Răspuns:
a) n∈[63,+∞)∩N
b) n∈(0; 63)∩N
Explicație pas cu pas:
1+2+3.... +n=n(n+1)/2
[tex]\frac{ 1+2+3.... +n}{2013} =\frac{n(n+1)}{2}*\frac{1}{2013}=\frac{n(n+1)}{4026} \\a)~supraunitara~daca~\frac{n(n+1)}{4026} >1,~deci~n(n+1)>4026.\\deoarece~63*64=4032>4026,~deci~n\geq 63\\b)~subunitara~daca~\frac{n(n+1)}{4026} <1,~deci~n(n+1)<4026.~deci~n<63[/tex]
Deci n∈(0; 63)∩N