (n 1)(n 2)(n 3)
−+ + + + 
n 1 22 (n n
(n 1)(n 2)(n 3)
+ + +
+ 
 Suy ra : 
2) 2−
+ +
n 1 2
0 1 n
n n n
1 1 1 2 (n n 2) 2C C ... C
3 4 n 3 (n 1)(n 2)(n 3)
+ + + −+ + + =+ + + + . 
PHAÏM ANG 
(Trung taâm Boài döôõng vaên hoùa vaø luyeän thi ñaïi hoïc Vónh Vieãn) 
 HOÀNG DANH - NGUYEÃN VAÊN NHAÂN - TRAÀN MINH QU