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2005+2*(1+2+3+......2004) este pătrat perfect

Răspuns :

[tex]\displaystyle 2005+2 \cdot (1+2+3+...+2004)=2005+2 \cdot \frac{2004(2004+1)}{2} = \\ \\ =2005+2 \cdot \frac{2004 \cdot 2005}{2} =2005+\not 2 \cdot \frac{4018020}{\not 2} = \\ \\ =2005+4018020=2005(1+2004)=2005 \cdot 2005=2005^2-p.p[/tex]