Page 103 - MATINF Nr. 7

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PROBLEME DE MATEMATICA PENTRU CONCURSURI 103

n−l
Dac˘a p = n, atunci definim x l = (−1) u n−l+1 , pentru l ∈ {1, . . . , n}. Evident, (x 1 , . . . , x n )
este solut , ie a sistemului s , i, sistemul fiind Cramer, solut , ia g˘asit˘a este unic˘a.

n p n ” —
P P n−i P n−i
Dac˘a p ≤ n − 1, atunci s k+i = (−1) u n−i+1 s k+i−1 + 1 + (−1) u n−i+1 s k+i−1 .
¨ i=p i=1 i=p+1
(−1) n−l u n−l+1 , dac˘a l ∈ {1, . . . , p}
Definim x l = n−l . Evident, (x 1 , . . . , x n ) este solut , ie
1 + (−1) u n−l+1 , dac˘a l ∈ {p + 1, . . . , n}
a sistemului s , i, sistemul fiind Cramer, solut , ia g˘asit˘a este unic˘a.
M 135. Demonstrat ,i c˘a s , irul (x n ) definit prin
n≥1

π 3π 5π (2n − 1)π
2C 2 cos + 3C 3 cos + 4C 4 cos + . . . +(n + 1)C n+1 cos
n+1 2n + 1 n+1 2n + 1 n+1 2n + 1 n+1 2n + 1
x n =
π 3π 5π (2n − 1)π
n
1
2
3
C sin + C sin + C sin + . . . + C sin
n
n
n
n
2n + 1 2n + 1 2n + 1 2n + 1
este convergent s , i calculat ,i limita sa.
Mihai Florea Dumitrescu, Potcoava s , i Costel B˘alc˘au, Pites , ti
k+1
k
Solut ,ie. Utilizˆand formula (k + 1)C n+1 = (n + 1)C , deducem c˘a x n = (n + 1)a n , unde
n
b n
π 3π (2n − 1)π π 3π
2
n
2
1
1
a n = C cos +C cos +. . .+C cos s , i b n = C sin +C sin +
n
n
n
n
n
2n + 1 2n + 1 2n + 1 2n + 1 2n + 1
(2n − 1)π π
n
. . . + C sin . Astfel notˆand t = s , i z = cos t + i sin t s , i folosind formula lui
n
2n + 1 2n + 1
2 n
(1 + z ) − 1
n 2n−1
1
2 3
Moivre s , i Binomul lui Newton, avem a n + ib n = C z + C z + . . . + C z = z =
n
n
n
n
n
n
(1 + cos 2t + i sin 2t) − 1 2 cos t(cos nt + i sin nt) − 1
n
n
= = 2 cos t[cos(n−1)t+i sin(n − 1)t]
cos t + i sin t cos t + i sin t
n
n
n
n
− cos t + i sin t, deci a n = 2 cos t cos(n − 1)t − cos t s , i b n = 2 cos t sin(n − 1)t + sin t,
π (n − 1)π π π 3π π
n
n
adic˘a a n = 2 cos n cos − cos = 2 cos n sin − cos s , i
2n + 1 2n + 1 2n + 1 2n + 1 4n + 2 2n + 1
π (n − 1)π π π 3π π
n
n
b n = 2 cos n sin + sin = 2 cos n cos + sin . Cum
2n + 1 2n + 1 2n + 1 2n + 1 4n + 2 2n + 1
π (1 ) lim n(cos π −1) lim (−2n sin 2 π ) (n + 1)a n
∞
0
lim cos n = e n→∞ 2n+1 = e n→∞ 4n+2 = e = 1, iar x n = =
•
n→∞ 2n + 1 ˜ b n
3π π n + 1 π
2 n (n + 1) sin cos n − n cos 3π 3π
‹
 4n + 2 2n + 1 2 2n + 1 s , i lim (n + 1) sin = , rezult˘a
3π π 1 π n→∞ 4n + 2 4
2 n cos cos n + sin
4n + 2 2n + 1 2 n 2n + 1
3π
c˘a x n → (deci este convergent).
4
∞ 2
X k 1
M 136. Ar˘atat ,i c˘a ≤ .
e 2k 4
k=1
Sladjan Stankovik, Macedonia de Nord
Solut ,ie (Leonard Mihai Giugiuc, Drobeta Turnu Severin). Pentru fiecare n ≥ 3 arbitrar
n n
P k x n+1 −1 0 P k−1 0
fixat not˘am f (x) = x = , x ∈ (0, 1). Avem f (x) = kx , deci xf (x) =
x−1
k=0 k=1