Question

A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

Tính A

Answer 1

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)

\(=>A=1-\frac{1}{2^{99}}\)

Answer 2

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{99}}\)

Vậy \(A=1-\frac{1}{2^{99}}\)

 

Answer 3

Ta có :

\(A=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)

\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^{98}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{99}}\)