Question

Thu gọn tổng sau:

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

Answer 1

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

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

\(\Rightarrow2A-A=1-\frac{1}{2^{99}}\)

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

Chúc bn học tốt !!!!!!!

Answer 2

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

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

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

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