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 k
nC ta ñaïo haøm hai veá trong khai trieån (a
+ x)n..
k
• Khi caàn chöùng minh ñaúng thöùc chöùa k(k – 1)
nC ta ñaïo haøm 2 laàn hai veá cuûa
khai trieån (a + x)n.
Baøi 136. Chöùng minh :
a)
... nC
n 1 n2 −
+
+
+ +
=
C 2C 3C3 2 n n
1 n
n n
n 1 −
b)
..
( 1)
nC
0
n n
2 C 2C 3C . − + n
1 n
3 n
n 1 −
n 3 −
n 1 −
− + − =
c)
( 1)
nC
n
.
n 1 −−
2 2 C 2 C 3.2 C ... n
1 n
3 n
n n
− + − + =
Giaûi
Ta coù nhò thöùc
2
n
n 1 −
−
(a + x)n =
.
x
... C x
1 C a C a x C a n
0 n n
2 n 2 n
n n
Ñaïo haøm 2 veá ta ñöôïc :
n 1
2
−
−
n 1 −
a
x 3C a
x
... nC x
n(a + x)n-1 = C −
+ + + +
1 n
2 n 2 2C a n
3 n 3 n
n n
+ + + +
a) Vôùi a = 1, x = 1, ta ñöôïc :
n 1 −
C 2C 3C ... nC
n2
1 n
3 n
2 n
n n
+ + + + =
b) Vôùi a = 1, x = –1, ta ñöôïc :
n 1 −
( 1)
nC
0
C 2C 3C ... +
1 n
3 n
2 n
n n
− + − − =
c) Vôùi a = 2, x = –1, ta ñöôïc :
n
n 1 −
n 3 −
n 1 −
n 1 −
.
( 1)
nC
n
2 2 C 2 C 3.2 C ... n
1 n
3 n
n =
− + − − +
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
100
99
k
k
... C x
100 C 2 .x ... C 2 − + +
( x) −
0 = C 2 100
1 100
k 100
100 100 100
− + +
a) ÖÙng vôùi k = 97 ta ñöôïc a97.
97
3
97
Vaäy
( 1)−
a97 =
100C 2
=
= –8.
− × × = – 1 293 600 100 ! 3!97! 8 100 99 98 × 6
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 :
′ f (x) = a1 + 2a2x + 3a3x2 + … + 100a100x99
Maët khaùc f(x) = (x – 2)100
′ f (x) = 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
//f (1)
b) Chöùng minh
+
+ +
−
=
2 2.1.C 3.2.C + n
3 n
4 n
n n
Ñaïi hoïc An ninh 1998
. 4.3.C ... n(n 1)C n(n 1)2n 2− −
Giaûi
a) T
′ f (x)
a co ù : f(x ) = (1 + x)n
⇒
= n(1 + x)n – 1
// (x
n – 2
⇒
f ) = n(n – 1)(1 + x)
//f (1) = n(n – 1)2n – 2 .
Vaäy
)
b Do khai trieån nhò
4
3
2
thöùc Newt on
nx
0C C+ 1 n
n
4 n
3 n
2 n
n n
3
n 1
x C x C x C x ... C f(x) = (1 + x)n = + + + + +
2 3x C
1 n
2 n
n n
3 n
4 n
2
n 2
′ f (x) = n(1 + x)n - 1 = C 2xC+ + + 4x C ... nx C− + + ⇒
3 n
2 n
4 n
n n
′′ f (x) = n(n – 1)(1 + x )n - 2 = n(n 1)x C− 2C 6xC 12x C ... + + + + − ⇒
Choïn x = 1 ta ñöôïc
n – 2 =
4 2C 6C 12C ... n(n 1)C n
3 n
2 n
n n
n(n – 1)2 . + + + + −
Baøi 139
. Chöùng minh
n 1 −
n 4 −
n 3 − .2 C
n 1 n3 − .
n 1 −+ 2 C 2 C 3
1 n
2 n
3 n
4 4.2 C ... nC n
n n
Ñaïi hoïc Kinh teá Quoác daân 2000
+ + + + =
Giaûi
n
0
2
3
n
n 1 −
n 2 −
n 3 −
Ta coù :
nC 2 +
1 C 2 n
2 n
3 n
n n
x C 2 x ... C x (2 + x)n = + x C 2 + + +
n 1 −
n 2 −
n 3 −
n 1 −
ha c Ñaïo øm 2 veá ta ñöôï
2 3x C 2
1 C 2 n
2 n
3 n
n n
2xC 2 ... nx C n(2 + x)n – 1 = + + + +
n 1 −
n 3 −
n 1 − 2 C 2 C 3C 2
Choï n x = 1 ta ñö ôïc
1 n
3 n
2 n
n n
n 1 −
n 2 −
n 3 −
n 1 −
... nC . n3n – 1 = + + + +
Baøi 140. Chöùng minh
+
+
+ +
=
1 C 3 n
2 2C 3 n
3 3C 3 n
n n
Ñaïi hoïc Luaät 2001
. ... nC n4
Giaûi
n
0
2
3
n
n 1 −
n 2 −
n 3 −
Ta coù :
+
+ +
nC 3 + 1
2 n
3 n
n n
(3 + x)n = x C 3 x ... C x x C 3 + C 3 n
n 1 −
n 2 −
n 3 −
n 1 −
n – 1 =
Ñaïo øm 2 veá ta ñöôïc ha
2 3x C 3
+
+
+ +
1 C 3 n
2 n
3 n
n n
2xC 3 ... nC x n(3 + x)
n 1 −
n 2 −
n 3 −
h C oïn x = 1
1 =
+
+
+ +
1 C 3 n
2 2C 3 n
3 3C 3 n
n n
n
n4n – ... n C . ⇒
Baøi 141. Tính A =
n 1 nC−
+ + −
−
+
−
n
2 n
3 n
1 n
4 n
Ñaïi hoïc Baùch khoa Haø Noäi 1999
C 2C 3C 4C ... ( 1)
Giaûi
2
3
n
Ta coù :
n 1) C x
0 n
n n
3 n
( (1 – x )n = − x C x − ... + + − + 2 1 C C x C n n
2
n 1 −
ña ñöôïc Laáy ïo haøm hai veá ta
n ( 1) nC x
2 C 2xC 3x C n
1 n
3 n
n n
... –n(1 – x)n – 1 = − + − + + −
n x ta coù : Choï = 1
n ( 1) nC
2 n
3 n
n n
C 3C ... + + − − 0 = − 1 C 2+ n
1 n
3 n
2 n
n n
⇒ A = nC C 2C 3C ... + + + − n 1 − ) ( 1 − = 0
Baøi 142. Ch
n
öùn g mi nh vôùi n ∈ N vaø n > 2
+ +
+
nnC ) <
2 (C 2C 3C ... + n
1 n
3 n
n! (*) 1 n
Giaûi
n
2 x C
+
+
+
0 n
1 xC n
2 n
n n
Ta coù : (1 + x)n = C ... x C+
n 1
1 =
ña eá ta ñöôïc : Laáy ïo haøm theo x hai v
+ +
+
1 n
2 n
n n
C 2xC ... nx C− n(1 + x)n –
n x Choï = 1 ta ñöôïc
+
1 n
n n
2 n
nC 2n – 1 = n C 2C + ... +
n 1 )−
(n.2 2n – 1 < n! (**) Vaäy (*) ⇔ < n! ⇔ 1 n
Keát quaû (**) seõ ñöôïc chöùng minh baèng qui naïp
u = 22 < 3! = 6 (**) ñ ùng khi n = 3. Thaät vaäy 4
k – 1
û ! > 2k – 1 G ö (**) ñuùng khi n = k vôùi k > 3 nghóa laø ta ñaõ coù : k iaû s
k – 1
Vaäy (k + 1)k! > (k + 1)2
k do k > 3 neân k + 1 > 4 )
⇔ (k + 1)! > 2 . 2
= 2 (
n – 1
Do ñoù (**) ñuùng khi n = k + 1.
Keát luaän : 2 < n! ñuùng vôùi ∀ n ∈ N vaø n > 2.
Baøi 143.
2
Chöùng minh
a)
+ +
+
=
n n
2 n
3 n
n 2 −
1.2C 2.3C ... (n 1)nC − n(n 1)2n− −
b)
+ + −
−
= 0
n n
2 n
3 n
n 1 −
n 2 −
n 2 −
1.2C 2.3C ... ( 1) (n 1)nC −
c)
n 4 − 3.4.2 C ...
+
+ +
=
−
2 n
3 n
n n
4 n
n
4
n 1 −
−
n 2 −
n(n 1)3 2 C 3.2 C + (n 1)nC −
d)
n 2 − 2 C 3.2 C 3.4.2
−
+
− + −
=
− .
3 n
n n
4 n
2 n
C ... ( 1) n(n 1) (n 1)nC −
Giaûi
2
n
n
n 1 −
−
Ta coù nhò thöùc
+
+
+
+
2 n 2 n
nC x
1 C a C a x C a n
0 n n
. (a + x)n = x ...
n
2
−
−
−
n – 2
Ñaïo haøm 2 veá 2 laàn , ta ñöôïc :
+
+ +
−
2 n 2 n
3 n 3 n
n n
n(n – 1)(a + x) = 1.2C a 2.3C a x ... (n 1)nC x
Vôùi a = 1, x = 1, ta ñöôïc :
a)
n 2 −
+ +
+
=
n n
2 n
3 n
1.2C 2.3C ... (n 1)nC − n(n 1)2 −
Vôùi a = 1, x = – 1, ta ñöôïc :
b)
n 2 −
+ + −
−
= 0
2 n
3 n
n n
1.2C 2.3C ... ( 1) (n 1)nC −
c) Vôùi a = 2, x = 1, ta ñöôïc :
n 2 −
n 2 −
n 3 −
+ +
+
=
3 n
n n
2 n
n 4 −
n 2 −
n 1 −
n 2 −
1.2.2 C 2.3.2 C ... (n 1)nC − n(n 1)3 −
+ +
+
+
=
3 2 C 3.2 C 3.4.2 C ... n
2 n
4 n
n n
⇔ (n 1)nC − n(n 1)3 −
d) Vôùi a = 2, x = –1, ta ñöôïc :
n 4 −
n 2 −
n 2 −
n 3 − 1.2.2 C 2.3.2 C 3.4.2 C ...
− + −
−
+
=
3 n
2 n
4 n
n n
n 1 −
n 2 −
n 4 −
n 2 −
( 1) n( n (n 1)nC − − 1)
−
− + −
+
=
− .
2 2 C 3 n
3 n
4 n
n n
⇔ .2 C 3.4.2 C ... ( 1) n(n 1) (n 1)nC −
aø
B i 144. Chöùng minh :
a)
n 1 (6−
+
+ +
=
+
1 4C ... n
n n
0 n
3C 2 n) . (n 3)C +
b)
n ( 1) (n
+ + −
−
+
= . 0
1 n
n n
0 n
3C 4C ... 3)C
Giaûi
2
n
−
n 1 − a C a x C a
+
+
+ +
0 nC n
2 n 2 n
1 n
n n
x ... C x Ta coù nhò thöùc (a + x)n =
0 n
3
5
−
−
n 1 4 C a x
ñöôïc : Nhaân 2 veá vôùi x3, ta
n 3 + .
+
+
+ +
nC a x
n n
1 n
2 n 2 C a n
x ... C x x3(a + x)n =
2
n 2 +
n 1 3 a x
Ñaïo haøm 2 veá, ta ñöôïc :
+
+
0 n 3C a x n
1 4C − n
n n
(n 3)C x . 3x2(a + x)n + nx3(a + x)n – 1 = ... + +
a = 1, x = 1, ta ñöôïc :
a) Vôùi
n
n 1 −
n 1 −
+
+ +
+
=
+
=
0 n
1 4C ... n
n n
3C (n 3)C 3.2 n2 2 . (6 n) +
a = , x = –1, ta ñöôïc :
b) Vôùi
1
−
n ) (n 3)C +
= . 0
1 n
0 n
n n
3C 4C ... ( 1 + + −
-- ------------- --------------------------
Daïng
3:
TÍCH PH
AÂN HAI VEÁ CUÛA NHÒ THÖÙC NEWT
ON ÑEÅ
CHÖÙNG MINH MO
ÄT ÑAÚNG THÖÙC
+ Vieát khai trieån Newton cuûa (ax + b)n.
+ 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. ta seõ ñöôï
Chuù yù :
k
• Caàn chöùng minh ñaúng thöùc ch
nC k 1 +
öùa ta laáy tích phaân vôùi caän thích hôïp hai veá
trong khai trieån cuûa (a + x)n.
k
• Caàn chöùn minh ña g th g
nC ta laáy tích phaân vôùi caän thích hôïp
+
ún öùc chöùa 1 k m 1
+ g khai trieån cu xm(a + x)n.
hai veá tron ûa
Baøi 145. Cho n
∈
≥
1
2
3 n
x (1 x ) dx
N vaø n 2.
a) Tính I =
+
0
∫
C
C
C
...
+
+
+
+
=
b) Chöùng minh :
0 n
1 C n
2 n
n n
1 3
1 6
1 9
3(
n 1 + − 2 1 3(n 1) +
1 n 1) +
Ñaïi hoïc Môû 1999
.
Giaûi
1
1
3
2 x (
3 n 1 x ) dx
3 n (1 x ) d(x
a) Ta coù : I =
+
+
+ 1)
0
∫
0∫
1 3
1
n 12 +
=
.
.
I =
⎡ ⎣
⎤− 1 ⎦
1 (1 x+ 3
1 3(n 1) +
n 1+
3 n 1 ) + ⎤ ⎥ ⎦
0
3
6
n
3n
=
b Ta coù : (1 + x3)n = )
... C
+
+
+ +
1 C C x C x n
2 n
0 n
nx
2
8
5
3
n
n 2 +
x2(1
+ x3)n =
⇒
x C x C x C ... x +
+ +
+
1 n
2 n
0 n
nC
Laáy tích phaân töø 0 ñeán 1 hai veá ta ñöôïc :
1
3
6
9
3n 3 +
I
=
C
C ...
+
+
+ +
0 n
1 C n
2 n
x 3
x 6
x 9
⎡ ⎢ ⎣
⎤ x ⎥+ 3n 3 ⎦
0
C
C ...
C
=
+
+
+ +
Vaäy :
0 n
1 C n
2 n
n n
1 3
1 6
1 9
+ −n 1 1 2 3(n 1) +
1 3n 3 +
Baøi 146.
Chöùng minh
=
k n C +∑ n k 1
n 12 + − 1 n 1 +
k 0=
Ñaïi hoïc Giao thoâng Vaän taûi 2000
Giaûi
2
n
Ta coù : (1 + x)n =
... C x
+
+
+ +
1 C C x C x n
2 n
0 n
n n
1
1
n
2
n
(1 x) dx
... C x
dx
=
Vaäy
+
+
+
+ +
1 C C x C x n
2 n
0 n
n n
)
(
0
0
∫
∫
1
1
2
3
n 1 +
n 1 +
⇔
1 C x C + n
0 n
2 n
n n
0
0
(1 = C ... C + + + x 2 x 3 x) + n 1 + x n 1 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⎤ ⎥ ⎦ ⎡ ⎢ ⎣
+
+
+ +
0 n
1 C n
2 n
n n
n 12 + − 1 n 1 +
n
= C C ... C ⇔ 1 2 1 3 1 + n 1
⇔
k C +∑ n k 1
n 12 + − 1 + n 1
k 0 =
2
3
n 1 +
=
2 1 2 1 2
Baøi 147. Tính :
+
+
+ + ...
0 n
1 C n
n1 C n
2 n
− 2
− 3
− + n 1
Tuyeån sinh Ñaïi hoïc khoái B 2003
. C C
Giaûi
2
3
n
+ C C x C x C x
+
+
+ +
2 n
3 n
n n
0 n
1 n
2
2
2
3
n
... C x
n dx =
dx
x)+
+
+
+
+ +
Ta coù : (1 + x)n = ... C x
1 C C x C x n
2 n
0 n
3 C x n
n n
Vaäy
)
(
1
1
(1∫
∫
2
2
2
3
4
+ n 1
+ n 1
⇔
+
+
+ +
1 + C x C n
0 n
2 n
3 n
n n
+ (1 x) + n 1
⎡ ⎢ ⎣
⎤ ⎥ ⎦
⎡ ⎢ ⎣
⎤ x ⎥+ n 1 ⎦
1
1
+ n 1
+ n 1
2
2
2
= C C ... C x 2 x 3 x 4
2
3
+ n 1
⇔
+
+
+ + ...
2 0 C [x] 1 n
1 ⎡ C x ⎣ n
⎤ ⎦
2 ⎡ C x ⎣ n
⎤ ⎦
n ⎡ C x ⎣ n
⎤ ⎦
1
1
1
+
+
2
3
+ n 1
+ n 1
+ n 1
2 = 3 − n 1 n 1 1 3 1 2 1 + n 1
+
+
+ +
⇔
0C n
1 C n
2 n
n n
− 2
− 3
− 1 + n 1
− 2 n 1+
2 1 2 2 3 1 = C ... C
Baøi 148.
nn
Chöùng minh :
3
+ n 1 2 C
−
+
+ +
=
n n
0 n
2 1 2 .C n
2 2 .C ... n
− ( 1) + n 1
− + ( 1) + n 1
Ñaïi hoïc Giao thoâng Vaän taûi 1996
1 2C 1 2 1 3
Giaûi
2
n
n ( 1) C x
−
+
+ + −
1 C n
2 n
n n
2
2
n
2
d
...
n ( 1) C
x =
dx
(1 x)−
−
+
+ + −
Ta coù : (1 – x)n x C x ... = C0 n
1 C C x C x n
2 n
0 n
n n
Vaäy
)nx
(
0
0
∫
∫
2
2
3
n
+ n 1
+ n 1
−
+
+ +
⇔
0 C x n
2 1 x C n
2 n
n n
− (1 x) + n 1
− ( 1) x + n 1
⎡ −⎢ ⎣
⎤ ⎥ ⎦
⎤ ⎥ ⎦
⎡ ⎢ ⎣
0
0
2
3
n
+ n 1
+ n 1
= C ... C 1 2 x 3
− ( 1)
−
−
+
+ +
⇔
0 n
1 C n
2 n
n C n
− + n 1
− ( 1) 2 + n 1
n
2
3
n
+ n 1
1 = 2C C ... 2 2 2 3
−
+
+ +
⇔
0 n
1 C n
2 n
n n
− ( 1) 2 + n 1
+ − 1 ( 1) n 1+
= 2C C ... C 2 2 2 3
Baøi 149. Chöùng minh :
n
n
− n 1
a)
+ −
+ +
=
0 − ( 1) C ( 1) n
1 n
n n
− ( n
n
C ... C 1 2 1 + n 1 1) + 1
b)
−
+ + −
=
0 n
1 n
n n
. C C ... ( 1) C 1 2 1 + n 1 1 + n 1
Giaûi
2
n
− 1
−
Ta coù nhò t höùc
n =
na
+
+
+ +
0 n 1 C a C n n
2 n 2 n
n n
1
1
n
n
− n 1
(a x) dx
C a C a x ... C x dx
+
+ +
+
(a + x) x C a x ... C x .
0 n n
1 n
n n
=
)
(
0
0
∫
1
1
−
+ n 1
Vaäy : ∫
n 1 2 C a x
⇔
+
+ + ...
0 n C a x n
1 n
n C x n
⎞ ⎟ ⎠
⎛ ⎜ ⎝
0
0
+ n 1
+ n 1
= 1 2 (a x) + n 1 + n 1+ 1 + n 1
+ (a 1)
− n 1
+
+ + ...
⇔
0 n C a n
1 C a n
n n
− n 1+
a = . C 1 2 1 + n 1
a)
n
+ n 1
− −
− n 1
V ôùi a = –1 , ta ñ öôïc :
n − ( 1) C
+ − ( 1 )
+ +
=
=
0 n
1 n
n n
− ( 1) + n 1
C ... C 1 2 1 + n 1 ( 1) + n 1
)
b Ta coù nhò thöùc
2
n
− n 1
−
n =
+
+
+ +
0 nC a n
2 n 2 n
1 n
n n
1 −
1 −
n
n
n 1 −
(a x) dx
C a C a x ... C x dx
(a + x) C a x C a x ... C x .
+
+ +
+
0 n n
1 n
n n
= Vaäy
(
)
0
0
∫
∫
1 −
1 −+ n 1
−
n 1 +
n 1 2 C a x
⇔
+
+
+
0 n C a x n
1 n
n C x n
+ (a x) n 1+
⎛ ⎜ ⎝
⎞ ⎟ ⎠
0
0
n 1 +
n 1 +
= ... 1 2 1 + n 1
− (a 1)
n 1 −
n 1 +
−
+
− + −
0 n C a n
1 C a n
n n
− n 1+
a = . ( 1) ... C ⇔ 1 2 1 + n 1
n 1 +
Vôùi a = 1, ta ñöôïc :
−
+
− + −
=
0 n
1 n
n n
− 1 + 1 n
n
. C C ... ( 1) C 1 2 1 + n 1
−
+ + −
=
0 n
1 n
n n
. ⇔ C C ... ( 1) C 1 2 1 + n 1 1 + n 1
1
19
x(1 x) dx
Baøi 150. Tính
−
0
∫
−
+
+ + ...
−
18 19
19 19
0 19
1 C 19
2 19
Ñaïi hoïc Noâng nghieäp Haø Noäi 1999
C C C C Ruùt goïn S = 1 2 1 3 1 4 1 20 1 21
G
iaûi
•
⇒
Ñaët t = 1 – x dt = –dx
Ñoåi caän
x 0 1
0
1
19
19
x(1 x) dx
t 1 0
−
(1 t)t −
( dt) −
0
1
∫
∫
1
1
20
21
20t
Vaäy I = =
)dt =
19 (t −
−
t
t
−
0
∫
1 20
1 21
⎤ ⎥⎦
0
1
2
= I = = ⇔ 1 20 1 21 1 420
Ta coù : •
+
+ +
−
2 19
18 18 19
− 19 C19
19 19 C x 19
2
3
19
20
(1 – x)19 = C0 x C x ... C x
−
+
+ +
−
0 19
1 C 19
2 C x 19
18 19 19
19C x
1
2
3
20
21
1
1 x(
19x) dx
x(1 – x )19 = xC x ... C x ⇒
−
−
+
+
−
0 19
1 C 19
18 19
19 19
0
∫
⎤ ⎥ ⎦
⎡ ⎢ ⎣
0
= C ... C C Vaä y I = x 2 x 3 x 20 x 21
−
+ + ...
−
0 19
1 C 19
18 19
19 19
= C C C ⇔ 1 420 1 2 1 3 1 20 1 21
. y S = Vaä 1 420
Baøi 151.
1
a) Tính
2 nx ) dx
x(1−
0
∫
n
C
C
C ...
C
b) Chöùng minh
−
+
−
+ +
=
0 n
1 C n
2 n
3 n
n n
1 2
1 4
1 6
1 8
( − 2n
2
1) +
1 2(n 1) +
Ñaïi hoïc Baùch khoa Haø Noäi 1997
Giaûi
1
1
2 n
x(1 x ) dx
a) T co I =
2 (1 x ) d(1 x )
−
2 n −
0
0
∫
a ù : =
∫
1
2 n 1 +
n 1 +
1 −− 2
−
⎡ −⎣ 0 1
⎤ ⎦
− 1 (1 x ) + n 1 2
⎡ ⎢ ⎣
⎤ ⎥ ⎦
0
= ⇔ I = − 1 + 2(n 1)
I = . ⇔ 1 2(n 1)+
b) Ta coù :
2
6
2n
4
n ( 1) C x
2 n 1 – x ) =
+ C C x C x C x
−
−
+ + −
3 n
n n
1 n
2 n
0 n
3
7
2n 1 +
( ...
⇒
n ( 1) C x
−
+
5 − C x C x
+ + −
0 n
1 C x n
3 n
n n
2 n
1
6
8
2
4
n
1
2n 2 +
x(1
2 n) dx =
x(1 ... – x2)n = xC
x−
+
−
+
−
+ C ...
0 n
1 C n
2 n
3 n
n n
0
∫
− ( 1) + 2n 2
⎡ ⎢ ⎣
⎤ ⎥ ⎦
0
n
C C x C Vaäy I = x 6 x 8 x 2 x 4
⇔
+
−
+ +
2 n
3 n
n n
0 C − n
1 C n
− ( 1) + 2n 2
= C C ... C 1 8 1 2 1 4 1 6 1 2(n 1)+
Baøi 152
* .Chöùng minh :
2
n 1 +
(n
n
C
C ...
+
+ +
=
0 n
1 n
1 3
1 4
C n3
n
1 +
2) 2 n 2 − + + (n 1)(n 2)(n 3) + +
+
.
Giaûi
a) Ta coù nhò thöùc
n
n 1 −
n
+
+ +
0 n C a n
1 n
n n
2
−
n 2 +
2(a + x)n =
C a x ... C x (a + x) =
n 1 3 + C x C a x
+ +
0 na n
1 n
n n
1
1
2
n
2
2
n
−
+
n 1 3 C a x
... C x
... C x Suy ra : x
x (a x) dx +
+
+ +
0 n C a x n
1 n
n n
= Vaäy
) dx
(
0
0
∫
∫
n 1 −
+
+
+
0 n C a n
1 C a n
n n
= ... C 1 3 1 4 1 + n 3
⇒
dt = dx Ñeå tính tích phaân ôû veá traùi, ñaët t = a + x
Ñoåi caän :
x 0 1
t a a + 1
1
a 1 +
2
n
2 n
(t a) t dt
Suy ra :
x (a x) dx +
−
0
a
∫
∫
a 1 +
n 3 +
n 2 +
2 n 1 +
a 1 +
n 2 +
n 1 +
(t
2at
2 n a t )dt
−
+
=
−
+
a
∫
⎛ ⎜ ⎝
⎞ ⎟ ⎠
a
n 2 +
n 2 +
n 1 +
n 1 +
n 3 +
n 3 +
= = t + n 3 2at + n 2 a t + n 1
2 a (a 1)
+
−
+
−
+ (a 1)
⎡ ⎣
⎤ ⎦
⎡ ⎣
⎤ ⎦
−
+
− ++ n 3
+ n 1
2a (a 1) a a a = n 2
n 3 +
n 2 +
n 1 +
1
Vôùi a = 1, ta ñöôïc :
2
n
x (a x) dx +
+
0
∫
− 2 + n 3
− + n 2
− 1 + n 1
1) 2 = 1 2(2 −
+ ⎛ n 1 ⎜ ⎝
2
n
n 1 +
2
−
= 2 − + + − − n 1 4 n 3 + 4 n 2 + 1 n 1 + 2 n 2 + 1 n 3 + 1 + ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠
2 (n 1)(n 2)(n 3) +
+
+
n 2 + + 2)(n 3) + +
(n 1)(n +
2
n 1 +
(n
=
2) 2− n 2 + + (n 1)(n 2)(n 3) + +
+
2
n 1 +
(n
C
C ...
C
=
+
+ +
=
0 n
1 n
n n
1 3
1 4
1 n 3 +
n 2) 2 2 + + (n 1)(n 2)(n 3) +
− +
+
. Suy ra :
