Question

Aflați suma tuturor elementelor a următoarei progresii geometrice finite:
1, 2, 4, 8, 16, 32, 64, . . . , 1024.

Answer 1

$S=1+2+2^2+...+2^{10}=2^{11}-1. \\ \\ Sau~fara~a~utiliza~formule: \\ \\ S=1+2+2^2+...+2^{10} \\ \\ 2S=2+2^2+2^3+...+2^{11} \\ \\ ------------- \\ \\ 2S-S=2^{11}-1 \Leftrightarrow S=2^{11}-1=2047.$