Question

Ma poate ajuta si pe mine cineva la exercitiul acesta?Va rog mult!!Dau coroana​

Answer 1

$\it a_1=\dfrac{2^1-1}{2^1+1}=\dfrac{2-1}{2+1}=\dfrac{1}{3}\\ \\ \\ a_2=\dfrac{2^2-1}{2^2+1}=\dfrac{4-1}{4+1}=\dfrac{3}{5}\\ \\ \\ a_3=\dfrac{2^3-1}{2^3+1}=\dfrac{8-1}{8+1}=\dfrac{7}{9}\\ \\ \\ .\\.\\.\\ \\ a_{100}=\dfrac{2^{100}-1}{2^{100}+1}$