pt n=0∈N
|1-3|-|5-1|=|-2|-|4|=2-4=-2
ptn=1
|2-3|-|5-2|=|-1|-|3|=1-3=-2
pt n=2
|1|-|1|=0
pt n≥3
|2^n-3|-|5-2^n|=2^n-3-(2^n-5)=2*n-3-2^n+5=-3+5=5-3=2
deci a∈{-2;0;2},∀n∈N
dar
|-2|=|2|=2
so
|0|=0
atunci
|a|∈{0;2},∀n∈N
C.C.T.D.
destul de greu pt gimnaziu
