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Compararea

Posted: Sat Nov 08, 2008 5:33 pm
by alex2008
Comparati numerele :

\( A=\frac{2+2^2+2^3+...+2^{1997}}{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1997}}} \) ;\( B=\frac{3+3^2+3^3+...+3^{1331}}{\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{1331}}} \)

Posted: Mon Aug 10, 2009 1:05 am
by Mateescu Constantin
Se observa ca \( x^{n+1}\left\(\frac 1x+\frac{1}{x^2}+...+\frac{1}{x^n}\right\)=x+x^2+...+x^n \) .

Deci \( A=2^{1998}=8^{666} \) si \( B=3^{1332}=9^{666}\ \Longrightarrow\ A\ <\ B \) .