Notes de cours - Mathématiques

o
Pn
S=1+2+3+4+5+6+7+8+9+10+11+12+13+14+15,
T= 1 + 1
4+1 9+1 16 +· · · +1 n2.
Σ
S=
15
X
k=1
k, T =
n
X
k=1
1
k2
x1, . . . , xn∈C
n
X
k=1
xk=x1+· · · +xn
•1p≤n
Pn
k=pxk=xp+· · · +xn
•(xi)i∈I
Pi∈Ixi
ln 1 + ln 3 + ln 5 + ln 7 + · · · + ln 19 = X
k∈[[1,19]]:k
ln k
1
2+1 3+1 5+1 7+1 11 +1 13 +1 17 +1 19 +1 23 =X
k∈[[1,25]]:k
1
k

IPi∈Iai= 0
Pn
k=0 xkk
k
kR1
0f(x)dx x
n
S=
n
X
i=1
i
n+i=1 n+ 1 +2 n+ 2 +3 n+ 3 +· · · +n 2n.
(ai)i∈I(bi)i∈I
λ∈C
Pi∈I(ai+bi) = Pi∈Iai+Pi∈Ibi,
Pi∈Iλai=λPi∈Iai.
I=I1∪I2I1I2
Pi∈Iai=Pi∈I1ai+Pi∈I2ai.
p≤q≤n
Pn
k=pak=Pq k=pak+Pn k=q+1 ak.
n
X
k=0
ak=X
k∈[[0,n]]:k
ak+X
k∈[[0,n]]:k
ak
ai≥0Pi∈Iai≥0
ai≥biPi∈Iai≥Pi∈Ibi
Pn
k=1
1
kPn
k=1
1
√k

(ak)p≤k≤n
q∈Z
Pn
k=pak=Pn+q k0=p+qak0−q
Pn
k=pak=Pn−p k0=0 an−k0
k0=k+q
k0
k0=n−k
k0k
•S=
n
X
k=3
k2−4k+ 4,•
n
X
k=1
k2+n2−2nk
akakbk+1 −bk
n
X
k=p
ak= ( bp+1 −bp)+( bp+2 − bp+1)+( bp+3 − bp+2) + · · · + (bn+1 − bn)
ak=bk+1 −bk(bk)
Pn
k=pak=bn+1 −bp
n
X
k=1
1
k(k+ 1) =
n
X
k=1 1
k−1 k+ 1= 1 −1 n+ 1.

λ∈C
n
X
k=1
λ=λn
k0n n + 1
n
X
k=0
λ=λ(n+ 1).
λ
k
n
X
k=1
n=n2,
n
X
k=1
cos x=ncos x,
n
X
k=1
aj=naj.
n
n≥1
n
X
k=1
k=n(n+ 1)
2
j=n+ 1 −k Sn=Pn
j=1(n+
1−j) = n(n+ 1) −SnSn
n
n≥1
n
X
k=1
k2=n(n+ 1)(2n+ 1)
6
Cn=Pn
k=1 k2En=Pn k=1 k(k+
1)3−k3= 3k2+ 3k+ 1
(n+ 1)3−1 = 3Cn+ 3En+n
Cn

00= 1
n≥0q∈C\{1}
n
X
k=0
qk=1−qn+1
1−q=qn+1 −1 q−1
p≤n
n
X
k=p
qk=qn+1 −qp
q−1.
Sn=Pn
k=0 qkqSn=
Pn+1
j=1 qj=Sn+qn+1 −1Sn
Pn
k=pqk=Sn−Sp−1
q= 1 q= 1
Pn
k=0 qk=n+ 1
n p
(xi,j )1≤i≤n,1≤j≤p
x1,1x1,2· · · · · · x1,p
x2,1x2,2
xn,1xn,2· · · · · · xn,p
S
S=X
1≤i≤n
1≤j≤p
xi,j .
p=n
S=X
1≤i,j≤n
xi,j .