Corrigés - Animath

n xn=xn+1
xn
3xn=x1xn=xn+1 +xn−1xn+1 = 3xn−xn−1
x0= 2 x1= 3
x2= 3x1−x0= 7
x3= 3x2−x1= 18
x4= 3x3−x2= 47
x5= 3x4−x3= 123
x6= 3x5−x4= 322
x7= 3x6−x5= 843
S= 1+2+3+· · ·+2012 = 2012×2013/2 = 1006×2013
±1±2± · · · ± 2012
−S S
a−S
S±1±2± · · · ± 2012 a=−S
a > −S b 0≤b<S
a=−S+ 2b k b ≤
2012
X
j=k
j
2012
X
j=k−1
j < b ≤
2012
X
j=k
j
`∈[1, k −1] b=`+
2012
X
j=k
j
S
a=−1−2· · ·−(`−1)+`−(`+1)−(`+2)−· · ·−(k−1)+k+(k+1)+(k+2)+· · ·+2012.
2012 ×2013
2+ 1
±1±2± · · · ± 2012
(−,−,...,−)
+
(+,−,−,...,−,−)
(−,+,−,...,−,−)
(−,−,+,...,−,−)
· · ·
(−,−,−,...,−,+)
+
(+,−,−,...,−,−,+)
· · ·

(−,−,−,...,−,+,+)
(+,+,...,+)
k+k−(k+ 1)
−k+ (k+ 1) −S S
−S S
n a1, . . . , an
n ai
a1= 1
k > 1ak100
pk10 pk
ak
2,3,5,7n−1≤4
n≤5
n= 5 1,4,9,25,49
n
ABC EF GH
G H G0H0GG0H0H
EF GH
E
F G
H
G0
H0
G H (GH)
H
EF
GG0
H0
H−−→ EG F

H
EF
G
F0
H0
ABC
E
A
B CE
F
G
H
E
A
B CE
F
G
H
G0
F0
E∈[B, C]G∈[C, A]A B
(EG)H
F
H(EG)
[A, B]H
F
F H [B, C] [C, A]
[B, C]

A
BC E
H
G
F
F0
H0
E=B
F∈[B, C]H∈[A, B]
A
BC E
G
F
H
F0
H0G0
G[A, C]
A
BC
G
H
BF
BC =AG AC =AH BH = 1 −BH AB .
ABC ABCD
S(EF GH) = BF
BC ×BH BA ×S(ABCD) = 2x(1 −x)
x=BF
BC x(1 −x) = x−x2=1 4−x−1
22≤1
4EF GH
1
2E=B F [B, C]
H[A, B]
ABC
1
2

g:Q→Qx, y ∈Qg(x+y) = g(x) + g(y)
a∈Qg(x) = ax x ∈Q
x=y= 0 g(0) = 0
y=nx n ∈Ng(nx +x) = g(nx) + g(x)
g(nx) = ng(x)
m, n ∈Nn6= 0
gm
n=1 nngm n=1 ngnm n=1 ng(m) = 1 nmg(1) = am n
a=g(1)
g(x) = ax x ∈Q+
x∈Q−g(−x) = a(−x) = −ax
0 = g(0) = g(x+ (−x)) = g(x) + g(−x) = g(x)−ax
g(x) = ax x ∈Q
y=−f(z)
f(x+f(0)) = f(x+z)−f(z).
x z 7→ f(x+z)−f(z)
z= 0 f(x)−f(0)
f(x+z)−f(z) = f(x)−f(0).
g(x) = f(x)−f(0) x, z ∈Q
g(x+z)−g(z) = (f(x+z)−f(0))−(f(z)−f(0)) = f(x+z)−f(z) = f(x)−f(0) = g(x).
a∈Qg(x) = ax x ∈Q
f(x) = ax +b b =f(0)
f x =z= 0
ab +b= 0 b= 0 a=−1
b= 0
a(x+ (a+ (y+az)) = y+a(x+z),
a2y+a3z=y+az y = 1 z= 0
a∈ {−1,1}a∈ {−1,1}
a2= 1 a3= 3
f(x) = x f(x) = −x
a=−1f(x) = b−x
f(x+f(y+f(z))) = b−(x+b−(y+b−z)) = b−x+y−z=y+f(x+z)
x7→ x x 7→ b−x
b∈Q
A1, . . . , A8
P1
[A1, A2]P1[A5, A6]P1
P2
P1, P2, . . . , Pki Pi