LUẬN VĂN THẠC SĨ TOÁN HỌC " MỘT SỐ BÀI TOÁN VỀ SỐ HỌC VÀ DÃY SỐ " docx

− − − − −⋆ −−−−−
u
n
u
1
, u
2
, u
3
,
u
1
, u
2
, u
3
,
u
i
i
u
1
, u
2
, u
3
,

u
n+1
> u
n
n = 1, 2,
u
n+1
≥ u
n
n = 1, 2
u
n+1
< u
n
n = 1, 2,
u
n+1
 u
n
n = 1, 2,
u
1
, u
2
, u
3
,
K u
n
< K n = 1, 2,

m u
n
> m n = 1, 2,
u
1
, u
2
, u
3
, N
o
u
n
= C n ≥ N
o
C
u
1
, u
2
, u
3
, n
k p = 1, 2,




















u
n
= u
n+kp
u
n+1
= u
n+1+kp

u
n+k−1
= u
n+k−1+kp
.
k
{u
n

} : u
1
, u
2
, u
3
, ··· {v
n
} : v
1
, v
2
, v
3
, ···
{u
n
+ v
n
} : u
1
+ v
1
, u
2
+ v
2
, u
3
+ v

3
, ···
{u
n
v
n
} : u
1
v
1
, u
2
v
2
, u
3
v
3
, ···
v
k
= 0 k = 1, 2, ···

u
n
v
n

:
u

1
v
1
,
u
2
v
2
,
u
3
v
3
···).
{u
n
} u
n
= 2n + 1
n = 0, 1, 2, ··· 1, 3, 5, 7, ···
u
0
u
0
, u
1
, u
2
, ···
{u

n
}, n = 0, 1, 2, ···





u
0
= u
1
= 1
u
n+1
= u
n−1
u
n+1
,
n = 1, 2, 3, ···
k n {u
j
}, {v
j
}(j = 1, 2, ··· , n)
k n u
1
v
1
j : (j = 2, 3, ··· , n) u

j
v
j
k + v
j−1
n u
j
v
j
k = nu
1
+ v
1
;
k + v
1
= nu
2
+ v
2
;
k + v
2
= nu
3
+ v
3
;
···
k + v

n−1
= nu
n
+ v
n
.
y = f(x) f(x) x
0
, x
0
+ h, x
0
+
2h, ··· , x
0
+ nh, ··· h y
0
, y
1
, y
2
, ··· , y
n
, ···
∆y
i
= y
i
− y
i−1

1 f i = 1, 2, ···
∆
2
y
i
= ∆y
i
−∆y
i−1
= (y
i
−y
i−1
) −(y
i−1
−y
i−2
) = y
i
−2y
i−1
+ y
i−2
2 f i = 1, 2, ···
•
∆
k
(f ± g) = ∆
k
(f) ± ∆

k
(g)
k n k > n
k = n k
k
n

i=1
∆y
i
= (y
1
− y
0
) + (y
2
− y
1
) + (y
3
− y
2
) + ···+ (y
n
− y
n−1
) = y
n
− y
0

.
•
y = f(x)
x
0
, x
0
+ h, x
0
+ 2h, , x
0
+ nh,
y
o
, y
1
, y
2
, , y
n
,
a
n
y
n+i
+a
n−1
y
n−1+i
+a

n−2
y
n−2+i
+···+a
1
y
1+i
+a
0
y
i
= 0
n a
0
, a
1
, a
2
, , a
n
n y
0
, y
1
, , y
n
y
n
, y
n+1

,
y
i
, y
i+1
, , y
i+n
y
′
i
, y
′
i+1
, , y
′
i+n
y
i
± y
′
i
, y
i+1
± y
′
i+1
, ··· , y
i+n
± y
′

i+n
y
i
= c
1
λ
i
1
+ c
2
λ
i
2
+ ··· + c
n
λ
i
n
c
1
, c
2
, c
3
, , c
n
λ
1
, λ
2

, , λ
n
n
a
n
λ
n
+ a
n−1
λ
n−1
+ ··· + a
1
λ + a
0
= 0.
λ
1
s
y
i
= c
1
λ
i
1
+ c
2
iλ
i

1
+ c
3
i
2
λ
i
1
+ ··· + c
s
i
s−1
λ
i
1
+ c
s+1
λ
i
s+1
+ ··· + c
n
λ
i
n
.
u
0
= 1; u
1

= −1; u
2
= −1; u
3
= 1; u
4
= 5;
u
5
= 11; u
6
= 19; u
7
= 29; u
8
= 41; u
9
= 55.
n = 0, 9.
y = f(x)
∆y
∆
2
y
ax
2
+ bx + c x
a, b, c x = 0, 1, 2
u
0

= y
0
= 1 = c;
u
1
= y
1
= −1 = a + b + c;
u
2
= y
2
= −1 = 4a + 2b + c.













c = 1
a + b + c = −1
4a + 2b + c = −1
a = 1; b = −3; c = 1

x
2
− 3x + 1
u
n
= n
2
− 3n + 1, n = 0, 9.
{u
n
}




















u
0
= 1
u
1
= 2

u
n
= 3u
n−1
− 2u
n−2
n = 2, 3,
u
n
.
u
n
− 3u
n−1
+ 2u
n−2
= 0
λ
2
−3λ+2 = 0 λ
1
= 1, λ
2

= 2
u
n
u
n
= c
1
λ
n
1
+ c
2
λ
n
2
u
n
= c
1
+ c
2
2
n
.
c
1
c
2
u
0

= 1 = c
1
+ c
2
u
1
= 2 = c
1
+ 2c
2





c
1
+ c
2
= 1
c
1
+ 2c
2
= 2,
c
1
= 0 c
2
= 1.

u
n
= 2
n
u
1
, u
2
, u
3
, d (d = 0)
u
n
= u
n−1
+ d n = 2, 3,
•
u
n
= u
1
+ (n − 1)d, n = 1, 2, 3,
u
k
=
u
k−1
+ u
k+1
2

,
k = 2, 3,
u
1
, u
2
, , u
n−1
, u
n
u
1
+ u
n
= u
2
+ u
n−1
= u
3
+ u
n−2
= ··· .
u
1
+ u
n
= u
k
+ u

n−k
k = 2, 3, , n −1.
•
u
1
, u
2
, d
S
n
= u
1
+ u
2
+ ··· + u
n−1
+ u
n
.
S
n
=
(u
1
+ u
n
)n
2
=
[2u

1
+ (n −1)d]n
2
.
S
1
= 1 + 2 + 3 + ···+ n;
S
2
= 1
2
+ 2
2
+ 3
2
+ ···+ n
2
;
S
3
= 1
3
+ 2
3
+ 3
3
+ ···+ n
3
.
S

1
=
n(n + 1)
2
S
2
=
n(n + 1)(2n + 1)
2
;
S
3
=
n
2
(n + 1)
2
4
.
u
1
, u
2
, u
3
, q (q = 0, q = 1)
u
n
= u
n−1

q n = 2, 3,
•
u
n
= u
1
q
n−1
n = 1, 2, 3,
u
2
k
= u
k−1
u
k+1
k = 2, 3,
•
u
1
, u
2
, u
3
, q S
n
= u
1
+ u
2

+ ··· + u
n
.
S
n
=
u
1
(q
n
− 1)
q −1
.
u
1
, u
2
,














u
1
= 1, u
2
= 1
···
u
n
= u
n−1
+ u
n−2
,
n = 3, 4,
•
u
n
− u
n−1
− u
n−2
= 0.
λ
2
− λ −1 = 0.
λ
1
=
1 +

√
5
2
λ
2
=
1 −
√
5
2
.
u
n
= c
1
λ
n
1
+ c
2
λ
n
2
= c
1

1 +
√
5
2


n
+ c
2

1 −
√
5
2

n
.
c
1
, c
2







u
1
= 1 = c
1
1 +
√
5

2
+ c
2
1 −
√
5
2
u
2
= 1 = c
1

1 +
√
5
2

2
+ c
2

1 −
√
5
2

2








c
1
1 +
√
5
2
+ c
2
1 −
√
5
2
= 1
c
1

1 +
√
5
2

2
+ c
2

1 −

√
5
2

2
= 1
⇒







c
1
=
1
√
5
c
2
= −
1
√
5
u
n
=
1

√
5

1 +
√
5
2

n
−
1
√
5

1 −
√
5
2

n
.
•
u
1
, u
2
, ···
u
1
+ u

3
+ u
5
+ ···+ u
2n−1
= u
2n
u
2
+ u
4
+ u
6
+ ···+ u
2n
= u
2n+1
− 1
u
2
1
+ u
2
2
+ ··· + u
2
n
= u
n
u

n+1
u
2
2n
= u
1
u
2
+ u
2
u
3
+ ··· + u
2n−1
u
2n
u
n+1
u
n+2
− u
n
u
n+3
= (−1)
n
u
2
n
− u

n−1
u
n+1
= (−1)
n+1
a b a b a b
b a k a = k.b a b
a  b a b
p > 1
p.
a, b a b a ≥ b
a
i
b i = 1, n (a
1
+ a
2
+ ···+ a
n
) b
a b b = 0
q r a = bq + r 0  r < b.
a b
a b (a, b)
(a, b) a b
a b (a, b)
[a, b] a b
n a
1
, a

2
, , a
n
d a
1
, a
2
, , a
n
• a
i
d i = 1, n.
•
d
′
a
i
d
′
, ∀i = 1, n d d
′
,
d = (a
1
, a
2
, , a
n
)
b a

1
, a
2
, , a
n
• b a
i
, ∀i = 1 , n
• b
′
b a
i
, ∀i = 1, n b
′
b.
b = [a
1
, a
2
, , a
n
].
a b (a, b) = (a, a + b)
m
(ma, mb) = m(a, b);
[ma, mb] = m[a, b].
(a, b) d

a
d

,
b
d

=
1
d
(a, b).
a b (a, b) = 1 a, b, c
ab c (a, c) = 1 b c.
a, b x, y
ax + by = (a, b).
a, b
x y ax + by = 1.
n (n > 1). n
n = p
α
1
1
p
α
2
2
···p
α
k
k
.
k, α
i

(i = 1, k) p
i
(i = 1, k)
1 < p
1
< p
2
< ··· < p
k
n
a, b p ab p
a p b p
a b m (m = 0)
a b m a ≡ b ( mod m).
a b m m
(a − b) m.
Z.
a ≡ b (mod m) c ≡ d (mod m)
a + c ≡ b + d (mod m),
a − c ≡ b −d (mod m),
ac ≡ bd (mod m).
p ab ≡ 0 (mod p) a ≡ 0 (mod p)
b ≡ 0 (mod p)
p a
(a
p
− a) ≡ p (a, p) = 1 a
p−1
≡ 1 (mod p)
m (a, m) = 1 a

φ(m)
≡ 1 (mod m)
φ(m) m m (φ(m)
p (p − 1)! + 1 p
p = 4k + 1 a, b
p = a
2
+ b
2
.
r s
a b N
N ≡ a (mod r) N ≡ b (mod s) N
rs
x x [x]
x
[x] = a ⇔ x = a + d a 0  d < 1
[x + y] = x x 0  y < 1
n [n + x] = n + [x]
[x + y] ≥ [x] + [y]
n

[x]
n

=

x
n


n n[x]  [nx]
n q q = 0 q

n
q

 n.
{u
n
} n = 0, 1, 2,
u
1
= 1;
u
m+n
+ u
m−n
=
1
2
(u
2m
+ u
2n
), ∀m ≥ n; m, n ∈ N
u
0
+ u
0
=

1
2
(u
0
+ u
0
) ⇒ 2u
0
= u
0
⇒ u
0
= 0.
m = 1, n = 0
u
1
+ u
1
=
1
2
(u
2
+ u
0
) ⇒ 2(u
1
+ u
1
) = u

2
(
u
0
= 0).
u
1
= 1 u
2
= 4 u
0
= 0
2
, u
1
= 1
2
, u
2
= 2
2
u
n
= n
2
n = k
m = k, n = 0
u
k
+ u

k
=
1
2
(u
2k
+ u
0
) ⇒ u
2k
= 4u
k
.
u
2k
= 4k
2
.
m = k, n = 1
u
k+1
+ u
k−1
=
1
2
(u
2k
+ u
2

).
u
k+1
=
1
2
u
2k
+
1
2
u
2
− u
k−1
u
2k
= 4 k
2
u
2
= 2
2
, u
k−1
= (k − 1)
2
u
k+1
=

1
2
4k
2
+
1
2
.4 − (k −1)
2
= 2k
2
+ 2 −k
2
+ 2k −1 = (k + 1)
2
.
n = k + 1
u
n
= n
2
, ∀n = 0, 1, 2,
{u
n
}






u
1
= 1; u
2
= −1
u
n
= −u
n−1
− 2u
n−2
; n ≥ 3.
a = 2
2006
− 7u
2
2004
v
n
= 2
n+1
− 7u
n−1
a = v
2005
= 2
2006
− 7u
2
2004

.
v
n
= (2u
n
+ u
n−1
)
2
, ∀ n = 2, 3,
n = 2 v
2
= 2
3
− 7u
2
1
= 8 −7 = 1
(2u
2
+ u
1
) = (−2 + 1)
2
= 1 v
2
= (2u
2
+ u
1

)
2
n = 2 n = k (k ≥ 2),
v
k
= (2u
k
+ u
k−1
)
2
.
n = k + 1
v
k+1
= 2
k+2
− 7u
2
k
.
{u
n
}
(2u
k+1
+ u
k
)
2

= [2(−u
k
− 2u
k−1
) + u
k
]
2
= (−u
k
− 4u
k−1
)
2
= u
2
k
+ 8u
k
u
k−1
+ 16u
2
k−1
=
= 2(4u
2
k
+4u
k

u
k−1
+u
2
k−1
)+14u
2
k−1
−7u
2
k
= 2(2u
k
+u
k−1
)
2
+14u
2
k−1
−7u
2
k
=
= 2v
k
+ 14u
2
k−1
−7u

2
k
= 2(2
k+1
−7u
2
k−1
)+ 14u
2
k−1
−7u
2
k
= 2
k+2
−7u
2
k
=
= v
k+1
.
v
k+1
= (2u
k+1
+ u
k
)
2

n = k + 1
∀n = 2, 3,
{v
n
}
a = 2
2006
− 7u
2
2004
{u
n
}





u
1
= 1; u
2
= 2
u
n+1
= u
n
(u
n
− 1) + 2; n = 2, 3,

{s
n
}
s
n
= (u
2
1
+ 1)(u
2
2
+ 1) (u
2
n
+ 1) −1, ∀n = 1, 2,
{s
n
}
s
k
= (u
k+1
+ 1)
2
.
k = 1 s
1
= (u
2
1

+ 1) −1 = u
2
1
= 1
2
= (2 −1)
2
= (u
2
− 1)
2
.
k = 1
k = n s
n
= (u
n+1
− 1)
2
k = n + 1
s
n+1
= (u
2
1
+ 1)(u
2
2
+ 1) (u
2

n
+ 1)(u
2
n+1
+ 1) −1 = (s
n
+ 1)(u
2
n+1
+ 1) −1 =
= [(u
n+1
− 1)
2
+ 1](u
2
n+1
+ 1) −1 = [(u
2
n+1
+ 1) − 2u
n+1
+ 1](u
2
n+1
+ 1) −1 =
= (u
n+1
+ 1)
2

− 2u
n+1
(u
2
n+1
+ 1) + (u
n+1
+ 1) − 1 =
= (u
2
n+1
+ 1)
2
− 2u
n+1
(u
2
n+1
+ 1) + u
2
n+1
=
= (u
2
n+1
+ 1 − u
n+1
)
2
.

u
2
n+1
+ 1 − u
n+1
= u
n+1
(u
n+1
− 1) + 1 = [u
n+1
(u
n+1
− 1) + 2] −1 = u
n+2
− 1.
s
n+1
= (u
2
n+1
+ 1 − u
n+1
)
2
= (u
n+2
− 1)
2
.

k = n + 1
k = 1, 2, u
1
, u
2
u
n
n. s
k
k = 1, 2, {s
n
}
{u
n
}





u
0
= 3, u
1
= 17
u
n
= 6u
n−1
− u

n−2
; n = 2, 3,
n = 0, 1, 2, v
n
=
u
2
n
− 1
2
x
2
− 6x + 1 = 0.
x
1
= 3 +
√
8 x
2
= 3 −
√
8
u
n
u
n
= a(3 +
√
8)
n

+ b(3 −
√
8)
n
.
u
0
= 3, u
1
= 17 a, b





a(3 +
√
8)
0
+ b(3 −
√
8)
0
= 3
(3 +
√
8)
1
+ b(3 −
√

8)
1
= 17
⇔





a + b = 3
a(3 +
√
8) + b(3 −
√
8) = 17
⇔





a =
1
2
(3 +
√
8)
b =
1
2

(3 −
√
8).
u
n
=
1
2

(3 +
√
8)
n+1
+ (3 −
√
8)
n+1

.
v
n
=
1
2
(u
2
n
− 1) =
1
2

·
1
4


(3 +
√
8)
n+1
+ (3 −
√
8)
n+1

2
− 4

=
=
1
2

(3 +
√
8)
n+1
− (3 −
√
8)
n+1

2

2
.
(3 ±
√
8)
n+1
= M ±N
√
2, M, N
1
2
(u
2
n
− 1) = N
2
{a
n
}





a
0
= 1, a
1

= 2
a
n+2
= 4a
n+1
− a
n
, ∀n ∈ N.
n a
n
− 1
t
2
= 4t −1
t
1
= 2 +
√
3, t
2
= 2 −
√
3 a
n
a
n
= α(2 +
√
3)
n

+ β(2 −
√
3)
n
(α, β ∈ R).
a
0
= 1, a
1
= 2 α, β





α + β = 1
2(α + β) +
√
3(α − β) = 2
⇔ α = β =
1
2
.
a
n
=
1
2
[(2 +
√

3)
n
+ (2 −
√
3)
n
].
(2 +
√
3) =
1
2
(
√
3 + 1)
2
=

√
3 + 1
√
2

2
.
(2 −
√
3) =
1
2

(
√
3 − 1)
2
=

√
3 −1
√
2

2
.
a
n
− 1 =
1
2


√
3 + 1
√
2

2n
+

√
3 −1

√
2

2n

− 1 =

(
√
3 + 1)
n
− (
√
3 − 1)
n
(
√
2)
n+1

2
.
a
n
− 1
A
n
:=
(
√

3 + 1)
n
− (
√
3 − 1)
n
(
√
2)
n+1
∈ Z.
n = 0 ⇒ A
1
= 0 ∈ Z.
n = 1 ⇒ A
2
= 1 ∈ Z.
n = 2k, k ∈ N
∗
,
(
√
3 + 1)
n
− (
√
3 − 1)
n
(
√

2)
n+1
=
(2 +
√
3)
k
− (2 −
√
3)
k
√
2
:= b
k
.
2 +
√
3; 2 −
√
3 t
2
= 4t − 1 {b
k
}
b
k+2
= 4b
k+1
− b

k
b
1
=
√
6 ⇒ b
k
∈ Q, ∀k ∈ N
∗
a
n
− 1
n = 2k, k ∈ N
∗
n = 2k + 1, k ∈ N
∗
(
√
3 + 1)
n
− (
√
3 −1)
n
(
√
2)
n+1
=
√

3 + 1
2

(
√
3 + 1)
n
− (
√
3 − 1)
n
(
√
2)
n+1

2
=
=
√
3 + 1
2
[(2 +
√
3)
k
− (2 −
√
3)
k

].
c
k
=
√
3 + 1
2
[(2+
√
3)
k
−(2−
√
3)
k
], k ∈ N
∗
{c
k
} c
k+2
= 4c
k+1
−c
k
c
1
= 5 ⇒ c
k
∈ Z, ∀ k ∈ N

∗
.
n = 2k + 1, k ∈ N a
n
− 1





u
1
= 1, u
2
= 3
u
n+2
= 2u
n+1
− u
n
+ 1, n = 1, 2, 3,
∀n A
n
= 4u
n
u
n+2
+ 1
u

n+2
= 2u
n+1
−u
n
+1 u
n+2
−u
n+1
= u
n+1
−u
n
+1.
u
n+2
− u
n+1
= v
n+2
v
n+2
= v
n+1
+ 1
d = 1.
u
n
= (u
n

−u
n−1
) +(u
n−1
−u
n−2
) +···+ (u
2
−u
1
) +u
1
= v
n
+ v
n−1
+ ···+ v
2
+ u
1
u
n
=
[2v
2
+ (n − 2)d](n −1)
2
+ 1 =
[2.2 + (n −2).1](n −1)
2

+ 1 =
n(n + 1)
2
.
u
n
=
n(n + 1)
2
n = 1, 2,
A
n
= 4
n(n + 1)
2
·
(n + 2)(n + 3)
2
+ 1 = n(n + 1)(n + 2)(n + 3) + 1 = (n
2
+ 3n + 1)
2
.
A
n
n
{u
n
}
u

n+2
+ u
n−1
= 2(u
n+1
+ u
n
), ∀n = 1, 2,
M n
n = 1, 2, M + 4u
n
u
n+1
u
n+2
+ u
n−1
= 2(u
n+1
+ u
n
)
u
n+2
− u
n+1
− u
n
= u
n+1

− u
n
− u
n−1
+ 2u
n
.
v
n
= u
n+2
− u
n+1
− u
n
, n = 1, 2, v
n
= v
n−1
+ 2u
n
, ∀n = 2, 3,
∀n = 2, 3,