Question

intr-o progresie geometrica bn=3^n-1 ,n>sau egal cu 1 .Calculati b1,r

Answer 1

$b_n = 3^{n-1} \\\\ b_1 = 3^{1-1} = 3^0 = 1\\ \\ b_2 = 3^{2-1}=3 \\ \\ q = \dfrac{b_2}{b_1} = \dfrac{3}{1} = 3$