Page 27 - RMGO 6

Rezolvarea problemelor din num˘arul anterior 27
√
2
n 2 n + 1 − 2n √ √
2
2
Obt , inem, succesiv: > √ ; 2 2n n + 1 > (n + 1)(n − 1) ;
2
n + 1 2 n + 1
2
2
2
2
4
2
2
2
2
8n (n + 1) > (n + 1) (n − 1) , (∗); (n + 2n + 1)(n − 2n + 1) < 8n (n + 1);
Å ã Å ã 2 Å ã
1 1 1 1
n + 2 + n − 2 + < 8 n + . Notˆand n + = t, inecuat , ia devine
n n n n
2
2
(t + 2)(t − 2) < 8t. Cum pentru t > 5 avem t + 2 > t s , i (t − 2) > 8, rezult˘a c˘a
a
t ≤ 5, deci n ∈ {1, 2, 3, 4}, care verific˘ (∗).
Clasa a XI-a
Ü ê
2 −1 0 0
2 5 0 0
MGO 191. Se consider˘a matricea A = .
0 0 3 −2
0 0 1 6
n
∗
Calculat ,i A , n ∈ N .
* * *
Å ã Å ã Å ã
2 −1 3 −2 B O 2
Solut ,ie. Notˆand B = s , i C = , avem A = s , i
2 5 1 6 O 2 C
Å B n ã
n
astfel A = O 2 .
n
O 2 C
2
a
Ecuat , ia caracteristic˘ a matricei B este B −7B+12I 2 = O 2 , cu forma numeric˘
a
n
n
n
2
r − 7r + 12 = 0 avˆand r˘ad˘acinile r 1 = 3 s , i r 2 = 4, deci B = 3 B 1 + 4 B 2 , unde
ß Å ã
B 1 + B 2 = I 2 2 1
a
. Obt , inem c˘ B 1 = 4I 2 − B = s , i B 2 = B − 3I 2 =
3B 1 + 4B 2 = B −2 −1
Å ã Å n n n n ã
−1 −1 2 · 3 − 4 3 − 4
n
, deci B = n n n n .
2 2 −2 · 3 + 2 · 4 −3 + 2 · 4
n
n
Å 2 · 4 − 5 n 2 · 4 − 2 · 5 n ã
n
Analog obt , inem C = n n n n s , i astfel
−4 + 5 −4 + 2 · 5
Ü n n n n ê
2 · 3 − 4 3 − 4 0 0
n
n
−2 · 3 + 2 · 4 n −3 + 2 · 4 n 0 0
n
A = n n n n .
0 0 2 · 4 − 5 2 · 4 − 2 · 5
n
n
0 0 −4 + 5 n −4 + 2 · 5 n
MGO 192. Ar˘atat ,i c˘a pentru orice a, b, c ∈ C, exact unul dintre sistemele
 
 ax + by + cz = 0  ax + 2y = 1
 
2x + 3y + 4z = 0 bx + 3y = 1
 
s , i
 x + y + z = 1  cx + 4y = 1
 
 
x, y, z ∈ C x, y ∈ C