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Soit S k = (−1)k
Alors,
n
(n − k)2
k
n =−
2
n−1
X
Sk
k=1
Ce est équivalent à
n =
2
n−1
X
(−1)
k+1
k=1
!
n
(n − k)2 .
k
Démonstration. Pour cela, on ajoute les termes S 0 et S n à la somme :
−
n−1
X
Sk = S0 + Sn −
n
X
S k.
k=0
k=1
On a :
!
n
S n = (−1)
(n − n)2 = 0.
n
!
0 n
S 0 = (−1)
(n − 0)2 = n2 .
0
n
La somme
!
n
X
k n
(−1)
(x − k)2
k
k=0
est égale à zéro, ce qui est facilement vérifiable pour n = 3 et pour tout x ∈ Z, incluant n.
Par induction,
n
!
!
n+1
X
X
n
+
1
n
+
1
(x − k)2 = x2 + (−1)k
(x − k)2 − (−1)n (x − n)2
(−1)k
k
k
k=0
k=1
!
!!
n
X
n
n
2
k
2
= x + (−1)
+
(x − k) − (−1)n (x − n)2
k
k−1
k=1
n
!
!
n
X
X
n
n
(x − k)2 − (−1)n (x − n)2
= x2 + (−1)k (x − k)2 − (−1)k
k
k−1
k=1
k=1
!
!
n
X
2
k n
k n
=
(−1)
(x − k) −
(−1)
(x − 1 − k)2
k
k
k=0
k=0
n
X
= 0 − 0 = 0.
Donc,
1
−
n−1
X
k=1
Sk = S0 + Sn −
n
X
S k = n2 + 0 − 0 = n2 .
k=0
Puisque l’induction a pour départ n = 3, cette relation n’est prouvée que pour n > 2
2
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