Sa ne ocupam de 2+4+6+8+........+198 mai intai :
Daca dam 2 factor comun :
2+4+...+198=2(1+2+3+...+99)
Observam ca in paranteza avem o suma gauss. Stim ca [tex]1+2+...+n=\frac{n(n+1)}{2}[/tex]. Deci :
2(1+2+3+...+99) = [tex]2\frac{99*100}{2} = 99*100[/tex]
Daca inlocuim 2+4+6+8+........+198 = 99*100 in expresia initiala :
a=[tex]\sqrt{100+99*100}[/tex]
Daca dam 100 factor comun :
a=[tex]\sqrt{100(1+99)} =\sqrt{100^{2} } = 100 = 10^{2}[/tex]
Deci a= 10², ceea ce inseamna ca a este patrat perfect
