Raisonnement par récurrence

F FF
n
n
un= 1 + 3 + 5 + 7 + ... + (2n−1).
n
n
un=1+3+5+7+... + (2n−3) + (2n−1).

u1u2u3u4
unn≥1
n
•
•
•
n
un=n2
(n+ 3)n=
n+2
P
k=3
kn
f
[0; +∞[→R,
x7→ (x+ 1) ln x+ 4
x+ 3−ln 2.
g: [0,+∞[→Rlim
x→+∞
g(x) = 0
g
f
x0
[4; 5] x∈[5; +∞[
0< f(x)
n
(n+ 3)n=
n+2
X
k=3
kn.
n= 1,2,3,4,5
BY:
$
\
C

n∈N?
n+2
X
k=3
kn+1 <(n+ 3)
n+2
X
k=3
kn.
n≥5 (n+ 3)n>
n+2
X
k=3
kn
BY:
$
\
C

f
f
[a, b]f(a)f(b)f
[a, b]
n= 4 n= 5
BY:
$
\
C

u1= 1 u2= 1 + 3 = 4 u3= 1 + 3 + 5 = 9 u4= 1 + 3 + 5 + 7 = 16
1 4 9 16 25 36
un=n2
1 4 9 16
1+3 1+3+5 1+3+5+7
1
n n = 1,2,3,4
n
n2n+ 1
n n n 1
du jour nau jour n+ 1du jour 5 au jour 6
n
n n2
n(Pn)un=n2
(Pn)n≥1
u1= 1 12= 1 u1= 12(P1)
n≥1 (Pn)
un=n2(Pn+1)un+1 =
(n+ 1)2
BY:
$
\
C