Page 99 - MATINF Nr. 7

˘
PROBLEME DE MATEMATICA PENTRU CONCURSURI 99

– É É ™
k − 12 2(k − 12)
Cazul 1. k ∈ [12, 18]. Atunci prin intersect , ie obt , inem a ∈ 2 + , 2 + ,
• 6 3 ˜
3(k − 4) p
2
p
2
2
2
s , i cum 2(b +c )−a = 2k−3a rezult˘a c˘a M k = 12 − 4 6(k − 12), − 2 6(k − 12) .
2
– É É ™
k − 18 2(k − 12)
Cazul 2. k ∈ [18, 36]. Atunci prin intersect , ie obt , inem a ∈ 3 + , 2 + ,
• ˜ 2 3
k
p p
s , i M k = 12 − 4 6(k − 12), − 9 2(k − 18) .
2
√
7
M 128. Ar˘atat ,i c˘a log 3 + log 5 + log 7 + . . . + log 15 > 7 2.
2
14
6
4
Dorin M˘arghidanu, Corabia
Solut ,ie. Vom demonstra, mai general, c˘a pentru orice n ∈ N, n ≥ 2, avem
È
log 3 + log 5 + log 7 + . . . + log (2n + 1) > n · 2n 1 + log (n + 1).
2 4 6 2n 2
ˆ
Intr-adev˘ar, pentru orice m ∈ N, m ≥ 2, avem
m + 1 m + 2 m + 2
log (m + 1) = 1 + log > 1 + log > 1 + log = log (m + 2). (∗)
m m m m+1 m+1
m m + 1 m + 1
Notˆand P n = log 3·log 5·log 7·. . .·log (2n+1) s , i Q n = log 4·log 6·log 8·. . .·log 2n+1 (2n+2)
2
4
2n
7
6
5
3
2
din (∗) rezult˘a c˘a P n > Q n . Prin urmare P > P n Q n = log 3·log 4·log 5·. . .·log 2n+1 (2n+2) =
4
2
n
3
p
log (2n + 2), deci P n > 1 + log (n + 1). Cum tot , i logaritmii considerat , i sunt pozitivi, prin
2 2 √
aplicarea Inegalit˘at ,ii mediilor avem log 3 + log 5 + log 7 + . . . + log (2n + 1) ≥ n · n P n >
È
2n
6
4
2
p p
n · n 1 + log (n + 1) = n · 2n 1 + log (n + 1). Pentru n = 7 obt , inem inegalitatea din enunt , .
2
2
n 3 2 k
X (3k + 9k + 11k + 4) C
∗
M 129. Calculat ,i suma 2k+1 , unde n ∈ N .
k + 2
k=1
Mih´aly Bencze, Bras , ov
2
3
2
2
(3k + 9k + 11k + 4) C k [(4k + 6)(k + k + 1) − (k + 2)(k − k + 1)] C k
Solut ,ie. Cum 2k+1 = 2k+1
k + 2 k + 2
2
k
(k + k + 1)(2k + 2)(2k + 3)C 2k+1
k+1
k
2
2
2
k
= −(k −k+1)C 2k+1 = (k +k+1)C 2k+3 −(k −k+1)C 2k+1 ,
(k + 1)(k + 2)
3
2
n
P (3k + 9k + 11k + 4) C k P
k+1
2
k
2
n
rezult˘a c˘a 2k+1 = (k + k + 1)C 2k+3 − (k − k + 1)C 2k+1 =
k=1 k + 2 k=1
1
2
2
(n + n + 1)C n+1 − C = (n + n + 1)C n+1 − 3.
2n+3 3 2n+3
M 130. Fie z 1 , z 2 , . . . , z 12 numere complexe distincte dou˘a cˆate dou˘a s , i avˆand modulele egale.
Dac˘a |z 1 − z 2 | = |z 4 − z 5 | = |z 7 − z 8 | = |z 10 − z 11 |, |z 2 − z 3 | = |z 5 − z 6 | = |z 8 − z 9 | = |z 11 − z 12 |,
|z 3 − z 4 | = |z 6 − z 7 | = |z 9 − z 10 | = |z 12 − z 1 | s , i punctele de afixe z 1 , z 2 , . . . , z 12 sunt, ˆın aceast˘a
ordine, vˆarfurile unui poligon convex, atunci ar˘atat ,i c˘a z 1 + z 2 + . . . + z 12 = 0.
Sorin Ulmeanu, Pites , ti