tính toán và đánh giá các tổng hữu hạn

S
p
(n) =
n

k=1
k
p
, L
p
(n) =
n

k=1
(2k − 1)
p
,
T
p
(n) =
n

k=1
(−1)
k
k
p
, F
p
(n) =

n

j=1
j!j
p
.
S
p
(n) =
n

k=1
k
p
,
S
1
(n) =
n(n + 1)
2
.
S
2
(n) S
3
(n).
S
2
(n).

k
3
− (k − 1)
3
= 3k
2
− 3k + 1 ⇔ 3k
2
= k
3
− (k − 1)
3
+ 3k − 1.
3
n

k=1
k
2
=
n

k=1
[k
3
− (k − 1)
3
+ 3k − 1] =
n


k=1
[k
3
− (k − 1)
3
] + 3
n

k=1
k −
n

k=1
1
= n
3
+ 3
n(n + 1)
2
− n =
n(n + 1)(2n + 1)
2
.
S
2
(n) =
n(n + 1)(2n + 1)
6
.
S

3
(n).
k
4
− (k − 1)
4
= 4k
3
− 6k
2
+ 4k − 1 ⇔ 4k
3
= k
4
− (k − 1)
4
+ 6k
2
− 4k + 1.
4
n

k=1
k
3
=
n

k=1
[k

4
− (k − 1)
4
+ 6k
2
− 4k + 1]
=
n

k=1
[k
4
− (k − 1)
4
] + 6
n

k=1
k
2
− 4
n

k=1
k + n
= n
4
+ 6S
2
(n) − 4S

1
(n) + n
= n
4
+ n(n + 1)(2n + 1) − 2n(n + 1) + n.
S
3
(n) =
n
2
(n + 1)
2
4
.
S
p
(n), p ∈ N, p ≥ 4.
(a + b)
p
=
p

i=0
C
i
p
a
p−i
b
i

, C
i
p
=
p!
i!(p − i)!
.
(k − 1)
p
=
p

i=0
C
i
p
k
p−i
(−1)
i
= k
p
− pk
p−1
+
p

i=2
C
i

p
k
p−i
(−1)
i
,
pk
p−1
= k
p
− (k − 1)
p
+
p

i=2
C
i
p
k
p−i
(−1)
i
.
k = 1, 2, , n,
pS
p−1
(n) = n
p
+

p

i=2
C
i
p
(−1)
i
S
p−i
(n)
S
p−1
.
S
4
(n) =
1
30
n(n + 1)(2n + 1)(3n
2
+ 3n − 1),
S
5
(n) =
1
12
n
2
(n + 1)

2
(2n
2
+ 2n − 1),
S
6
(n) =
1
42
n(n + 1)(2n + 1)(3n
4
+ 6n
3
− 3n + 1),
S
7
(n) =
1
24
n
2
(n + 1)
2
(3n
4
+ 6n
3
− n
2
− 4n + 2),

S
8
(n) =
1
90
n(n + 1)(2n + 1)(5n
6
+ 15n
5
+ 5n
4
− 15n
3
− n
2
+ 9n − 3),
S
9
(n) =
1
20
n
2
(n + 1)
2
(n
2
+ n − 1)(2n
4
+ 4n

3
− n
2
− 3n + 3).
L
p
(n) =
n

k=1
(2k − 1)
p
, p ∈ N
∗
.
L
1
(n) = 1 + 3 + 5 + + (2n − 1) = n
2
.
p > 1
L
p
(n) = 1
p
+ 3
p
+ 5
p
+ + (2n − 1)

p
= [1
p
+ 2
p
+ 3
p
+ 4
p
+ + (2n − 1)
p
+ (2n)
p
]
− [2
p
+ 4
p
+ 6
p
+ + (2n)
p
]
= S
p
(2n) − 2
p
S
p
(n).

L
p
(n) = S
p
(2n) − 2
p
S
p
(n),
S
p
(n)
L
2
(n) = S
2
(2n) − 2
2
S
2
(n) =
4
3
n
3
−
1
3
n,
L

3
(n) = S
3
(2n) − 2
3
S
3
(n) = 2n
4
− n
2
,
L
4
(n) = S
4
(2n) − 2
4
S
4
(n) =
16
5
n
5
−
8
3
n
4

+
7
15
n,
L
5
(n) = S
5
(2n) − 2
5
S
5
(n) =
16
3
n
6
−
20
3
n
4
+
7
3
n
2
,
L
6

(n) = S
6
(2n) − 2
6
S
6
(n) =
64
7
n
7
− 16n
5
+
28
3
n
3
−
31
21
n,
L
7
(n) = S
7
(2n) − 2
7
S
7

(n) = 16n
8
−
112
3
n
6
+
98
3
n
4
−
31
3
n
2
,
L
8
(n) = S
8
(2n) − 2
8
S
8
(n) =
256
9
n

9
−
256
3
n
7
+
1568
15
n
5
−
496
9
n
3
+
127
15
n,
L
9
(n) = S
9
(2n) − 2
9
S
9
(n) =
256

5
n
10
− 192n
8
+
1568
5
n
6
− 248n
4
+
381
5
n
2
.
p ∈ N
∗
.
P
p
(n) =
n

k=1
k
p
(k + 1)

2
.
P
p
(n)
P
p
(n) =
n

k=1
(k
p+2
+ 2k
p+1
+ k
p
) = S
p+2
(n) + 2S
p+1
(n) + S
p
(n),
S
p
(n)
P
1
(n) =

1
4
n
4
+
7
6
n
3
+
7
4
n
2
+
5
6
n,
P
2
(n) =
1
5
n
5
+ n
4
+
5
3

n
3
+ n
2
+
2
15
n,
P
3
(n) =
1
6
n
6
+
9
10
n
5
+
5
3
n
4
+
7
6
n
3

+
1
6
n
2
−
1
15
n,
P
4
(n) =
1
7
n
7
+
5
6
n
6
+
17
10
n
5
+
4
3
n

4
+
1
6
n
3
−
1
6
n
2
−
1
105
n,
P
5
(n) =
1
8
n
8
+
11
14
n
7
+
7
4

n
6
+
3
2
n
5
+
1
8
n
4
−
1
3
n
3
+
1
21
n,
P
6
(n) =
1
9
n
9
+
3

4
n
8
+
38
21
n
7
+
5
3
n
6
+
1
30
n
5
−
7
12
n
4
+
1
18
n
3
+
1

6
n
2
−
1
105
n,
P
7
(n) =
1
10
n
10
+
13
18
n
9
+
15
8
n
8
+
11
6
n
7
−

14
15
n
5
+
5
24
n
4
+
4
9
n
3
−
1
15
n
2
−
1
15
n.
p ∈ N
∗
.
Q
p
(n) =
n


k=1
k(k + 1)
p
.
Q
p
(n) i = k + 1
Q
p
(n) =
n+1

i=2
(i − 1)i
p
=
n+1

i=1
(i − 1)i
p
=
n

i=1
(i − 1)i
p
+ n(n + 1)
p

,
Q
p
(n) =
n

i=1
i
p+1
−
n

i=1
i
p
+ n(n + 1)
p
= S
p+1
(n) − S
p
(n) + n(n + 1)
p
,
S
p
(n)
Q
1
(n) =

1
3
n
3
+ n
2
+
2
3
n,
Q
2
(n) =
1
4
n
4
+
7
6
n
3
+
7
4
n
2
+
5
6

n,
Q
3
(n) =
1
5
n
5
+
5
4
n
4
+
17
6
n
3
+
11
4
n
2
+
29
30
n,
Q
4
(n) =

1
6
n
6
+
13
10
n
5
+
47
12
n
4
+
17
3
n
3
+
47
12
n
2
+
31
30
n,
Q
5

(n) =
1
7
n
7
+
4
3
n
6
+ 5n
5
+
115
12
n
4
+
59
6
n
3
+
61
12
n
2
+
43
42

n,
Q
6
(n) =
1
8
n
8
+
19
14
n
7
+
73
12
n
6
+
29
2
n
5
+
473
24
n
4
+
91

6
n
3
+
73
12
n
2
+
41
42
n.
S
k
(n) = 1
k
+ 2
k
+ + n
k
=
n

m=0
m
k
,
S
k
(n)


S
k
(0) = 0,
S
k
(n + 1) − S
k
(n) = (n + 1)
k
, (n = 0, 1, 2 ).
b
0
, b
1
, b
2
, ,

b
0
= 1,
C
1
k+1
b
k
+ C
2
k+1

b
k−1
+ + C
k
k+1
b
1
= k,
b
0
= 1, b
1
=
1
2
, b
2
=
1
6
, b
3
= 0, b
4
= −
1
30
, b
5
= 0,

b
6
=
1
42
, b
7
= 0, b
8
= −
1
30
, b
9
= 0, b
10
=
5
66
.
b
m
= 1 −
m−1

k=0
C
k
m
b

k
m − k + 1
, b
0
= 1, (m = 1, 2, ).
b
0
= 1, b
1
=
1
2
, b
1
= −
1
2
, b
2k+1
= 0, (k = 0, 1, ).
B
k
(x)





B
0

(x) = 1,
B

n
(x) = nB
n−1
(x),

1
0
B
n
(x)dx = 0, n ≥ 1.
B
0
(x) = 1,
B
1
(x) = x −
1
2
,
B
2
(x) = x
2
− x +
1
6
,

B
3
(x) = x
3
−
3
2
x
2
+
1
2
x,
B
4
(x) = x
4
− 2x
3
+ x
2
−
1
30
,
B
5
(x) = x
5
−

5
2
x
4
+
5
3
x
3
−
1
6
x,
B
6
(x) = x
6
− 3x
5
+
5
2
x
4
−
1
2
x
2
+

1
42
,
B
7
(x) = x
7
−
7
2
x
6
+
7
2
x
5
−
7
6
x
3
+
1
6
x.
B
m
(x) =
m


k=0
C
k
m
(−1)
k
b
k
x
m−k
,
B
m
(x) =
m

n=0
1
n + 1
n

k=0
C
k
n
(−1)
k
(x + k)
m

.
n
B
n
(x + 1) − B
n
(x) = nx
n−1
, B
n
(1) = b
n
,
b
n
1
k
+2
k
+ +n
k
=
1
k + 1
(n
k+1
+C
1
k+1
b

1
n
k
+C
2
k+1
b
2
n
k−1
+ + C
k
k+1
b
k
n).
S
k
(n) k + 1 n
(k + 1)(1
k
+ 2
k
+ + n
k
) = n
k+1
+ C
1
k+1

α
1
n
k
+ C
2
k+1
α
2
n
k−1
+ + C
k
k+1
α
k
n.
(k + 1)(1
k
+ 2
k
+ + n
k
) = (n + α)
k+1
− α
k+1
,
α
k

= α
k
. n n + 1
(k + 1)[(1
k
+ 2
k
+ + (n + 1)
k
] = (n + 1 + α)
k+1
− α
k+1
.
,
(k + 1)(n + 1)
k
= (n + 1 + α)
k+1
− (n + α)
k+1
.
n = 0
(α + 1)
k+1
− α
k+1
= k + 1.
α
k

= α
k
C
1
k+1
α
k
+ C
2
k+1
α
k−1
+ + C
k
k+1
α
1
+ α
0
= k + 1.
α
0
= 1 α
k
= b
k
S
k
(n) =
1

k + 1
k

m=1
C
m
k+1
b
m
n
k+1−m
.
S
p
(m) =
B
p+1
(m + 1) − B
p+1
(1)
p + 1
.
1.2
1
p
=
1
p + 1
[B
p+1

(2) − B
p+1
(1)],
2
p
=
1
p + 1
[B
p+1
(3) − B
p+1
(2)],
,
m
p
=
1
p + 1
[B
p+1
(m + 1) − B
p+1
(m)].
S
p
(m) =
B
p+1
(m + 1) − B

p+1
(1)
p + 1
.
n
k+1
k + 1
< S
k
(n) <
(n + 1)
k+1
− 1
k + 1
.
(1 + x)
k
≥ 1 + kx, (x ≥ −1, k > 0).

1 +
1
j

k+1
> 1 +
k + 1
j
,

1 −

1
j

k+1
> 1 −
k + 1
j
.
j
k+1
.
j
k+1
− (j − 1)
k+1
k + 1
< j
k
<
(j + 1)
k+1
− j
k+1
k + 1
, (j = 1, 2 , n).
j = 1, 2, , n
T
p
(n) =
n


k=1
(−1)
k
k
p
, p ∈ N
∗
.
T
p
(n) =
n

k=1
(−1)
k
k
p
p ∈ N
∗
, m, n ≥ 0.
T
p
(n) = (−1)
n
(n + 1)
p
− 1 −
p


m=0
C
m
p
T
m
(n).
k = j + 1,
T
p
(n) =
n−1

j=0
(−1)
j+1
(j + 1)
p
= −1 −
n−1

j=1
(−1)
j
(j + 1)
p
= −1 −
n


j=1
(−1)
j
(j + 1)
p
+ (−1)
n
(n + 1)
p
= (−1)
n
(n + 1)
p
− 1 −
n

j=1
(−1)
j
p

m=0
C
m
p
j
m
= (−1)
n
(n + 1)

p
− 1 −
p

m=0
C
m
p
n

j=1
(−1)
j
j
m
= (−1)
n
(n + 1)
p
− 1 −
p

m=0
C
m
p
T
m
(n),
T

m
(n) =
n

j=1
(−1)
j
j
m
.
n ≥ 1
a)
2n

k=1
(−1)
k+1
sin kx =
sin
x
2
− sin(2n +
1
2
)x
2 cos
x
2
.
b)

2n+1

k=1
(−1)
k+1
sin kx =
sin
x
2
+ sin(2n +
3
2
)x
2 cos
x
2
.
c)
2n

k=1
(−1)
k+1
cos kx =
cos
x
2
− cos(2n +
1
2

)x
2 cos
x
2
.
d)
2n+1

k=1
(−1)
k+1
cos kx =
cos
x
2
+ cos(2n +
3
2
)x
2 cos
x
2
.
2 cos
x
2
(sin x −sin 2x+sin 3x − −sin nx + −sin 2nx) = sin
x
2
−sin(2n +

1
2
)x.
2 sin x cos
x
2
= sin
x
2
+ sin
3x
2
,
− 2 sin 2x cos
x
2
= − sin
3x
2
− sin
5x
2
,
,
− 2 sin 2nx cos
x
2
= − sin(2n −
1
2

)x − sin(2n +
1
2
)x.
2 cos
x
2
(cos x−cos 2x+cos 3x− −cos nx+ −cos 2nx) = cos
x
2
−cos(2n+
1
2
)x.
2 cos x cos
x
2
= cos
x
2
+ cos
3x
2
,
− 2 cos 2x cos
x
2
= − cos
3x
2

− cos
5x
2
,
,
− 2 cos 2nx cos
x
2
= − cos(2n −
1
2
)x − cos(2n +
1
2
)x.
b, d a, c
(1.1)
i
1
)
2n

k=1
(−1)
k+1
k = −n,
i
2
)
2n


k=1
(−1)
k
k
2
= n(2n + 1),
i
3
)
2n

k=1
(−1)
k
k
3
= n
2
(4n + 3),
i
4
)
2n

k=1
(−1)
k+1
k
4

= n(−8n
2
− 8n + 1),
i
5
)
2n

k=1
(−1)
k+1
k
5
= −16n
5
− 20n
4
+ 5n
2
,
i
6
)
2n

k=1
(−1)
k
k
6

= 32n
6
+ 48n
5
− 20n
3
+ 3n,
i
7
)
2n

k=1
(−1)
k
k
7
= 64n
7
+ 112n
6
− 70n
4
+ 21n
2
,
i
8
)
2n


k=1
(−1)
k+1
k
8
= −128n
8
− 265n
7
+ 224n
5
− 112n
3
+ 17n,
i
9
)
2n

k=1
(−1)
k+1
k
9
= −256n
9
− 576n
8
+ 672n

6
− 504n
4
+ 153n
2
,
i
10
)
2n

k=1
(−1)
k
k
10
= 512n
10
+ 1280n
9
− 1920n
7
+ 2016n
5
− 1020n
3
− 155n.
a) 1 −
3
2

+
5
4
−
7
8
+ + (−1)
n−1
2n − 1
2
n−1
.
b) 3 −
5
2
+
7
4
−
9
8
+ + (−1)
n−1
2n + 1
2
n−1
.
P (x) =
n


k=1
(2k − 1)x
k−1
= 2
n

k=1
kx
k−1
−
n

k=1
x
k−1
=
2nx
n
(x − 1) − (x + 1)(x
n
− 1)
(x − 1)
2
.
Q(x) =
n

k=1
(2k + 1)x
k−1

= 2
n

k=1
kx
k−1
+
n

k=1
x
k−1
=
(2n + 1)x
n+1
− (2n + 3)x
n
− x + 1
(x − 1)
2
.
x = −
1
2
,
1 −
3
2
+
5

4
−
7
8
+ + (−1)
n−1
2n − 1
2
n−1
=
2
n
+ (−1)
n+1
(6n + 1)
9.2
n−1
.
3 −
5
2
+
7
4
−
9
8
+ + (−1)
n−1
2n + 1

2
n−1
=
(−1)
n+1
(6n + 7) + 3.2
n
9.2
n−1
.
F
p
(n) =
n

j=1
j!j
p
, p ∈ N
∗
.
F
p
(n) =
n

j=1
j!j
p
, p ∈ N

∗
, m, n ≥ 0.
F
p
(n) = 1 − n!(n + 1)
p+1
+
p+1

m=0
C
m
p+1
F
m
(n).
j = k + 1,
F
p
(n) =
n−1

k=0
(k + 1)!(k + 1)
p
=
n−1

k=0
k!(k + 1)

p+1
= 1 +
n−1

k=1
k!(k + 1)
p+1
= 1 +
n

k=1
k!(k + 1)
p+1
− n!(n + 1)
p+1
= 1 − n!(n + 1)
p+1
+
n

k=1
k!
p+1

m=0
C
m
p+1
k
m

= 1 − n!(n + 1)
p+1
+
p+1

m=0
C
m
p+1
n

k=1
k!k
m
= 1 − n!(n + 1)
p+1
+
p+1

m=0
C
m
p+1
F
m
(n),
F
m
(n) =
n


k=1
k!k
m
.
1.2
1) F
1
(n) = (n + 1)! − 1,
2) F
2
(n) + F
0
(n) = n(n + 1)!,
3) F
3
(n) − F
0
(n) = (n + 1)!(n
2
− 2) + 2,
4) F
4
(n) − 2F
0
(n) = (n + 1)!(n
3
− 3n + 3) − 3,
5) F
5

(n) + 9F
0
(n) = (n + 1)!(n
4
− 4n
2
+ 6n + 4) − 4.
1234567
∈ {1, 2, , 7}
6!
S = (6!.1 + + 6!.7) + (6!.1 + + 6!.7)10 + + (6!.1 + + 6!7)10
6
= 6!(1 + 2 + + 7)(1 + 10 + + 10
6
)
= 720.28.1111111
= 22399997760.
m, n ∈ N,
S
m,n
= 1 +
m

k=1
(−1)
k
(n + k + 1)!
n!(n + k)
m!, m, n ∈ N S
m,n

m!(n + 1)
m ∈ N, m ∈ N.
S
m,n
= (−1)
m
(n + m)!
n!
, m = 1
S
1,n
= 1 −
(n + 2)!
n!(n + 1)
= 1 − (n + 2) = −
(n + 1)!
n!
.
m ∈ N
m + 1,
S
m+1,n
= S
m,n
+ (−1)
m+1
(n + m + 2)!
n!(n + m + 1)
= (−1)
m

(n + m)!
n!
+ (−1)
m+1
(n + m)!(n + m + 2)
n!
= (−1)
m+1
(n + m)!
n!
(−1 + n + m + 2) = (−1)
m+1
(n + m + 1)!
n!
.
m + 1.
S
m,n
= (−1)
m
(n + m)!
n!m!
m! = (−1)
m
C
m
n+m
m! m! C
m
n+m

∈ N.
n = 2, m = 3 S
m,n
= 60 m!(n + 1) = 18.
S
n
=
n

k=0
(a + kd)q
k
.
n

k=0
(a + kd)q
k
= a
n

k=0
q
k
+ d
n

k=1
kq
k

,
a
n

k=0
q
k
= a(1 + q + q
2
+ + q
n
),
d
n

k=1
kq
k
= d[(q + q
2
+ + q
n
) + (q
2
+ q
3
+ + q
n
) + + (q
n

)].
S
n
= a(1 + q + q
2
+ + q
n
) + d[(q + q
2
+ + q
n
)+
+ (q
2
+ q
3
+ + q
n
) + + (q
n
)]
= a
q
n+1
− 1
q − 1
+ d.

q.
q

n
− 1
q − 1
+ q
2
.
q
n−1
− 1
q − 1
+ + q
n
q − 1
q − 1

= a
q
n+1
− 1
q − 1
+ d

q
n+1
− q + q
n+1
− q
2
+ + q
n+1

− q
n
q − 1

= a
q
n+1
− 1
q − 1
+ d

n.q
n+1
− (q + q
2
+ + q
n
)
q − 1

=
1
q − 1

a(q
n+1
− 1) + d(nq
n+1
−
q

n+1
− q
q − 1
)

.
S
n
=
1
q − 1

a(q
n+1
− 1) + d(nq
n+1
−
q
n+1
− q
q − 1
)

.
(n + 1)2
0
+ n.2
1
+ + 2.2
n−1

+ 1.2
n
.
n

k=0
(n + 1 − k)2
k
= (n + 1)2
0
+ n.2
1
+ + 2.2
n−1
+ 1.2
n
.
a = n + 1, d = −1, q = 2,
n

k=0
(n + 1 − k)2
k
=
1
2 − 1

(n + 1)(2
n+1
− 1) − (n2

n+1
−
2
n+1
− 2
2 − 1
)

= (n + 1)2
n+1
− (n + 1) − n2
n+1
+ 2
n+1
− 2
= 2
n+2
− (n + 3).
G
n
=
n

k=1
kp
k
, p = 0, p = 1.
n

k=1

p
k
=
n

k=0
p
k
− 1 =
p
n+1
− 1
p − 1
− 1 =
p
n+1
− p
p − 1
.
n

k=1
kp
k−1
=
np
n+1
− (n + 1)p
n
+ 1

(p − 1)
2
.
G
n
=
n

k=1
kp
k
=
np
n+2
− (n + 1)p
n+1
+ p
(p − 1)
2
.
x
0
, x
1
, , x
n
, x
0
= 2000,
x

n
=
−2000
n
n−1

k=0
x
k
, n ≥ 1.
S =
2000

n=0
2
n
x
n
.
x
0
= 2000, x
1
= −2000x
0
,
x
2
=
−2000

2
(x
0
+ x
1
) =
−1
2
2000x
1
−
1
2
2000x
0
= −
1
2
(2000 − 1)x
1
.
x
n
=
−2000
n−1
n
n−1

k=0

x
k
= −
1
n
(2000 − n + 1)x
n−1
= (−1)
n
2000(2000 − 1) (2000 − n + 1)
1.2 n
x
0
= (−1)
n
C
n
2000
x
0
.
S =
2000

n=0
2
n
x
n
= x

0
2000

n=0
(−1)
n
2
n
C
n
2000
= x
0
(1 − 2)
2000
= x
0
= 2000.
E n n ≥ 1 k ∈ N, 0  k  n.
k n E k E.
C
k
n

n
k

k n,
C
k

n
=

n
k

=
n!
k!(n − k)!
.
1) C
0
n
= C
n
n
= 1,
2) C
1
n
= n,
3) C
k
n
= C
n−k
n
,
4) C
k

n
+ C
k−1
n
= C
k
n+1
,
5) C
0
n
+ C
1
n
+ C
2
n
+ + C
n
n
= 2
n
.
(a + b)
n
=
n

k=0
C

k
n
a
n−k
b
k
.
a = 1, b = x
(1 + x)
n
=
n

k=0
C
k
n
x
k
.
x = 1, x = −1
n

k=0
C
k
n
= 2
n
,

n

k=0
(−1)
k
C
k
n
= 0.
x
n(1 + x)
n−1
=
n

k=0
kC
k
n
x
k−1
=
n

k=1
kC
k
n
x
k−1

.
S = 1
2
C
1
n
+ 2
2
C
2
n
+ + P
2
C
p
n
+ + n
2
C
n
n
.
f(x) = (1 + x)
n
= C
0
n
+ C
1
n

x + + C
n
n
x
n
.
g(x) = x(1 + x)
n
= C
0
n
x + C
1
n
x
2
+ + C
n
n
x
n+1
.
x
f

(x) = n(1 + x)
n−1
= C
1
n

+ 2xC
2
n
+ + nx
n−1
C
n
n
.
g”(x) = 2n(1 + x)
n−1
+ n(n − 1)x(1 + x)
n−2
= 2C
1
n
+ 3.2xC
2
n
+ 4.3x
2
C
3
n
+ + (n − 1)nx
n−1
C
n
n
.

x = 1
f

(1) = n2
n−1
= C
1
n
+ 2C
2
n
+ + nC
n
n
.
g”(1) = 2n2
n−1
+ n(n − 1)2
n−2
= 2C
1
n
+ 3.2C
2
n
+ 4.3C
3
n
+ + (n + 1)nC
n

n
.
g

(1) − f

(1)
S = 1
2
C
1
n
+ 2
2
C
2
n
+ + P
2
C
p
n
+ + n
2
C
n
n
= n(n − 1)2
n−2
.

a) C
k
n
+ 4C
k−1
n
+ 6C
k−2
n
+ 4C
k−3
n
+ C
k−4
n
= C
k
n+4
, (4  k  n).
b) C
0
m
C
k
n
+ C
1
m
C
k−1

n
+ + C
m
m
C
k−n
n
= C
k
m+n
, (m  k  n).
C
r
n
= C
r
n−1
+ C
r−1
n−1
, (0 ≤ r ≤ n)
V T = C
k
n
+ C
k−1
n
+ 3(C
k−1
n

+ C
k−2
n
) + 3(C
k−2
n
+ C
k−3
n
) + C
k−3
n
+ C
k−4
n
= C
k
n+1
+ 3C
k−1
n+1
+ 3C
k−2
n+1
+ C
k−3
n+1
= C
k
n+1

+ C
k−1
n+1
+ 2(C
k−1
n+1
+ C
k−2
n+1
) + C
k−2
n+1
+ C
k−3
n+1
= C
k
n+2
+ 2C
k−1
n+2
+ C
k−2
n+2
= C
k
n+2
+ C
k−1
n+2

+ C
k−1
n+2
+ C
k−2
n+2
= C
k
n+3
+ C
k−1
n+3
= C
k
n+4
= V P
x, n, m
(1 + x)
m+n
= (1 + x)
m
(1 + x)
n
,
(1 + x)
m+n
=
m+n

k=0

C
k
m+n
x
k
.
(1 + x)
m
(1 + x)
n
=
m

p=0
C
p
m
x
p
n

p=o
C
p
n
x
p
=
m+n


k=0
k

p=0
(C
p
m
C
k−p
n
)x
k
.
C
k
m+n
=
k

p=0
C
p
m
C
k−p
n
= C
0
m
C

k
n
+ C
1
m
C
k−1
n
+ + C
m
m
C
k−n
n
.
S =
1
2
C
1
2n
−
1
3
C
2
2n
+ + (−1)
k
1

k
C
k−1
2n
+ + (−1)
2n+1
1
2n + 1
C
2
2n
n,
C
k
n
∀k = 2, 3, , 2n + 1
1
k
C
k−1
2n
=
1
k
(2n)!
(k − 1)!(2n − k + 1)!
=
1
2n + 1
(2n + 1)!

k!(2n + 1 − k)!
=
C
k
2n+1
2n + 1
.
S =
2n+1

k=2
(−1)
k
C
k
2n+1
2n + 1
=
1
2n + 1
2n+1

k=0
(−1)
k
C
k
2n+1
− C
0

2n+1
+ C
1
2n+1
=
1
2n + 1
[(1 − 1)
2n+1
− 1 + 2n + 1]
=
2n
2n + 1
.
A = C
0
n
+
2
2
− 1
2
C
1
n
+
2
3
− 1
3

C
2
n
+ +
2
n+1
− 1
n + 1
C
n
n
.

n + 1
k + 1

C
k
n
= C
k+1
n+1
,
n k k ≤ n
A =
1
n + 1

(n + 1)C
0

n
+ (2
2
− 1)
(n + 1)C
1
n
2
+ + (2
n+1
− 1)
(n + 1)C
n
n
n + 1

=
1
n + 1

(2 − 1)C
1
n+1
+ (2
2
− 1)C
2
n+1
+ + (2
n+1

− 1)C
n+1
n+1

=
1
n + 1


2C
1
n+1
+ 2
2
C
2
n+1
+ + 2
n+1
C
n+1
n+1

−

C
1
n+1
+ + C
n+1

n+1


=
1
n + 1

(2 + 1)
n+1
− 1 − (2
n+1
− 1)

=
3
n+1
− 2
n+1
n + 1
.
n
3
n
C
0
n
− 3
n−1
C
1

n
+ 3
n−2
C
2
n
+ + 3C
n
n
= 2
n
.
x n
(1 − x)
n
= C
0
n
− C
1
n
x + C
2
n
x
2
+ + (−1)
n
C
n

n
x
n
.
x =
1
3

1 −
1
3

n
= C
0
n
−
1
3
C
1
n
+
1
3
2
C
2
n
+ + (−1)

n
1
3
n
C
n
n
2
n
3
n
= C
0
n
−
1
3
C
1
n
+
1
3
2
C
2
n
+ + (−1)
n
1

3
n
C
n
n
3
n
C
0
n
− 3
n−1
C
1
n
+ 3
n−2
C
2
n
+ + 3C
n
n
= 2
n
.
n ∈ N
n

k=0

(2n)!
(k!)
2
.(n − k!)
2
= (C
n
2n
)
2
.
(1 + x)
2n
= (1 + x)
n
(1 + x)
n
.
C
0
2n
+ C
1
2n
x + + C
2n
2n
x
2n
= (C

0
n
+ C
1
n
x + + C
n
n
x
n
)(C
0
n
+ C
1
n
x + + C
n
n
x
n
).
x
n
C
k
n
= C
n−k
n

, (k = 0, 1, , n)
C
n
2n
= (C
0
n
)
2
+ (C
1
n
)
2
+ + (C
n
n
)
2
.
n

k=0
(2n)!
(k!)
2
(n − k!)
2
=
(2n)!

n!n!
=
n

k=0
(n!)
2
(k!)
2
(n − k!)
2
= C
n
2n
[(C
0
n
)
2
+ (C
1
n
)
2
+ + (C
n
n
)
2
] = (C

n
2n
)
2
.
S = C
0
2000
+ 2C
1
2000
+ 3C
2
2000
+ + 2001C
2000
2000
.
(1 + x)
2000
= C
0
2000
+ C
1
2000
x + C
2
2000
x

2
+ + C
2000
2000
x
2000
.
x
x(1 + x)
2000
= C
0
2000
x + C
1
2000
x
2
+ C
2
2000
x
3
+ + C
2000
2000
x
2001
.
x

C
0
2000
+2C
1
2000
x +3C
2
2000
x
2
+ +2001C
2000
2000
x
2000
= (1+x)
2000
+2000x(1+x)
1999
.
x = 1
S = 2
2000
+ 2000.2
1999
= 2
1999
(2 + 2000) = 2002.2
1999

.
a) C
0
n
+ C
3
n
+ C
6
n
+ =
1
3

2
n
+ 2 cos
nπ
3

.
b) C
1
n
+ C
4
n
+ C
7
n

+ =
1
3

2
n
+ 2 cos
(n − 2)π
3

.
c) C
2
n
+ C
5
n
+ C
8
n
+ =
1
3

2
n
+ 2 cos
(n − 4)π
3


.
a)
(1 + x)
n
= C
0
n
+ C
1
n
x + C
2
n
x
2
+ + C
n−1
n
x
n−1
+ C
n
n
x
n
.