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chủ đề 1: toán tính tổng theo quy luật
Bai 1: Tính tổng
1000.999
1
4.3
1
3.2
1
2.1
1
++++=S
Giải:
Ta có
1
11
)1()1(
1
)1(
)1(
)1(
1
+
=
+
+
+
=
+
+
=
+ kkkk
k
kk
k
kk
kk
kk
(1)
áp dụng đẳng thức (1) ta có:
1000
999
1000
1
1
1000
1
999
1
999
1
998
1
4
1
3
1
3
1
2
1
2
1
1
1000.999
1
4.3
1
3.2
1
2.1
1
1000
1
999
1
1000.999
1
999
1
998
1
999.998
1
.
4
1
3
1
4.3
1
3
1
2
1
3.2
1
2
1
1
1
2.1
1
==
++++=++++=
=
=
=
=
=
S
S
Bài 2: Tính tổng
)1(
1
4.3
1
3.2
1
2.1
1
+
++++=
nn
S
Giải:
áp dụng đẳng thức (1) ta có:
11
1
1
1
11
4
1
3
1
3
1
2
1
2
1
1
)1.(
1
4.3
1
3.2
1
2.1
1
1
11
)1.(
1
.
.
4
1
3
1
4.3
1
3
1
2
1
3.2
1
2
1
1
1
2.1
1
+
=
+
=
+
++++=
+
++++=
+
=
+
=
=
=
n
n
n
S
nnnn
S
nnnn
Bµi 3: TÝnh tæng:
99.97
4
9.7
4
7.5
4
5,3
4
++++=Q
Gi¶i:
Ta cã:
99
64
99
32
.2
99
133
.2
99
1
3
1
2
99
1
97
1
9
1
7
1
7
1
5
1
5
1
3
1
2
99.97
4
9.7
4
7.5
4
5,3
4
______________________________________________________
99
1
97
1
.2
99.97
4
9
1
7
1
.2
9.7
4
7
1
5
1
.2
7.5
4
5
1
3
1
.2
5.3
4
==
−
=
−=
−++−+−+−=++++=
−=
−=
−=
−=
Q
Q
Bµi 4: TÝnh tæng
100.97
1
11.8
1
8.5
1
5.2
1
++++=
S
Gi¶i:
Ta cã:
300
49
100
49
.
3
1
100
150
3
1
100
1
2
1
3
1
100
1
97
1
11
1
8
1
8
1
5
1
5
1
2
1
3
1
100.97
1
11.8
1
8.5
1
5.2
1
100
1
97
1
3
1
100.97
1
11
1
8
1
3
1
11.8
1
8
1
5
1
3
1
8.5
1
5
1
2
1
3
1
5.2
1
==
−
=
−=
−++−+−+−=++++=
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
−=
−=
−=
−=
S
S
Bµi 5: TÝnh tæng
)1.().1(
1
5.4.3
1
4.3.2
1
3.2.1
1
+−
++++=
nnn
S
Gi¶i:
Ta cã:
)1.(
1
).1(
1
)1.().1(
1
)1.().1(
1
)1.().1(
)1()1(
)1.().1(
2
+
−
−
=
+−
+
−
+−
+
=
+−
−−+
=
+− nnnnnnn
n
nnn
n
nnn
nn
nnn
Suy ra:
)1(
)1(
1
).1(
1
2
1
)1.().1(
1
+
−
−
=
+− nnnnnnn
¸p dông ®¼ng thøc (1) ta cã:
+
++++−
−
++++=
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
+
−
−
=
+−
−=
−=
−=
)1(
1
5.4
1
4.3
1
3.2
1
2
1
).1(
1
4.3
1
3.2
1
2.1
1
2
1
)1.(
1
.).1(
1
2
1
)1.().1(
1
5.4
1
4.3
1
2
1
5.4.3
1
4.3
1
3.2
1
2
1
4.3.2
1
3.2
1
2.1
1
2
1
3.2.1
1
nnnn
S
nnnnnnn
Ta cã:
n
n
n
S
nnnn
S
nnnn
11
1
1
1
1
4
1
3
1
3
1
2
1
2
1
1
)1.(
1
4.3
1
3.2
1
2.1
1
1
1
1
).1(
1
.
.
4
1
3
1
4.3
1
3
1
2
1
3.2
1
2
1
1
1
2.1
1
1
1
−
=−=
−
−
++−+−+−=
+
++++=
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
−
−
=
−
−=
−=
−=
11
1
1
1
11
4
1
3
1
3
1
2
1
2
1
1
)1.(
1
4.3
1
3.2
1
2.1
1
1
11
)1.(
1
.
.
4
1
3
1
4.3
1
3
1
2
1
3.2
1
2
1
1
1
2.1
1
2
2
+
=
+
−=
+
−++−+−+−=
+
++++=
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
+
−=
+
−=
−=
−=
n
n
n
S
nnnn
S
nnnn
VËy S =
2
1
(S
1
- S
2
)=
)1(2
1
)1(
1
2
1
1
1
2
1
22
+
−=
+
−−
=
+
−
−
nnnn
nn
n
n
n
n
Bµi 6: TÝnh tæng
)2)(1(
1
4,3,2
1
3.2.1
1
++
+++=
nnn
S
Gi¶i:
Ta cã:
)2)(1(2
1
)2)(1(2
122
)2)(1(2
)1()2(
)2(2
1
)1(2
)2(2
1
2
1
.
2
1
)2)(1(
1
4.3
1
3.2
1
2
1
)1(21
.
2
1
)1(
1
3.2
1
2.1
1
2
1
)2)(1(
1
4.3
1
3.2
1
2
1
)1(
1
3.2
1
2.1
1
2
1
)2)(1(
1
)1(
1
2
1
)2)(1(
1
4.3
1
3.2
1
2
1
4.3.2
1
3.2
1
2.1
1
2
1
3.2.1
1
22
2
21
2
1
++
−=
++
−−−+
=
++
+−+
=
+
+
−
+
=−=
+
+
=
+
+
=
++
+++=
+
=
+
=
+
++=
++
+++−
+
++=
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
++
−
+
=
++
−=
−=
nnnn
nnnn
nn
nnn
n
n
n
n
SSS
n
n
n
n
nn
S
n
n
n
n
nn
S
nnnn
S
nnnnnnn
(2)
Bµi 7: TÝnh tæng
)3)(2)(1(
1
5.4.3.2
1
4.3.2.1
1
+++
+++=
nnnn
S
Gi¶i:
Ta có
)1(
)3)(2)(1(
1
)2)(1(
1
3
1
)3)(2)(1(
)3(
.
3
1
)3)(2)(1(
1
+++
++
=
+++
+
=
+++ kkkkkkkkkk
kk
kkkk
áp dụng đẳng thức (1) ta có:
)3)(2)(1(.3.4.3.2.1
4.3.2.1)3)(2)(1(
)3)(2)(1(
1
4.3.2.1
1
3
1
)3)(2)(1(
1
5.4.3
1
4.3.2
1
4.3.2
1
3.2.1
1
3
1
)3)(2)(1(
1
5.4.3
1
4.3.2
1
3
1
5.4.3.2
1
4.3.2
1
3.2.1
1
3
1
4.3.2.1
1
+++
+++
=
+++
=
+++
++=
+++
=
=
nnnn
nnnn
nnnn
S
nnnnnnnn
Bài 8: Dạng tổng theo quy luật
S = a
1
+ a
2
+ a
3
+ + a
n
Với d = a
2
a
1
= a
4
a
3
= =a
n
a
n-1
Thì a
n
= a
1
+ (n - 1).d
2
).(
1
naa
S
n
+
=
Bài 9: Tính tổng S = 1 + 2 + 3 + 4 + n
Giải
Cách 1:Ta có: S = 1+ 2+ 3+ 4+ +n
+ S = n + (n-1) + (n - 2) + (n - 3) + 1
2S =
1)(n 1) (n 1) (n ) 1(n ++++++++
= (n +1).n
n lần
2
).1( nn
S
+
=
Cách 2: Chọn hàm số g(x) = x
Ta xác định hàm số f(x) bậc 2 có dạng f(x) = ax
2
+ bx + c thoả mãn:
g(x) = f(x) f(x -1)
<=> x = ax
2
+ bx + c a(x-1)
2
b(x-1) - c
<=> x = ax
2
+ bx + c ax
2
+ 2ax - a bx + b - c
<=> x = 2ax a + b
=
=
=
=
2
1
2
1
0
12
b
a
ab
a
cxxxf ++=
2
1
2
1
)(
2
(c tuỳ ý)
Mặt khác:
2
).1(
2
1
2
1
0.
2
1
0.
2
1
2
1
2
1
4321
)0()()( )4()3()2()1(
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
222
nn
nnccnnnS
fnfnggggg
nfnfng
ffg
ffg
ffg
ffg
+
=+=
++++=++++=
=+++++
=
=
=
=
=
Bài 10: Tính tổng: S = 1 + 3 + 5 + + (2n - 1)
Giải:
Đặt g(x) = 2x 1
Ta chọn hàm f(x) bậc 2 có dạng: f(x) = ax
2
+ bx + c sao cho
g(x) = f(x) f(x - 1)
<=> 2x - 1 = ax
2
+ bx + c a(x-1)
2
b(x-1) - c
<=> 2x - 1 = ax
2
+ bx + c ax
2
+ 2ax - a bx + b - c
<=> 2x 1 = 2ax a + b
=
=
=
=
0
1
1
22
b
a
ab
a
vậy f(x) = x
2
+ c (c tuỳ ý)
Ta có
( )
2222
0)12 (531
)0()()( )4()3()2()1(
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
nccnccnnS
fnfnggggg
nfnfng
ffg
ffg
ffg
ffg
=+=++=+++=
=+++++
=
=
=
=
=
Bài 11: Tính tổng S = 2 + 4 + 6 + + 2n
Giải:
Đặt g(x) = 2x
Chọn hàm số f(x) có bậc 2 có dạng f(x) = ax
2
+ bx +c sao cho:
g(x) = f(x) f(x -1)
<=> 2x = ax
2
+ bx + c a(x-1)
2
b(x-1) - c
<=> 2x = ax
2
+ bx + c ax
2
+ 2ax - a bx + b - c
<=> 2x = 2ax a + b
=
=
=
=
1
1
0
22
b
a
ab
a
suy ra: f(x) = x
2
+ x + c (c tuỳ ý)
nnccnnnS
fnfnggggg
nfnfng
ffg
ffg
ffg
ffg
+=++−++=++++=
−=+++++
−−−−−−−−−−−−−−−−−−−−−−−−−−
−−=
−=
−=
−=
−=
222
)00(2 8642
)0()()( )4()3()2()1(
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
Bµi 12: TÝnh tæng S = 1
2
+ 3
2
+ 5
2
+ + (2n - 1)
2
Gi¶i:
§Æt g(x) = (2x -1)
2
Chän hµm sè f(x) cã bËc 3 cã d¹ng f(x) = ax
3
+ bx
2
+ cx + d sao cho:
g(x) = f(x) – f(x-1)
<=> (2x -1)
2
= ax
3
+ bx
2
+ cx + d – a(x-1)
3
– b(x-1)
2
– c(x-1) - d
<=> 4x
2
– 4x + 1 = ax
3
+ bx
2
+ cx + d – ax
3
+ 3ax
2
– 3ax + a – bx
2
+ 2bx – b – cx + c - d
<=> 4x
2
– 4x + 1 = 3ax
2
+ (2b – 3a)x + a – b + c
=+−
−=−
=
⇔
1
432
43
cba
ab
a
−=
=
=
⇔
3
1
0
3
4
c
b
a
nªn f(x) =
dxx +−
3
1
3
4
3
(d tuú ý)
3
4
3
1
3
4
0.
3
1
0.
3
4
3
1
3
4
)12( 7531
)0()()( )4()3()2()1(
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
3
33322222
nn
nnddnnnS
fnfnggggg
nfnfng
ffg
ffg
ffg
ffg
−
=−=
+−−
+−=−+++++=
−=+++++
−−−−−−−−−−−−−−−−−−−−−−−−−−
−−=
−=
−=
−=
−=
Bµi 13: TÝnh tæng: S = 1
2
+ 2
2
+ 3
2
+ n
2
Gi¶i:
§Æt g(x) = x
2
Chän hµm sè f(x) cã bËc 3 cã d¹ng f(x) = ax
3
+ bx
2
+ cx + d sao cho:
g(x) = f(x) – f(x-1)
<=> x
2
= ax
3
+ bx
2
+ cx + d – a(x-1)
3
– b(x-1)
2
– c(x-1) - d
<=> x
2
= ax
3
+ bx
2
+ cx + d – ax
3
+ 3ax
2
– 3ax + a – bx
2
+ 2bx – b – cx + c - d
<=> x
2
= 3ax
2
+ (2b – 3a)x + a – b + c
=
=
=
⇔
=+−
=−
=
⇔
6
1
2
1
3
1
0
032
13
c
b
a
cba
ab
a
f(x)=
dxxx +++
6
1
2
1
3
1
23
(d tuú ý)
6
)12)(1(
6
32
6
1
2
1
3
1
0.
6
1
0.
2
1
0.
3
1
6
1
2
1
3
1
4321
)0()()( )4()3()2()1(
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
23
23232322222
++
=
++
=
++=
+++−
+++=+++++=
−=+++++
−−−−−−−−−−−−−−−−−−−−−−−−−−
−−=
−=
−=
−=
−=
nnnnnn
S
nnnddnnnnS
fnfnggggg
nfnfng
ffg
ffg
ffg
ffg
Bµi 14: TÝnh tæng S = 2
2
+ 4
2
+ 6
2
+ + (2n)
3
Gi¶i:
§Æt g(x) = (2x)
2
Chän hµm sè f(x) cã bËc 3 cã d¹ng f(x) = ax
3
+ bx
2
+ cx + d sao cho:
g(x) = f(x) – f(x-1)
<=> 4x
2
= ax
3
+ bx
2
+ cx + d – a(x-1)
3
– b(x-1)
2
– c(x-1) - d
<=> 4x
2
= ax
3
+ bx
2
+ cx + d – ax
3
+ 3ax
2
– 3ax + a – bx
2
+ 2bx – b – cx + c - d
<=> 4x
2
= 3ax
2
+ (2b – 3a)x + a – b + c
=
=
=
⇔
=+−
=−
=
⇔
3
2
2
3
4
0
032
43
c
b
a
cba
ab
a
f(x)=
dxxx +++
3
2
2
3
4
23
(d tuú ý)
3
)132(2
3
264
3
2
2
3
4
0.
3
2
0.20.
3
4
3
2
2
3
4
)2( 8642
)0()()( )4()3()2()1(
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
2
23
23232322222
++
=
++
=
++=
+++−
+++=+++++=
−=+++++
−−−−−−−−−−−−−−−−−−−−−−−−−−
−−=
−=
−=
−=
−=
nnn
nnn
S
nnnddnnnnS
fnfnggggg
nfnfng
ffg
ffg
ffg
ffg
Bµi 15: TÝnh tæng: S = 1
3
+ 2
3
+ 3
3
+ + n
3
Gi¶i:
§Æt g(x) = x
3
Chän hµm sè f(x) cã bËc 4 cã d¹ng: f(x) = ax
4
+ bx
3
+ cx
2
+ dx + e sao cho
g(x) = f(x) – f(x-1)
<=> x
3
= ax
4
+ bx
3
+ cx
2
+ dx + e – a(x-1)
4
- b(x-1)
3
– c(x -1)
2
– d(x -1) – e
<=> x
3
= ax
4
+ bx
3
+ cx
2
+ dx + e – a(x
4
– 4x
3
+ 6x
2
– 4x + 1) – b(x
3
– 3x
2
+ 3x - 1)
- c(x
2
– 2x + 1) – dx + d – e
<=> x
3
= ax
4
+ bx
3
+ cx
2
+ dx + e – ax
4
+ 4ax
3
– 6ax
2
+ 4ax – a – bx
3
+ 3bx
2
– 3bx + b
- cx
2
+ 2cx – c – dx + d - e
<=> x
3
= 4ax
3
+ (3b – 6a)x
2
+ (4a – 3b + 2c)x – a + b – c + d
=
=
=
=
⇔
=+−+−
=+−
=−
=
⇔
0
4
1
2
1
4
1
0
0234
063
14
d
c
b
a
dcba
cba
ab
a
f(x) =
exxx +++
234
4
1
2
1
4
1
(e tuú ý)
−−−−−−−−−−−−−−−−−−−−−−−−−−
−−=
−=
−=
−=
−=
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
nfnfng
ffg
ffg
ffg
ffg
2
2222
234
23423423433333
2
)1(
4
)1(
4
)12(
4
2
4
1
2
1
4
1
0.
4
1
0.
2
1
0.
4
1
4
1
2
1
4
1
4321
)0()()( )4()3()2()1(
+
=
+
=
++
=
++
=
++=
+++−
+++=+++++=
−=+++++
nnnnnnn
nnn
S
nnneennnnS
fnfnggggg
Bµi 16: TÝnh tæng: S = 1
3
+ 3
3
+ 5
3
+ + (2n - 1)
3
Gi¶i:
§Æt g(x) = (2x - 1)
3
Chän f(x) bËc 3 cã d¹ng: f(x) = ax
4
+ bx
3
+ cx
2
+ dx + e sao cho:
g(x) = f(x) – f(x -1)
<=> (2x -1)
3
= ax
4
+ bx
3
+ cx
2
+ dx + e – a(x-1)
4
- b(x-1)
3
– c(x -1)
2
– d(x -1) – e
<=> 8x
3
– 12x
2
+ 6x -1 = ax
4
+ bx
3
+ cx
2
+ dx + e – a(x
4
– 4x
3
+ 6x
2
– 4x + 1) – b(x
3
–
3x
2
+ 3x - 1) - c(x
2
– 2x + 1) – dx + d – e
<=>8x
3
– 12x
2
+ 6x -1 = ax
4
+ bx
3
+ cx
2
+ dx + e – ax
4
+ 4ax
3
– 6ax
2
+ 4ax – a – bx
3
+
3bx
2
– 3bx + b - cx
2
+ 2cx – c – dx + d - e
<=> 8x
3
– 12x
2
+ 6x -1 = 4ax
3
+ (3b – 6a)x
2
+ (4a – 3b + 2c)x – a + b – c + d
=
−=
=
=
⇔
−=+−+−
=+−
−=−
=
⇔
0
1
0
2
1
6234
1263
84
d
c
b
a
dcba
cba
ab
a
f(x) = 2x
4
– x
2
+ e (e tuú ý)
−−−−−−−−−−−−−−−−−−−−−−−−−−
−−=
−=
−=
−=
−=
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
nfnfng
ffg
ffg
ffg
ffg
( ) ( )
24
24242433333
2
200.22)12( 7531
)0()()( )4()3()2()1(
nnS
nneennnS
fnfnggggg
−=
−=+−−+−=−+++++=
−=+++++
Bµi 17: TÝnh tæng S = 2
3
+ 4
3
+ 6
3
+ 8
3
+ + (2n)
3
Gi¶i
§Æt g(x) = (2x)
3
Chän f(x) bËc 3 cã d¹ng: f(x) = ax
4
+ bx
3
+ cx
2
+ dx + e sao cho:
g(x) = f(x) – f(x -1)
<=> 8x
3
= ax
4
+ bx
3
+ cx
2
+ dx + e – a(x-1)
4
- b(x-1)
3
– c(x -1)
2
– d(x -1) – e
<=> 8x
3
= ax
4
+ bx
3
+ cx
2
+ dx + e – a(x
4
– 4x
3
+ 6x
2
– 4x + 1) – b(x
3
– 3x
2
+ 3x - 1) -
c(x
2
– 2x + 1) – dx + d – e
<=>8x
3
= ax
4
+ bx
3
+ cx
2
+ dx + e – ax
4
+ 4ax
3
– 6ax
2
+ 4ax – a – bx
3
+ 3bx
2
– 3bx + b -
cx
2
+ 2cx – c – dx + d - e
<=> 8x
3
= 4ax
3
+ (3b – 6a)x
2
+ (4a – 3b + 2c)x – a + b – c + d
=
=
=
=
⇔
=+−+−
=+−
=−
=
⇔
0
2
4
2
0
0234
063
84
d
c
b
a
dcba
cba
ab
a
f(x) = 2x
4
+ 4x
3
+ 2x
2
+ e (e tuú ý)
−−−−−−−−−−−−−−−−−−−−−−−−−−
−−=
−=
−=
−=
−=
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
nfnfng
ffg
ffg
ffg
ffg
( ) ( )
2222
23423423433333
)1(2)12(2
2420.20.40.2242)2( 8642
)0()()( )4()3()2()1(
+=++=
++=+++−+++=+++++=
−=+++++
nnnnnS
nnneennnnS
fnfnggggg
Bµi 18: TÝnh tæng S = 1
4
+ 2
4
+ 3
4
+ + n
4
Gi¶i:
§Æt g(x) = x
4
Chän f(x) bËc 5 cã d¹ng: f(x) = ax
5
+ bx
4
+ cx
3
+ dx
2
+ ex + g sao cho:
g(x) = f(x) – f(x -1)
<=> x
4
= ax
5
+ bx
4
+ cx
3
+ dx
2
+ ex + g – a(x- 1)
5
– b(x-1)
4
– c(x-1)
3
– d(x-1)
2
– e(x-1) – g
<=> x
4
= ax
5
+ bx
4
+ cx
3
+ dx
2
+ ex – a(x
5
-5x
4
+ 10x
3
-10x
2
+ 5x -1) – b(x
4
– 4x
3
+ 6x
2
– 4x +
1) – c(x
3
– 3x
2
+ 3x – 1) – d(x
2
-2x +1) – e(x -1)
<=> x
4
= ax
5
+ bx
4
+ cx
3
+ dx
2
+ ex – ax
5
+ 5ax
4
– 10ax
3
+10ax
2
-5ax + a – bx
4
+ 4bx
3
- 6bx
2
+
6bx – b – cx
3
+ 3cx
2
– 3cx + c – dx
2
+ 2dx – d – ex + e
<=> x
4
= 5ax
4
+ (4b – 10a)x
3
+ (10a -6b + 3c)x
2
+ (6b – 5a - 3c + 2d)x + a – b + c – d + e
=
−=
=
=
=
⇔
=+−+−
=+−−
=+−
=−
=
⇔
15
8
2
1
3
1
2
1
5
1
0
02356
03610
0104
15
e
d
c
b
a
edcba
dcab
cba
ab
a
f(x) =
gxxxxx
++−++
15
8
2
1
3
1
2
1
5
1
2345
(g tuú ý)
−−−−−−−−−−−−−−−−−−−−−−−−−−
−−=
−=
−=
−=
−=
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
nfnfng
ffg
ffg
ffg
ffg
nnnnnS
ggnnnnnnS
fnfnggggg
15
8
2
1
3
1
2
1
5
1
0.
15
8
0.
2
1
0.
3
1
0.
2
1
0.
5
1
15
8
2
1
3
1
2
1
5
1
4321
)0()()( )4()3()2()1(
2345
2345234544444
+−++=
++−++−++−++=+++++=
−=+++++
Bµi 19: TÝnh tæng S = 1.2 + 2.5 + 3.8 + + n(3n - 1)
Gi¶i:
§Æt g(x) = x(3x -1)
Chän f(x) bËc 3 cã d¹ng: f(x) = ax
3
+ bx
2
+ cx + d sao cho:
g(x) = f(x) – f(x -1)
<=> x(3x-1) = ax
3
+ bx
2
+ cx + d – a(x-1)
3
– b(x-1)
2
– c(x-1) - d
<=> 3x
2
- x = ax
3
+ bx
2
+ cx + d – ax
3
+ 3ax
2
– 3ax + a – bx
2
+ 2bx – b – cx + c - d
<=> 3x
2
- x = 3ax
2
+ (2b – 3a)x + a – b + c
=
=
=
⇔
=+−
−=−
=
⇔
0
1
1
0
132
33
c
b
a
cba
ab
a
f(x) = x
3
+ x
2
+ d (d tuú ý)
−−−−−−−−−−−−−−−−−−−−−−−−−−
−−=
−=
−=
−=
−=
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
nfnfng
ffg
ffg
ffg
ffg
( )
)1(d n n)13( 8.35.22.1
)0()()( )4()3()2()1(
22323
+=+=−++=−++++=
−=+++++
nnnndnnS
fnfnggggg
Bµi 20: TÝnh tæng S = 1.2 + 2.3 + 3.4 + + n(n+1)
Gi¶i:
§Æt g(x) = x(x +1)
Chän f(x) bËc 3 cã d¹ng: f(x) = ax
3
+ bx
2
+ cx + d sao cho:
g(x) = f(x) – f(x -1)
<=> x(x+1) = ax
3
+ bx
2
+ cx + d – a(x-1)
3
– b(x-1)
2
– c(x-1) - d
<=> x
2
+ x = ax
3
+ bx
2
+ cx + d – ax
3
+ 3ax
2
– 3ax + a – bx
2
+ 2bx – b – cx + c - d
<=> x
2
+ x = 3ax
2
+ (2b – 3a)x + a – b + c
=
=
=
⇔
=+−
=−
=
⇔
3
2
1
3
1
0
132
13
c
b
a
cba
ab
a
f(x) =
3
1
x
3
+ x
2
+
3
2
x + d (d tuú ý)
−−−−−−−−−−−−−−−−−−−−−−−−−−
−−=
−=
−=
−=
−=
)1()()(
)3()4()4(
)2()3()3(
)1()2()2(
)0()1()1(
nfnfng
ffg
ffg
ffg
ffg
3
)2)(1(
3
)2)(1(
3
)2)(1(
3
)1(2)1(
3
22
3
23
3
2
3
1
3
2
3
1
)1( 5.44.33.22.1
)0()()( )4()3()2()1(
22
22323
23
23
++
=
++
=
++
=
+++
=
+++
=
++
=++=
−
+++=+++++=
−=+++++
nnn
S
nnnnnnnnnn
nnnnnnn
nnnS
ddnnnnnS
fnfnggggg
Bµi 21: TÝnh tæng S = 1 + 3 + 3
2
+ 3
3
+ + 3
n
Gi¶i:
Ta cã:
S = 1 + 3 + 3
2
+ 3
3
+ + 3
n
3S = 3 + 3
2
+ 3
3
+3
4
+ 3
n
+ 3
n+1
3S – S = 3
n+1
– 1 <=> 2S = 3
n + 1
– 1
S =
2
13
1
−
+n
Bµi 22: TÝnh tæng
S =
222222222222
2008
1
2007
1
1
1
2007
1
2006
1
1
1
4
1
3
1
1
1
3
1
2
1
1
1
++++++++++++
Gi¶i:
Ta cã:
++
−
++=++⇒
++
+++=+++++=
++
abc
cba
cba
cba
abc
cba
cba
bcacab
cba
cba
2
111111
2
111222111111
2
222
222222
2
NÕu a + b + c = 0 th×
22
222
1110
2
111111
++=
−
++=++
cbaabccba
cba
cbacba
cba
111111111
2
222
++=
++=++
(*)
¸p dông (*) ta cã:
S
2008
1
2008
2008
1
2007
1
1
2007
1
2006
1
1
4
1
3
1
1
3
1
2
1
1
2008
1
2007
1
1
1
2007
1
2006
1
1
1
4
1
3
1
1
1
3
1
2
1
1
1
)2008(
1
2007
1
1
1
)2007(
1
2006
1
1
1
)4(
1
3
1
1
1
)3(
1
2
1
1
1
2008
1
2007
1
1
1
2007
1
2006
1
1
1
4
1
3
1
1
1
3
1
2
1
1
1
222222222222
222222222222
−=
−++−+++−++−+=
−++−+++−++−+=
−
+++
−
++++
−
+++
−
++=
++++++++++++=
Bµi 23: TÝnh tæng S = 5 + 15 + 20 + + 2005
Gi¶i: Ta cã:
S = 5 + 10 + 15 + 20 + + 2005
S = 2005 + 2000 + 1995 +1990 + + 5
______________________________________
2S = 2010 + 2010 + 2010 + 2010 + + 2010 = 401.2010
S =
403005
2
2010.401
=
Bµi 24: TÝnh tæng S =
1 1 1
1 2 2 3 2004 2005
+ + +
× × ×
Gi¶i: Ta cã:
2005
2004
2005
1
1
2005
1
2004
1
3
1
2
1
2
1
1
1
20052004
1
32
1
21
1
2005
1
2004
1
20052004
1
3
1
2
1
32
1
2
1
1
1
21
1
=−=
−++−+−=+++=
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
−=
−=
−=
+
S
xxx
S
x
x
x
Bµi 25: TÝnh tæng S
n
=
1 1 1
1 2 3 2 3 4 2004 2005 2006
+ + +
× × × × × ×
.
Gi¶i: Ta cã:
−=
−++−+−=+++=
−=
−=
−=
+
20062005
1
21
1
2
1
20062005
1
20052004
1
43
1
32
1
32
1
21
1
2
1
200720062005
1
432
1
321
1
__________________________________________________________
20062005
1
20052004
1
2
1
200620052004
1
43
1
32
1
2
1
432
1
32
1
21
1
2
1
321
1
xx
S
xxxxxxxxxxxx
S
xxxx
xxxx
xxxx
n
n
Bµi 26: TÝnh tæng : A = 3+ 3
2
+ 3
3
+ + 3
30
Gi¶i: Ta cã:
A = 3 + 3
2
+ 3
3
+ +3
29
+ 3
30
3A = 3
2
+ 3
3
+ 3
4
+ +3
30
+ 3
31
________________________________
3A - A = 3
31
– 3
<=> 2A = 3
31
– 3
<=> A =
2
33
31
−
Bµi 27: Chøng minh r»ng A = 3+ 3
2
+ 3
3
+3
4
+3
5
+ 3
6
3
28
+ 3
29
+ 3
30
chia hÕt cho 13
Gi¶i:
Ta cã : A = 3(1 + 3 + 3
2
) + 3
4
(1 + 3 + 3
2
) + + 3
28
(1 + 3 + 3
2
)
)(13
1313.3)331.(3
1313.3)331(3
1313.3)331(3
28228
424
2
dfcmA
−−−−−−−−−−−−−−−−−−−−−−−
=++
=++
=++
Bµi 28: TÝnh A =
1 1 1 1
1 2 1 2 3 1 2 3 4 1 2 3 4 2010
+ + + ××××+
+ + + + + + + + + +××××+
XÐt biÓu thøc:
( )
( )
1 1 2 1 1
2
1
1 2 3 1 1
2
n n
n n n n n
= = = × −
÷
+
+ + +××××+ + +
v i ớ
*
n N∈
Víi n = 2 ta cã:
1 1 1
2
1 2 2 3
= × −
÷
+
Víi n = 3 ta cã:
1 1 1
2
1 2 3 3 4
= × −
÷
+ +
Víi n = 4 ta cã:
1 1 1
2
1 2 3 4 4 5
= × −
÷
+ + +
Víii n = 100 ta cã:
1 1 1
2
1 2 3 4 2010 2010 2011
= × −
÷
+ + + +×××+
VËy A =
1 1 2009 2009
2 2
2 2011 2 2011 2011
× − = × =
÷
×
Bµi 31: B=
21
1
+
+
32
1
+
+
43
1
+
+…+
20092008
1
+
+
20102009
1
+
Gi¶i:
12010
2009201020082009 342312
20092010
20092010
20082009
20082009
34
34
23
23
12
12
−=
−+−++−+−+−=
−
−
+
−
−
++
−
−
+
−
−
+
−
−
=
B
B
B
Bµi 32: So s¸nh A vµ B, sè nµo lín h¬n?
A = 2
0
+ 2
1
+ 2
2
+ 2
3
+ + 2
50
B = 2
51
Gi¶i: Ta cã:
A = 2
0
+ 2
1
+ 2
2
+ 2
3
+ + 2
50
2A = 2
1
+ 2
2
+ 2
3
+ 2
4
+ + 2
51
2A – A = 2
51
- 2
0
<=> A = 2
51
-1 < 2
51
= B
VËy A < B
Bµi 33: TÝnh tæng:
)100)(99(
1
)3)(2(
1
)2)(1(
1
)1(
1
++
++
++
+
++
+
+
=
xxxxxxxx
S
Gi¶i:
Ta cã:
100
1
99
1
99
1
2
1
2
1
1
1
1
11
)100)(99(
1
)3)(2(
1
)2)(1(
1
)1(
1
________________________________________________________________
100
1
99
1
)100)(99(
1
3
1
2
1
)3)(2(
1
2
1
1
1
)2)(1(
1
1
11
)1(
1
+
−
+
+
+
−+
+
+
+
−
+
+
+
−=
++
++
++
+
++
+
+
=
+
−
+
=
++
+
−
+
=
++
+
−
+
=
++
+
−=
+
+
xxxxxxxx
S
xxxxxxxx
S
xxxx
xxxx
xxxx
xxxx
)100(
100
100
11
+
=
+
−=
xxxx
S
Bµi 34: TÝnh tæng
)5)(4(
1
)4)(3(
1
)3)(2(
1
)2)(1(
1
)1(
1
209
1
127
1
65
1
23
11
22222
++
+
++
+
++
+
++
+
+
=
++
+
++
+
++
+
++
+
+
=
xxxxxxxxxx
S
xxxxxxxxxx
S
)5(
5
5
11
5
1
4
1
1
1
1
11
)5)(4(
1
)4)(3(
1
)3)(2(
1
)2)(1(
1
)1(
1
_____________________________________________________
5
1
4
1
)5)(4(
1
2
1
1
1
)2)(1(
1
1
11
)1(
1
+
=
+
−=
+
−
+
++
+
+
+
−=
++
+
++
+
++
+
++
+
+
=
+
−
+
=
++
+
−
+
=
++
+
−=
+
+
xxxx
S
xxxxx
S
xxxxxxxxxx
S
xxxx
xxxx
xxxx
Bµi 35: TÝnh tæng:
2006
20062006
2006
2006
1
2006
2006
2005
2005
4
4
3
3
3
3
2
2
2
2
1
2006.2005
2006200520052006
12
4334
6
3223
2
2112
2006.2005
2006200520052006
3648
4334
1218
3223
24
2112
2006200520052006
1
4334
1
3223
1
2112
1
−
=−=
−++−+−+−=
−
++
−
+
−
+
−
=
−
++
−
−
+
−
−
+
−
−
=
+
++
+
+
+
+
+
=
A
A
A
A
A
Bµi 36: Chøng tá
+++++=
2006
1
2005
1
3
1
2
1
1
1
2006.2005 3.2.1A
chia hÕt cho 2007
)(2007 )2006.204 3.12005.3.2.1(2007
1.2.3 2005.2006
) 2006.2004 4.3.12005.3.2.1(2007
2006.2005 3.2.1
1004.1003
2007
2004.3
2007
2005.2
2007
2006.1
2007
2006.2005 3.2.1
1003
1
1004
1
2004
1
3
1
2005
1
2
1
2006
1
12006.2005 3.2.1
dfcmA
A
A
A
++=
+++
=
++++=
++
++
++
+=
Bµi 37: TÝnh tæng S = 2 + 22+ 222+ +
n
2222222 2222222
(*)10 101010
2
9
10 101010
2
9
)1 111(10 101010
2
9
)110 ()110()110()110(
2
9
999.9.999999 999999
2
9
11111 11111 111111
2
11111 11111111.2 111.211.21.2
321
321
321
321
n
n
n
n
n
n
n
n
n
S
n
S
S
S
S
S
S
++++=+
−++++=
++++−++++=
−+−+−+−=
+++=
++++=
++++=
§Æt S
1
= 10
1
+ 10
2
+ 10
3
+ +10
n
10S
1
= 10
2
+ 10
3
+ 10
4
+ + 10
n+1
10S
1
– S
1
= 10
n+1
– 10 <=> 9S
1
= 10
n+1
-10 => S
1
=
9
1010
1
−
+n
Thay vµo (*) ta ®îc:
81
)10910(2
9
91010
2
9
9
1010
2
9
1
11
−−
=⇔
−−
=⇔
−
=+
+
++
n
S
nS
n
S
n
nn
Bµi 38: TÝnh tæng S = 1+ 2x +3x
2
+ 4x
3
+ +nx
n-1
Gi¶i:
Ta cã: S = 1+ 2x +3x
2
+ 4x
3
+ +nx
n-1
xS = x + 2x
2
+ 3x
3
+ 4x
4
+ + nx
n
S – xS = 1 + x +x
2
+ x
3
+ x
n-1
- nx
n
<=> (1-x)S = 1 + x +x
2
+ x
3
+ x
n-1
- nx
n
<=> S =
(*)
11
1
1
132
x
nxS
x
nxxxxx
n
nn
−
−
=
−
−+++++
−
TÝnh: S
1
= 1 + x + x
2
+ x
3
+ + x
n-1
xS
1
= x + x
2
+ x
3
+ x
4
+ + x
n
xS
1
– S
1
= x
n
– 1
<=> (x -1)S
1
= x
n
– 1
<=> S
1
=
1
1
−
−
x
x
n
Thay S
1
vµo (*) ta ®îc:
S =
2
1
1
)1(
1)1(
1
1
1
1
1
1
−
−−+
−=
−
−
+−−
=
−
−
−
−
+
+
x
nxxn
x
x
nxnxx
x
nx
x
x
nn
nnn
n
n
Bµi 39: TÝnh N = 75(4
1993
+ 4
1992
+ + 4
2
+5) +25
Gi¶i:
Ta cã: N = 75(4
1993
+ 4
1992
+ + 4
2
+5) +25
N = 75(4
1993
+ 4
1992
+ + 4
2
+4
1
+1) +25
N = 75.S +25 (*)
TÝnh S = 4
1993
+ 4
1992
+ + 4
2
+4
1
+1
4S = 4
1994
+4
1993
+ + 4
3
+ 4
2
+ 4
4S - S = 4
1994
– 1
3S = 4
1994
– 1
S =
3
14
1994
−
Thay vµo (*) ta ®îc:
N = 75.
3
14
1994
−
+25= 25(4
1994
-1 + 1) = 25.4
1994
Bµi 40: TÝnh B =
9932
2
1
2
1
2
1
2
1
++
+
+
1009932
2
1
2
1
2
1
2
1
2
+
++
+
=
B
99100100
2
1
1
2
1
2
1
22
1
2
1
2
−=⇒
−=⇔−
=− B
B
B
B
Bµi 41: Chøng minh bÊt ®¼ng thøc:
72,1
!
1
!3
1
!2
1
171,1 <++++<
n
víi n
5
≥
Chøng minh:
§Æt
!
1
!3
1
!2
1
1
n
S
n
++++=
, ta cã ngay víi n
5≥
th×:
S
n
>S
5
=
71,1
60
143
>
V©y, bÊt ®¼ng thøc thø nhÊt ®îc chøng minh.
B©y giê ta chøng minh bÊt ®¼ng thøc cßn l¹i. HiÓn nhiªn ta cã:
.
!6
1
!
1
!6
1
!
1
)!1(
1
!8
1
!7
1
!7
1
!6
1
!
1
!
!8
1
!8
8
!7
1
!7
7
!
1
!8
7
!7
6
!
1
!8
1
!7
1
<−=−
−
+−+−=
−++−+−=
−
+++<+++
nnn
nn
n
n
n
n
Tõ ®ã suy ra:
)(72,1
120
204
!6
2
120
202
!6
1
!6
1
!5
1
!4
1
!3
1
!2
1
1
!
1
!3
1
!2
1
1 dfcm
n
<=+=++++++<++++
Bµi 42: Chøng minh r»ng:
4
11
3
1
2
1
333
<+++
n
Chøng minh:
Ta cã:
(*)
)1.().1(
1
)1(
111
233
+−
=
−
=
−
<
kkk
kkkkk
¸p dông (*) ta cã:
(**)
)1.().1(
1
5.4.3
1
4.3.2
1
3.2.1
11
3
1
2
1
_________________________________________________________________
)1.().1(
11
5.4.3
1
4
1
4.3.2
1
3
1
3.2.1
1
2
1
333
3
3
3
3
+−
+++<+++
+−
<
<
<
<
+
nnn
n
nnn
n
MÆt kh¸c:
*)*(*
4
1
)1(2
1
4
1
)1(
1
2.1
1
2
1
)1.().1(
1
5.4.3
1
4.3.2
1
3.2.1
1
)1(
1
).1(
1
5.3
1
4.3
1
4.3
1
3.2
1
3.2
1
2.1
1
2
1
)1.().1(
1
5.4.3
1
4.3.2
1
3.2.1
1
____________________________________________________
)1.(
1
).1(
1
2
1
)1.().1(
1
5.4
1
4.3
1
2
1
5.4.3
1
4.3
1
3.2
1
2
1
4.3.2
1
3.2
1
2.1
1
2
1
3.2.1
1
<
+
−=
+
−<
+−
+++
+
−
−
++−+−+−<
+−
+++
+
−
−
=
−−
−=
−=
−=
+
nnnnnnn
nnnnnnn
nnnnnnn
KÕt hîp (**) vµ (***) ta cã:
)(
4
11
3
1
2
1
4
1
)1.().1(
1
5.4.3
1
4.3.2
1
3.2.1
11
3
1
2
1
333
333
dfcm
n
nnn
n
<+++
<
+−
+++<+++
Bµi 43: Chøng minh r»ng:
40
1
2004
1
1
6
1
5
1
65
1
3333
<++++<
n
Chøng minh:
Ta cã:
(*)
)1.().1(
1
)1(
111
233
+−
=
−
=
−
<
kkk
kkkkk
(**)
)2).(1.(
1
)23(
1
23
11
2233
++
=
++
=
++
>
kkk
kkkkkkk
¸p dông (*) ta cã:
*)*(*
2005.2004.2003
1
)1.().1(
1
7.6.5
1
6.5.4
1
2004
1
1
6
1
5
1
_________________________________________________________________
2005.2004.2003
1
2004
1
)1.().1(
11
7.6.5
1
6
1
6.5.4
1
5
1
3333
3
3
3
3
+
+−
+++<+++++
<
+−
<
<
<
+
nnn
n
nnn
n
MÆt kh¸c:
**)*(*
40
1
2005.2004.2
1
40
1
2005.2004
1
5.4
1
2
1
2005.2004.2003
1
)1.().1(
1
7.6.5
1
6.5.4
1
____________________________________________________
2005.2004
1
2004.2003
1
2
1
2005.2004.2003
1
)1.(
1
).1(
1
2
1
)1.().1(
1
7.6
1
6.5
1
2
1
7.6.5
1
6.5
1
5.4
1
2
1
6.5.4
1
<−=
−<++
+−
++
−=
+
−
−
=
−−
−=
−=
+
nnn
nnnnnnn
KÕt hîp (***) vµ (****) ta cã:
40
1
2004
1
1
6
1
5
1
3333
<++++
n
(I)
¸p dông (**) ta cã:
)(
65
1
390.2
12
390
1
30
1
2
1
2006.2005
1
6.5
1
2
1
2004
1
1
6
1
5
1
________________________________________________________________
2006.2005
1
2005.2004
1
2
1
2006.2005.2004
1
2004
1
)2)(1(
1
2
1
)2)(1(
11
8.7
1
7.6
1
2
1
8.7.6
1
6
1
7.6
1
6.5
1
2
1
7.6.5
1
5
1
3333
3
3
3
3
II
n
nnnnnn
n
==
−>
−>++++
−=>
++
=
++
>
−=>
−=>
+
Tõ (I) Vµ (II) ta cã ®iÒu ph¶i chøng minh:
40
1
2004
1
1
6
1
5
1
65
1
3333
<++++<
n
Bµi 44: T×m tØ sè cña A vµ B biÕt r»ng:
2005.1980
1
)25(
1
27.2
1
26.1
1
2005.25
1
)1980(
1
1980.2
1
1981.1
1
++
+
+++=
++
+
+++=
mm
B
nn
A
Trong ®ã, A cã 25 sè h¹ng vµ B cã 1980 sè h¹ng.
Gi¶i:
++++−
++++=
−++
+
−+−+−=
++
+
+++=
2005
1
1983
1
1982
1
1981
1
25
1
3
1
2
1
1
1
1980
1
2005
1
25
1
1980
11
1982
1
2
1
1981
1
1
1
1980
1
2005.25
1
)1980(
1
1982.2
1
1981.1
1
A
nn
A
nn
A
++++−
++++=
+++
+
++−
+
−+++++=
−++
+
−++−+−=
++
+
+++=
2005
1
1983
1
1982
1
1981
1
25
1
3
1
2
1
1
1
25
1
2005
1
1981
1
1980
1
25
1
27
1
26
1
1980
1
25
11
26
1
.
25
1
3
1
2
1
1
1
25
1
2005
1
1980
1
25
11
27
1
2
1
26
1
1
1
25
1
2005.1980
1
)25(
1
27.2
1
26.1
1
B
mmm
B
mm
B
mm
B
396
5
1980
25
2005
1
1983
1
1982
1
1981
1
25
1
3
1
2
1
1
1
25
1
2005
1
1983
1
1982
1
1981
1
25
1
3
1
2
1
1
1
1980
1
==
++++−
++++
++++−
++++
=
B
A
Bài 45: Chứng minh rằng bất đẳng thức sau với
2, nNn
22
1
3
1
2
1
32 <+++< n
n
n
Giải:
Ta có:
1)(**);1(2
1
)1(2
1
221
0)(*);1(2
1
)1(2
1
221
>=
+
=
+
<
+
=
>+=
+
+
=
++
>
+
=
kkk
kk
kk
kkkkk
kkk
kk
kk
kkkkk
áp dụng (*) và (**) ta có:
)(22
1
3
1
2
1
32
22
1
3
1
2
1
221222232
22
1
3
1
2
1
2212222
22
1
3
1
2
1
2212
)1(2
1
3
1
2
1
)21(2
)1 2312(2
1
3
1
2
1
)1 3423(2
_________________________________________________________
)1(2
1
)1(2
)23(2
3
1
)34(2
)12(2
2
1
)23(2
dfcmn
n
n
n
n
nnn
n
n
nn
n
n
n
n
n
n
nn
n
nn
nn
n
nn
<+++<
<+++<+<<
<+++<+<
<+++<+
<+++<+
+++<+++<++++
<<+
<<
<<
+
Bài 46: Chứng minh rằng
1 1 1
2
2 1 3 1 ( 1)n n
+ + + <
+
với mọi n
N*
Chứng minh:
Ta có:
(*)
1
11
2
2
.
1
11
11
1
11
1
11
1
11
1
11
).1(
)1(
1
+
=
+
=
=
+
+
<
+
+
+
=
+
=
+
=
+
kkkkk
k
kkk
k
k
kkkk
k
kk
k
kk
k
kk
áp dụng (*) ta có:
)(2
1
2
2
1
1
12
)1(
1
34
1
23
1
12
1
______________________________________________
1
11
2
)1(
1
4
1
3
1
2
34
1
3
1
2
1
2
23
1
2
1
1
1
2
12
1
dfcm
nnnn
nnnn
<
+
=
+
<
+
++++
+
<
+
<
<
<
+
Bài 47: : Rút gọn biểu thức sau:
A =
1 1 1 1
1 2 2 3 3 4 1n n
+ + +ììì+
+ + + +
(với
1n
)
Giải:
A =
1
1
34
34
23
23
12
12
+
++
+
+
nn
nn
A =
1 342312 ++++ nn
A =
1n
Bài 48: Chứng minh rằng với mọi số nguyên ta luôn có:
a.
1 1 1 1
1
1.2 2.3 3.4 ( 1)n n
+ + + + <
+
Chứng minh:
Biến đổi vế trái:
)(1
1
1
1
1
11
)1.(
1
4
1
3
1
4.3
1
3
1
2
1
3.2
1
2
1
1
1
2.1
1
dfcm
n
VT
nnnn
<
+
=
+
=
+
=
=
=
+
b.
1 1 1 1
2
2 2 2 2
1 2 3 n
+ + + + <
Giải:
Ta có:
(*)
)1(
111
22
=
<
kk
kkk
áp dụng (*) ta có:
)(2
1
2
1
1
1
1
11
4
1
3
1
2
1
1
1
__________________________________
1
1
1
)1(
11
4
1
3
1
4.3
1
4
1
3
1
2
1
3.2
1
3
1
2
1
1
1
2.1
1
2
1
222222
2
2
2
2
dfcm
nn
n
nnnn
n
<=+<+++++
=
<
=<
=<
=<
+
Bài 49:
a. Hãy tính
420
418
48
46
.
44
42
4
4
4
4
4
4
+
+
+
+
+
+
Giải:
T a có: a
4
+4=a
4
+4a
2
+4-4a
2
=(a
2
+2)
2
-(2a)
2
=[(a-1)
2
+1][(a+1)
2
+1] (*)
áp dụng (*) ta có :
)121)(119) (19)(17)(15)(13(
)119)(117) (17)(15)(11)(13(
420
418
48
46
.
44
42
222222
222222
4
4
4
4
4
4
++++++
++++++
=
+
+
+
+
+
+
221
1
442
2
121
11
2
2
==
+
+
=
b> Cho xyz = 1. Hãy tính tổng sau:
zxzyzyxyx ++
+
++
+
++ 1
1
1
1
1
1
Giải:
Ta có :
xxy
x
xyzxyx
x
yzy
++
=
++
=
++
11
1
(do xyz=1)
xyx
xy
yzxxyzxy
xy
zxz
++
=
++
=
++
11
1
2
( do xyz=1)
do đó :
1
111
1
1
1
1
1
1
1
=
=+
+
++
+
++
=
++
+
++
+
++
xyx
xy
xyx
x
xyxzxzyzyxyx
Bài 50: Cho các số a
1
, a
2
, , a
2009
đợc xác định bởi công thức sau:
( )
1)12(
2
+++
=
nnn
a
n
với n = 1, 2, 3, , 2009.
Chứng minh rằng:
2010
2008
2009321
<++++ aaaa
Chứng minh:
