Question

1-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{8}\)-...-\(\frac{1}{1024}\)

Answer 1

Đặt \(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(\Rightarrow A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(2A=2-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\)

\(2A+A=\left(2-1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\right)+\left(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)

\(\Rightarrow3A=2-\frac{1}{1024}\)

\(\Rightarrow3A=\frac{2048}{1024}-\frac{1}{1024}\)

\(\Rightarrow3A=\frac{2047}{1024}\)

\(\Rightarrow A=\frac{2047}{1024}:3\)

\(\Rightarrow A=\frac{2047}{3072}\)

 

 

Answer 2

gọi A=\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)

2xA=1+\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)

2xA‐A=﴾1+\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)﴿‐﴾\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)﴿

A=1‐\(\frac{1}{1024}\)

= \(\frac{1023}{1024}\)

vậy A=\(\frac{1023}{1024}\)

Answer 3

Cái này là Violympic ak.