Exercices : Sommes, Produits, Récurrences - 1ère

Σ
S1= 32+ 33+ 34+· · · + 315
S2=1
2+2 4+3 8+4 16 +· · · +10 1 024
S3=a+a2
2+a3 3+· · · +an n
S4= 2 −4+6−8 + · · · + 50
k=n
X
k=1
(2k+ 1) 4.
k=n
X
k=1
(−1)k7.
k=n
X
k=1
32k
k=2009
X
k=945
3 5.
k=n
X
k=1
k(2k2−1) 8.
k=n
X
k=1
2k+k2+ 2
k=n
X
k=1
(6k2+ 4k+ 1) 6.
k=18
X
k=1
1
3k9.
k=n
X
k=1
2k
3k+1
a b c ∀k>2k−5
k(k2−1) =a k−1+b k+c k+ 1
k=n
X
k=2
k−5
k(k2−1)
n∈N
k=n
X
k=1
(k+ 1)3−
k=n
X
k=1
k3
(k+ 1)3
k=n
X
k=1
(k+ 1)3
k=n
X
k=1
k2
PROF : ATMANI NAJIB
1er BAC Sciences Mathématiques BIOF
http://xriadiat.e-monsite.com

X
16i,j6n
ij X
16i6j6n
ij X
16i,j6n
i
jX
16i,j6n
|i−j|X
16i,j6n
i2j
1.
k=n
Y
k=2 1−1
k2.
k=n
Y
k=2 1−1
k23.
k=n
Y
k=1
(6k−3)
∀n>4 2n6n!
∀n>1n(2n+ 1)(7n+ 1) 6
∀n>1
k=n
X
k=1
k×k! = (n+ 1)! −1
∀n∈N
k=2n
X
k=1
1
k>1 + n 2
(un)u0= 0 ∀n∈Run+1 = 2un+ 1
un
(un)u0= 1 ∀n∈Nun+1 =1
3(un+ 4n+ 6) ∀n∈N
un= 2n+1
3n
(un)u0=u1= 0 u2= 2 ∀n∈Nun+3 = 3un+2 −3un+1 +un
un
n Pnn
n= 1
P1Pnn
Pn+1 n+ 1
n
Pn+1
n n
PROF : ATMANI NAJIB
1er BAC Sciences Mathématiques BIOF
http://xriadiat.e-monsite.com

S1=
k=15
X
k=2
3k
S2=
i=10
X
i=1
i
2i
S3=
k=n
X
k=1
ak
k
S4=
i=25
X
i=1
−2i(−1)i
k=n
X
k=1
(2k+ 1) = 2
k=n
X
k=1
k+
k=n
X
k=1
1 = n(n+ 1) + n=n(n+ 2)
k=2009
X
k=945
3 = 3 ×1 065 = 3 195
k=n
X
k=1
(6k2+ 4k+ 1) = 6
k=n
X
k=1
k2+ 4
k=n
X
k=1
k+
k=n
X
k=1
1 = n(n+ 1)(2n+1)+2n(n+ 1) + n
=n((n+ 1)(2n+ 1) + 2(n+ 1) + 1) = n(2n2+ 5n+ 4)
k=n
X
k=1
(−1)k=
k=n−1
X
k=0
(−1)k+1 =−1−(−1)n
1−(−1) =(−1)n−1 2
k=n
X
k=1
k(2k2−1) = 2
k=n
X
k=1
k3−
k=n
X
k=1
k=n2(n+ 1)2
2−n(n+ 1) 2=n(n+ 1)(n(n+ 1) −1) 2
=n(n+ 1)(n2+n−1)
2
k=18
X
k=1
1
3k=
k=18
X
k=1 1
3k
=
k=18
X
k=0 1
3k
−1 = 1−1
319
1−1
3
−1 = 3
21−1 319 −1 = 1 21−1 318 
k=n
X
k=1
32k=
k=n
X
k=1
9k=
k=n
X
k=0
9k−1 = 1−9n+1
1−9−1 = 9n+1 −1 8−1 = 9n+1 −9 8
k=n
X
k=1
2k+k2+ 2 =
k=n
X
k=1
2k+
k=n
X
k=1
k2+
k=n
X
k=1
2 =
k=n
X
k=0
2k−1 + n(n+ 1)(2n+ 1)
6+ 2n
= 2n+1 −2+2n+n(n+ 1)(2n+ 1)
6= 2(2n+n−1) + n(n+ 1)(2n+ 1) 6
k=n
X
k=1
2k
3k+1 =1 3
k=n
X
k=1 2
3k
=2
9
k=n
X
k=1 2
3k
=2
9
k=n−1
X
k=0 2
3k
=2
9
1−(2
3)n
1−2
3
=2
31−2 3n
PROF : ATMANI NAJIB
1er BAC Sciences Mathématiques BIOF
http://xriadiat.e-monsite.com
3

a
k−1+b k+c k+ 1 =ak(k+ 1) + b(k−1)(k+ 1) + ck(k+ 1) (k−1)k(k+ 1) =ak2+ak +bk2−b+ck2+ck k(k2−1)
a+b+c= 0 a+c= 1 −b=−5b= 5 a=−2
c=−3
k=n
X
k=2
k−5
k(k2−1) =
k=n
X
k=2
−2
k−1+5 k+−3 k+ 1 =−2
k=n
X
k=2
1
k−1+ 5
k=n
X
k=2
1
k−3
k=n
X
k=2
1
k+ 1 =
−2
k=n−1
X
k=1
1
k+ 5
k=n
X
k=2
1
k−3
k=n+1
X
k=3
1
k=−2
k=n−1
X
k=3
1
k−2−1 + 5
k=n−1
X
k=3
1
k+5 2+5 n−3
k=n−1
X
k=3
1
k−3 n−3 n+ 1 =
−1
2+2 n−3 n+ 1
k=n
X
k=1
(k+ 1)3−
k=n
X
k=1
k3=
k=n+1
X
k=2
k3−
k=n
X
k=1
k3= (n+ 1)3−1
(k+ 1)3=k3+ 3k2+ 3k+ 1
k=n
X
k=1
(k+ 1)3=
k=n
X
k=1
k3+ 3
k=n
X
k=1
k2
k=n
X
k=1
k+
k=n
X
k=1
1
k=n
X
k=1
(k+ 1)3−
k=n
X
k=1
k3= 3
k=n
X
k=1
k2+ 3
k=n
X
k=1
k+
k=n
X
k=1
1
3
k=n
X
k=1
k2+3
k=n
X
k=1
k+
k=n
X
k=1
1 = (n+1)3−1 3
k=n
X
k=1
k2+3 n(n+ 1)
2+n= (n+1)3−1 = n3+3n2+3n
3
k=n
X
k=1
k2=n3+3n2+3n−3
2n2−3 2n−n=n3+3 2n2+1 2n=
n(2n2+ 3n+ 1)
2=n(n+ 1)(2n+ 1) 2
k=n
X
k=1
k2=n(n+ 1)(2n+ 1)
6
•X
16i,j6n
ij =
i=n
X
i=1
i
j=n
X
j=1
j=
i=n
X
i=1
in(n+ 1)
2=n2(n+ 1)2 4
•X
16i6j6n
ij =
j=n
X
j=1
j
i=j
X
i=1
i=
j=n
X
j=1
jj(j+ 1)
2=1 2
j=n
X
j=1
j3+j2=n2(n+ 1)2
8+n(n+ 1)(2n+ 1) 12
•X
16i,j6n
i
j=
j=n
X
j=1
1
j
i=n
X
i=1
i=n(n+ 1)
2
j=n
X
j=1
1
j
X
16i6j6n
i
j
•X
16i,j6n
|i−j|=
j=n
X
j=1
(
i=j
X
i=1
(j−i)+
i=n
X
i=j+1
(i−j)) =
j=n
X
j=1
(j2−j(j+ 1)
2+n(n+ 1) 2−j(j+ 1) 2−(n−j)j) =
j=n
X
j=1 j2−(n+ 1)j+n(n+ 1)
2=n(n+ 1)(2n+ 1) 6−n(n+ 1)2 2+n2(n+ 1) 2
=n(n+ 1)
22n+ 1 3−(n+ 1) + n=n(n+ 1)(2n−2) 6=(n−1)n(n+ 1) 3
PROF : ATMANI NAJIB
1er BAC Sciences Mathématiques BIOF
http://xriadiat.e-monsite.com
4

•X
16i,j6n
i2j=
i=n
X
i=1
i
j=n
X
j=1
2j=n(n+ 1)
2×(
j=n
X
j=0
2j−1) = n(n+ 1)
2×(2n+1 −1−1) = n(n+ 1)(2n−1)
k=n
Y
k=2 1−1
k=
k=n
Y
k=2
k−1
k=
k=n
Y
k=2
k−1
k=n
Y
k=2
k
=
k=n−1
Y
k=1
k
k=n
Y
k=2
k
=1
n
k=n
Y
k=2 1−1
k2=
k=n
Y
k=2
k2−1
k2=
k=n
Y
k=2
(k−1)(k+ 1)
k2=
k=n
Y
k=2
(k−1)
k=n
Y
k=2
(k+ 1)
k=n
Y
k=2!2=
k=n−1
Y
k=1
k
k=n
Y
k=2
k
×
k=n+1
Y
k=3
k
k=n
Y
k=2
k
=
1
n×n+ 1 2=n+ 1 2n
k=n
Y
k=1
(6k−3) =
k=n
Y
k=1
3(2k−1) = 3n
k=n
Y
k=1
(2k−1) = 3n
k=2n
Y
k=1
k
k=n
Y
k=1
2k
= 3n(2n)!
2n×n!=3 2n(2n)! n!
Pn2n6n!n
4n= 4 24= 16 4! = 24
Pn2n6n!
2n+1 62n!6(n+ 1)n!=(n+ 1)! 2 n+ 1 n
4Pn+1 Pn
n4
Pnn(2n+ 1)(7n+ 1) 6 n= 1
1×(2 + 1) ×(7 + 1) = 24 6 P1
Pnan=n(2n+ 1)(7n+ 1) an+1 −an=
(n+ 1)(2n+ 3)(7n+ 8) −n(2n+ 1)(7n+ 1) = (2n2+ 5n+ 3)(7n+ 8) −(2n2+n)(7n+ 1) =
14n3+ 16n2+ 35n2+ 40n+ 21n+ 24 −14n3−2n2−7n2−n= 42n2+ 60n+ 24 = 6(7n2+ 10n+ 4)
an+1 −an6an6
an+1 6 6
Pn+1 Pnn
1
Pn
k=n
X
k=1
k×k! = (n+1)!−1n= 1
k=1
X
k=1
k×k! = 1×1! = 1
2! −1 = 2 −1 = 1 P1Pnn
k=n+1
X
k=1
k×k! =
k=n
X
k=1
k×k!+(n+1)(n+1)! = (n+1)!−1+(n+1)(n+1)! = (n+1)!(1+n+1)−1 =
(n+ 2)! −1Pn+1 Pnn
1
Pn
k=2n
X
k=1
1
k>1 + n 2n= 0
1 1 Pn
k=2n+1
X
k=1
1
k=
k=2n
X
k=1
1
k+
k=2n+1
X
k=2n+1
1
k>1 + n 2+
k=2n+1
X
k=2n+1
1
k
2n1
2n+1
PROF : ATMANI NAJIB
1er BAC Sciences Mathématiques BIOF
http://xriadiat.e-monsite.com
5