Question

A = 1/2 + 1/2^{2 +} 1/2³ +..........1/2¹⁰¹⁹

 

Answer 1

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2019}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2018}}\)

Lấy 2A trừ A theo vế ta có :

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^{2018}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2019}}\right)\)

\(A=1-\frac{1}{2^{2019}}\)

Vậy \(A=1-\frac{1}{2^{2019}}\)