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Theodoraerusoaica
a fost răspuns

Simplificati fractia: 2+(2^2013+2^2012+2^2011+...+2) supra 4^1008

Răspuns :

[tex]Notam:\\ S=2+2^2+...+2^{2013}|\cdot 2\\ 2S=2^2+2^3+...+2^{2014}\\ ------------\\ 2S-S=2^{2014}-2\\ S=2^{2014}-2\\ Revenind: \frac{2+2+2^2+...+2^{2013}}{4^{1008}}=\frac{2+2^{2014}-2}{2^{2016}}=\frac{2^{2014}}{2^{2016}}=\frac{1}{4}[/tex]

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