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2014+2(1+2+3+..+2014) este patrat perfect?



Răspuns :

[tex]\displaystyle 2014+2(1+2+3+..+2014)=2014+\not2 \cdot \frac{2014(2014+1)}{\not2} = \\ \\ =2014+2014 \cdot 2015=2014(1+2015)=2014 \cdot 2016=4060224 \not= p.p.[/tex]