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(un)n≥0 un+1 = aun + b
• un+1 =
un + b ⇒ (un) ⇒ un
= u0 + nb
•
⇒ un + 1 = aun (un)
⇒
un = an.u0
•
a = 1
vn = un + b
a − 1 , ∀n ≥ 0
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vn+1 = un+1 + b
a − 1
= aun + b +
b
a − 1= a(vn − b
a − 1 ) + b
a − 1 + b
= avn − aba − 1 +
ab
a − 1= avn.
(vn)n ≥ 0 vn =
anv0
⇒ un = a
n(u0 + b
a− 1)
−
b
a − 1,
∀n
≥ 0
un =
u0 + nb a = 1
an(u0 + ba−1)− ba−1 a =
1
S (n) =
n
i=0 ui
un+1 + S (n) = un+1 +n
i=0 ui
= aun + b +n
i=0(aui + b)
= u0 +n
i=0(aui + b)
= u0 + an
i=0 ui + nb
= u0 + aS (n) + nb
⇒ S (n) =
un+1−nb−u0a−1
un+1 un
un
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u0 = 2, u1 = 3
un+1 = 3un − 2un−1
un+1 = 3un − 2un−1 ⇔ un+1 − un
= 2(un −un−1) vn = un −
un −1 v1 = u1 − u0 = 3− 2 =
1vn+1 = un+1 − un = 2(un −un−1) =
vn⇒ (vn)n≥0 v1 = 1
q = 2⇒ vn = v02n
un = (un −un−1) + (un−1 − un−2)
+ ...(u1 −u3) + u0
= vn + vn−1 + ... + v1 +
2
=n
i=1
vi + 2
=
2n
−1
2 − 1 + 2= 2n − 1 + 2
= 2n + 1
un = 2n + 1
u01 = 1, u2 = 2
un+1 − 2un + un−1 = 1
vn = un−un−1
un = 1
2(n2 − n + 2)
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u1 =
13
un+1 = n+13n
un
vn = un
n
un u1 = 1
un+1 = 6un − 1 , ∀n ≥ 1
vn = un − 15
(vn)
(vn) (un)
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u1 = 1un+1 = un + 2n− 1 ,
∀n ≥ 1
vn = un+1 − un
un
u1 = 1un+1 = un + n ,
∀n ≥ 1
vn = un+1 − un
un
u1 = 1un+1 =
u2n + 2 , ∀n ≥ 0
S =
u21 + u22 + ... + u
21001
vn = u2n
vn+1 = u2n+1 = u
2n + 2 = vn + 2
⇒ (vn)
(un)
S = 3n − 1
3n−1
(un)
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un+2 =
aun+1 + bun(∗), ∀n ≥ 0
a, b ∈ R
Da,b
Da,b R
•
Da,b
(0n)n≥0 ∈ Da,b ⇒ Da,b =
Ø
(un)n≥0, (vn)n≥0 ∈ Da,b
xun+2 + yvn+2 = x(aun+1 + bun)
+ y(avn+1 + bvn)
= a(xun+1 + yvn+1)
+ b(xun + yvn),∀n ≥ 0
x(un)n≥ + y(vn)n≥0 ∈ Da,b
⇒ Da,b R
• Da,b
(U n)n≥0, (V n)n≥0 ∈ Da,b
U 0 = 1, U 1 = 0
V 0 = 0, V 1 = 1
x(U n)n≥0 + y(V n)n≥0 =
0
⇒
xU 0 + yV 0 = 0
xU 0 + yV 1 = 0
⇒ x = y = 0
{(U n)n≥0, (V n)n≥0} Da,b
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•
Da,b
(un)n≥0
∈ Da,b
un = u0U n + u1V n
n = 0 n = 1
n ≤ k + 1 n = k
n = k + 1 uk =
u0U k + u1V k
uk+1 = u0U k+1 + u1V k+1
uk+2 = auk+1 + buk
= a(u0U k+1 + u1V k+1)
+ b(u0U k + u1V k)
= u0(aU k+1 + bU k)
+ u1(aV k+1 + bV k)
= u0U k+2 + u1V k+2
{(U n)n≥0, (V n)n≥0} Da,b
⇒ D
a,b
t
t2 − at − b = 0 ∆ = a2 + 4b
•
∆ = a2 + 4b > 0
x1 x2 {(xn1 )n≥0, (xn2 )n≥}
Da,b
⇒ un = αx
n1 + βx
n2 ,∀n ≥ 0 α β
•
∆ = a2 + 4b = 0
x
{(xn)n≥0, (xn−1)n≥}
Da,b
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⇒
un = αxn + βnxn−1, ∀n ≥ 0
α
β
•
∆ = a2 + 4b
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(f n)n
≥0
f 0 = 0, f 1 = 1
f n+2 =
f n+1 + f n,∀n ≥ 0
t2 − t − 1 = 0
∆ = 5
⇒
1±√ 5
2
⇒
f n = x(1 +
√ 5
2 )n + y(
1 −√ 52
)n, ∀n ≥ 0
f 0 = 0 f 1 = 1
⇒
x + y = 0
x1+√ 5
2 + y1−√ 52 = 1
⇔
x =√ 55
y = −√ 55
f n =√ 5
5 ( 1 + √ 5
2 )n − √ 5
5 ( 1 −√ 5
2 )n,∀n ≥ 0
(un)n≥0 u0 = u1 = 1
un+2 = 6un+1 −9un, ∀n ≥ 0
t2 − 6t + 9 = 0 ∆ = 0
t = 3
un = x3n + y3n−1
u0 = u1 = 1 x = 1
3x + y = 1
x = 1
y = −2
un = 3n − 2.3n−1, ∀n ≥ 0
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(un)n≥0 u0 = 0; u1 = 1
2un+2 = 2un+1
−un
∀n
≥ 0
un+2 = un+1 − 12un t2 −
t + 12 = 0 ∆ = −3 < 0
z = 1+i
√ 3
2
⇒
r = |z| =√ 22
ϕ = Argz = π4
un = (
√ 2
2 )n( p cos n
π
4 − q sin n π
4), ∀n ≥ 0
u0 = 0; u1 = 1
⇒
p = 0
q =√ 22
( p√ 22 − q
√ 22
)
⇔
p = 0
q = −2
un = 2(
√ 2
2 )n sin n
π
4, ∀n ≥ 0
(un)n≥0 u0 = a > 0
: u1 = b > 0
un+2 = 3
u2nun+1,∀n ≥ 0
u3n+2 = u2n.un+1
⇒ ( un+2un+1
)3 = ( un
un+1)2
⇒ 3(ln un+2 − ln un+1) = 2(ln un − ln un+1)
= −2(ln un+1 − ln un)
vn = ln un − ln
un−1 ⇒ 3vn+1 = −2vn
vn+1 = −2
3 vn
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(vn)n≥0 q = −23
(vn)n≥0
(un)n≥0
(un)n≥0 u0 = a > 0;
u1 = b > 0
un+2 = 2unun+1un + un+1
, ∀n ≥ 0
2
un+2=
1
un+1+
1
un
vn = 1
un
(un)n≥0 u1 = u2 =
u3 = 1
un+3 = 1 + un+1un+2
un, ∀n ∈ Z+
(un)n≥0
(un)n≥0 u0 = 2
un+1 = 3un +
8u2n + 1, ∀n ≥ 0
un+1 = 3un +
8u2n + 1
⇔ u2n+1 −6un+1un + u2n = 0
n + 1
n
⇒ u2n −6unun−1 + u2n−2 = 0
⇒ (un+1 −un−1)(un+1 + un−1 −6un)
= 0
un+1 = 3un +
8u2n + 1 > 3un
un+1 > 3un > 9un > un−1
un+1 = 6un − un−1
(un)n≥0
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un+1 = f (un)
f
f : M −→ M
u0 = M
f
• f
un+1 − un =
f (un) − f (un−1) f
un+1 − un un − un−1
n
un+1 − un
u1 −u0
u0 ≤ u1 (un)n≥
u0 ≥ u1 (un)n≥
•
f
f 2 = f ◦ f
u0 ≤ u2 (u2n)n≥
(u2n+1)n≥
u0 ≥ u2 (u2n)n≥
(u2n+1)n≥
f R
(un)n≥0 α α ∈ M
f ⇒ f (un)n≥0
f (α) f (α) = α α
f (x) = x
(un)n≥0 u0 = a > 0
un+1 = 1
6(unn + 8), ∀n ≥ 0
f (x) = 1
6(unn + 8)
f (x) = x
2
4
f
[0;∞) −→ [0;∞)
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f ([0; 2]) ⊂ [0; 2]; f ([2;
4]) ⊂ [2; 4]; f ([4;∞)) ⊂ [4;∞))
• a ∈ [0; 2] u1 ≥ a
= u0
• a ∈ [2; 4]
u1 ≤ a = u0
• a = 4
• a ∈ [4;∞)
b > 4 b
f (x) = x
(un)n≥0
u0 = a > 0 un+1 =
16
(un + b2
un), ∀b ≥ 0
u0 = 1 un+1 = 1− 2un u0 =
a un+1 =
3√
7un − 6 u0 = a ≥ 0
un+1 = 62+u2
n
u0 =√
2 un+1 =√
2 + un,∀n ≥ 1
u0 = √ 2 un+1 = √ 2
+ un,∀n ≥ 1
a, b
u0 = b
un+1 = u2n + (1 − 2a)un + a2
un+3 = aun+2 + bun+1 + cun(∗),
∀n ≥ 0
a,b,c ∈R
• Da,b,c = {(un)n≥0 :
un+3 =
aun+2 + bun+1 + cun|∀n ≥ 0;
a,b,c ∈R}
•
Da,b,c R
Da,b,c
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(0n)n≥0 ∈ Da,b,c ⇒ Da,b,c =
Ø
(un)n≥0, (vn)n≥0 ∈ Da,b,c
xun+3 + yvn+3 =
x(aun+2 + bun+1 + cun)
+ y(avn+5 + bvn+1 + cvn)
= a(xun+2 + yvn+2)
+ b(xun+1 + yvn+1 + c(xun + yvn),
∀n ≥ 0
x(un)n≥ + y(vn)n≥0 ∈ Da,b,c
⇒ Da,b,c R
Da,b,c
(U n)n≥0, (V n)n≥0,
(Z n)n≥0 ∈ Da,b,c
U 0 = 1, U 1 = 0, U 2 = 0
V 0 = 0, V 1 = 1,
V 2 = 0
Z 0 = 0, Z 1 = 0, Z 2 = 1
x(U n)n≥0 + y(V n)n≥0 + c(Z n)n≥0 =
0
⇒
xU 0 + yU 1 + zU 2 =
0
xV 0 + xV 1 + zV 2 =
0
xZ 0 + yZ 1 + zZ 2 =
0
⇒ x = y = z = 0
{(U n)n≥0, (V n)n≥0, (Z n)n≥0}
•
Da,b,c
(un)n≥0 ∈ Da,b,c.
un =
u0U n + u1V n + u2Z n
n = 0
n = 1
n ≤ k + 1
n = k, n = k + 1
n = k + 2
uk = u0U k +
u1V k + u2Z k
uk+1 =
u0U k+1 + u1V k+1 + u2Z k+1
uk+2 =
u0U k+2 + u1V k+2 + u2Z k+2
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uk+3 = auk+2 + buk+1 + cuk
= a(u0U k+1+u1V k+2+u2Z k+2)+b(u0U k+1+u1V k+1+u2Z k+1)+c(u0U k
+u1V k +u2Z k)
= u0(aU k+2 + bU k+1 + cU k)
+ u1(aV k+2 + bV k+1 + cV k)
+
u2(aZ k+2 + bZ k+1 + cZ k)
= u0U k+3 + u1V k+3 + u2Z k+3
{(U n)n≥0, (V n)n≥0, (Z n)n≥0}
Da,b,c
⇒ Da,b,c
•
t3 − at2 − bt − c = 0(∗∗)
t1, t2, t3
{(t1)n, (t2)n, (t3)n} Da,b,c
un = xtn1 + yt
n2 + zt
n3 , ∀n ≥ 0
x ,y,z ∈ R
t1, t1, t2
{(t1)n, (t1)n−1, (t2)n} Da,b,c
un = xtn1 + ynt
n−11 + zt
n2 ,∀n ≥ 0
x ,y,z ∈ R
t1 = t2 = t3 = t
{(t)n, (t)n−1, (t)n−2} Da,b,c
un = xtn + yntn−1
+ zn2tn−2,∀n ≥ 0
x ,y,z ∈ R
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β
t1, t2
un = xβ n + rn( p cos nα− q sin
nα), ∀n ≥ 0, ∀( p, q ) ∈R2
r = |t1|, α = Argt1
un+k
= a1un+k−1 + a2un+k−2 + .. + akuk
a1, a2, ...ak ∈ R, k ≥ 3
Da1,a2,..,ak
Da1,a2,..,ak R
{(u1n)n≥0, (u2n)n≥0, ..., (ukn)n≥0}
uij =
1 i = j
0 i = j
i, j = 1,...,k
tk − a1tk−1 − a2tk−2 − ...− ak = 0
t1, t2,..,tk
un =k
i=1bit
ni , b1, b2,...,bk ∈ R
t1, t2,..,tk u1, u2,...,uk
t1, t2,..,tl s1, s2,...,sl (s1 + s2 +
... + sl = k)
un =l
i=1
(
sl−1j=1
bijnjt
n−ji )
bij ∈ R
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(un)n≥0
u0 = 0, u1 = 1, u2 = 2
un+3 = (√
2− 1)un+2 + (√
2 − 1)un+1 +√
2un, ∀n ≥ 0
un+1 = q nun + f n
∀n ≥ 0 u0 = a
f n = 0 un =
q nun
vn+1 = q nvn.
un = vn + u
∗n (vn) vn+1 =
q nvn
vn = C
n−1i=1 q i
u∗n
un = C nvn
C n+1vn+1 =
q nC nvn + f n
⇒ C n+1vnq n − q nC nvn =
f n
⇒ vnq n(C n+1 − C n) =
f n
⇒ (C n+1 −C n) =
f nvnq n
C n = C 0 +n−1i=0
f i
vi+1
u∗n = C n.vn = (C 0 +n−1
i=0f i
vi+1)vn
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un = vn + (C 0 +
n−1i=0
f i
vi+1)vn
un = vn + C n.vn
(un)n≥0 u1 = 98
un+1 = nun + n.n!
(vn) vn+1 = nvn
(vn) vn = C.(n− 1)!
u∗n = C n.vn = C nC (n−
1)!
C n+1Cn! = C nCn!
+ nn! ⇒ C n+1 −C n =
nC
⇒ C n = C 0 +
n(n−1)2C ⇒
un = C.(n− 1)!(1 + C 0 + n(n−
1)2C
) = C.(n− 1)! + C.C 0(n− 1)!
+ 12
(n− 1)n!
u1 = 9
8 C + C.C 0 =
98
⇒ un = 98(n− 1)! + 12 (n− 1)n!
un = 98(n− 1)! + 12(n− 1)n!
10.000
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30
30 − 4 2001
