Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(\Rightarrow A=1-\frac{1}{1024}=\frac{1023}{1024}\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{1024}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{4}+.......+\frac{1}{512}\)
=> \(2A-A=1-\frac{1}{1024}\)
=> \(A=\frac{1023}{1024}\)
Ta có:
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(\Rightarrow\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow\frac{2^{10}-1}{2^{10}}=\frac{1023}{1024}\)
cách làm khác là;
1/2+1/4+1/8 .....+1/1024
=1-1/2+1/2-1/4+1/4-1/8+.....+1/512-1/1024
=1+(1/2-1/2)+(1/4-1/4)+(1/8-1/8)+........+(1/512-1/512)-1/1024
=1+0+0+0+...+0-1/1024
=1-1/1024
=1023/1024
Đặt \(F=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
Ta có: 4 : 2 = 2 ; 8 : 4 = 2
\(\Rightarrow2F=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(\Rightarrow2F=1-\frac{1}{1024}\)
\(=\frac{1024}{1024}-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
Vậy F = \(\frac{1023}{1024}\)
