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a fost răspuns

Se considera n=2015+2·(1+2+3...+2014)
a.Calculati suma 1+2+3...+2014
b.Aratati ca n este patrat perfect.

Răspuns :

0000000
[tex]n=2015+2*(1+2+3...+2014) \\ n=2015+2*[\frac{2014(2014+1)}{2}] \\ n=2015+2014*2015 \\ n=2015(2014+1) \\ n=2015^2=p.p[/tex]
Bunicaluiandrei
S = 1+2+3+...........+2014 = 2015·2014/2 = 2015 ·1007 = 2029105
n = 2015 + 2·2014·2015/2 = 2015(1+2014) = 2015² = p.p.

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