Răspuns:
13. N = ( (-1)^n (2n - 1) + 1 ) / 4
a. n = 2k
N = ( 2n - 1 + 1 ) / 4
N = 2n / 4
N = n / 2
Cum n-par => N € Z
b. n = 2k + 1
N = ( 1 - 2n + 1 ) / 4
N = ( 2 - 2n ) / 4
N = 2(1 - n)/4
N = (1-n)/2
Cum n-impar => 1-n = par => N € Z
14. k+1 / 4 € Z =>
4 | k+1 (I)
k-2 / 3 € Z
3 | k-2
Dar 3 | 3
Rezulta
3 | k-2+3
3 | k+1 (II)
Din I si II => 3*4 | k+1
12 | k+1
k € M12 - 1
k € {-1, 11, 23, 35, 47, ...}
