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Hantatucalexandra
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[tex](4 + 8 + 12 + ........ + 2012) \div (2 + 4 + 6 + ........ + 1006) [/tex]
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Răspuns :

 

[tex]\displaystyle\bf\\(4+8+12+...+2012):(2+4+6+...+1006)=\\\\=\frac{(4+8+12+...+2012)}{(2+4+6+...+1006)}=(Dam~factor~comun~pe~2~la~numarator)\\\\\\=\frac{2(2+4+6+...+1006)}{(2+4+6+...+1006)}=\boxed{\bf2}[/tex]

 

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