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Aratati ca numarul 2014+2×(1+2+......+2013) este patrat perfect.

Răspuns :

[tex]\displaystyle 2014+2 \cdot (1+2+...+2013) \\ \\ 1+2+3+...+2013= \frac{2013(2013+1)}{2} = \frac{2013 \cdot 2014}{2} \\ \\ 2014+\not2 \cdot \frac{2013\cdot 2014}{\not2} =2014+2013 \cdot 2014=2014(1+2013 \cdot 1)= \\ \\ =2014(1+2013)=2014 \cdot 2014=2014^2-p.p[/tex]
=2014  + 2 · ( 1 + 2013 ) · 2013 / 2 =
=2014 + 2014 · 2013 
 = 2014 · ( 1 + 2013 ) 
= 2014 · 2014 = 
 = 2014²