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a fost răspuns

Aflati x din : 1+2+3+..+2004 supra 2+4+6...+2004 =2x-1 supra x .

Răspuns :

[tex] \frac{\sum\limits_{k=1}^{2004}{k}}{2\cdot\sum\limits_{k=1}^{1002}{k}} = \frac{2x-1}{x} \Longleftrightarrow \frac{2004\cdot2005}{2}\cdot \frac{1}{1002\cdot1003} = \frac{2x-1}{x} \Longleftrightarrow \frac{2005}{1003} = \frac{2x-1}{x}[/tex]
[tex]x=1003[/tex]

[tex]\frac{\sum\limits_{k=1}^{2004}{k}}{2\cdot\sum\limits_{k=1}^{1002}{k}}= \frac{1+2+...+2004}{2(1+2+...+1002)} [/tex]