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Arătați ca 2018+2(1+2+...+2017) e pătrat perfect

Răspuns :

=2018+2(2017×2018:2); =2018+2017×2018=2018×1+2017×2018=2018(1+2017)= 2018×2018=2018²;=> e pătrat perfect.
Eu daca as avea acest exercitiu , asa as face :
2018+2(1+2+...+2017)=
2018+2[2017 x (2017+1) : 2 ]
2018+2(2017 x 2018 : 2 )
2018+2 x 2035153
2018+ 4070306
4072324      ⇒ 4=patrat perfect 
Ultimele cifre ale unui patrat perfect pot  fi 0,1,4,5,6,9
Sper sa fie bun ! :*