Compararea

[alex2008]

Comparati numerele :

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

[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 \( B=3^{1332}=9^{666}\ \Longrightarrow\ A\ <\ B \) .