2A=1+1/2+1/4+1/8+.........+1/512

2A‐A=﴾1+1/2+1/4+1/8+....+1/512﴿‐﴾1/2+1/4+1/8+.....+1/1024﴿

A=1‐1/1024 =1023/1024

vậy A=1023/1024

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

Ta có:  2A=\(2+1+\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{512}\) (2)

Từ (1) và (2) \(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...........+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{1024}\right)\)

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

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

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

\(=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

=> \(\frac{1}{2}A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\)

=> \(\frac{1}{2}A-A=\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

=> \(-\frac{1}{2}A=\frac{1}{2^{11}}-1=\frac{1-2048}{2048}=-\frac{2047}{2048}\)

=> \(A=-\frac{2047}{2048}:\left(-\frac{1}{2}\right)=\frac{2047}{1024}\)

Minh Hiền Trần~ t vs bà cách khác nhau nhưng kết quả lại giống nhau  

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

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

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

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

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

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

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

học ngu vl