Algèbre vide. est stable par combinaison linéaire.

ER
un+2 =un+1 + 2 un∀n∈N(R)
(R)F
unvnF a b
(wn) = a.(un) + b.(vn)(R)F
FE(E, +, .)
R(F, +, .) (E, +, .)
(un)(u0)q
F(R)
u0qn+2 =u0qn+1 + 2 u0qn∀n∈N
u0= 0 q
q2−q−2 = 0
q1= 2 q2=−1
F(2nu0) ((−1)nu0)
u0
u0= 1 (2n) ((−1)n)
nF
F(un)
nun=a2n+b(−1)nF
(un)F
α20+β(−1)0=u0
α21+β(−1)1=u1
(α, β)
(α, β) = u0+u1
3,2u0−u1 3
nun=α2n+β(−1)n(un) = α.(2n) + β.((−1)n)
{(2n),((−1)n)}(F, +, .)
α20+β(−1)0= 0
α21+β(−1)1= 0
(α, β) = (0,0)
{(2n),((−1)n)}(F, +, .)
((2n),((−1)n)) (F, +, .)
R

(xn) (yn)n x0= 1 y0= 0
xn+1 = 2 ynyn+1 =xn+yn
(yn)
y0= 0
x0= 1 y0= 0 y1= 1
x1= 0 y1= 1 y2= 1
x2= 2 y2= 1 y3= 3
x3= 2 y3= 3 y4= 5
x4= 6 y4= 5 y5= 11
x5= 10
k
xk+1 = 2 yk
yk+1 =xk+yk
yn+2
yn+2 =yn+1 +xn+1
xn+1 yn+1
yn+2 =yn+1 + 2 yn
RF(yn)F
x2=x1+ 2 x0
k xk= 2 yk−1
xn+2 =xn+1 + 2 xn1≤n
Rn= 0 n≥1n
(yn)F
(xn) (yn)(F, +, .)
α β α.(xn) + β.(yn) = (0)
α x0+β y0= 0
α x1+β y1= 0
1α+ 0 β= 0
0α+ 1 β= 0
(α, β) = (0,0)
(xn)(yn)
{(xn),(yn)}(F, +, .)
((xn),(yn))

(a, b)(xn) ((2n),((−1)n))
a20+b(−1)0=x0
a21+b(−1)1=x1
a+b= 1
2a−b= 0
(xn) = 1
3.(2n) + 2 3.((−1)n)
(yn)
((2n),((−1)n))
a+b= 0
2a−b= 1
(yn) = 1
3.(2n)−1 3.((−1)n)
((2n),((−1)n)) ((xn),(yn))
P=
1
3
1
3
2
3−1 3
A=
111
100
100
A2=
311
111
1 1 1
A3=
5 3 3
311
311
(α, β)α.A +β.A2= 0
1α+ 3 β= 0
1α+ 1 β= 0
0α+ 1 β= 0
(α, β) = (0,0)
A A2
(α, β)α.A +β.A2=A3
1α+ 3 β= 5
1α+ 1 β= 3
0α+ 1 β= 1
(α, β) = (2,1) A3= 2.A +A2

(α, β)AA2
(α, β)6= (α0, β0) (α−α0).A + (β−β0).A2= 0
(α−α0, β −β0)6= (0,0)
A= 1.A + 0.A2
A2= 0.A + 1.A2A1=x0.A +y0.A2A2=x1.A +y1.A2
A3= 2.A + 1.A2
A3=x2.A +y2.A2
1≤p≤n−1
Ap=xp−1.A +yp−1.A2
p=n−1
An−1=xn−2.A +yn−2.A22≤n
A
An=xn−2.A2+yn−2.A32≤n
A3
An=xn−2.A2+yn−2.(2.A +A2)
An= 2 yn−2.A + (xn−2+yn−2).A2
xn−1yn−1
An=xn−1.A +yn−1.A2
p= 1 p= 2 p= 3 n−1
n
n
An=xn−1.A +yn−1.A2
n= 13 x12 y12
x12 =1
3212 +2 3(−1)12
= 1366
y12 =1
3212 −1 3(−1)12
= 1365
A13
A13 = 1366.A + 1365.A2=
5461 2731 2731
2731 1365 1365
2731 1365 1365