\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{128}\)
Đặt : \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{128}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{64}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(\Rightarrow A=1-\frac{1}{128}\)
\(\Rightarrow A=\frac{127}{128}\)
Vậy : \(A=\frac{127}{128}\)
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128.
Gọi biểu thức là A ta có A.2
A.2=1+1/2+1/4+1/8+1/16+1/32+1/64
A.2-A=(1+1/2+1/4+1/8+1/16+1/32+1/64)-(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128)
A=1-1/128
A=127/128
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(\Rightarrow A=1-\frac{1}{128}\)
\(\Rightarrow A=\frac{127}{128}\)
Vậy \(A=\frac{127}{128}\)
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128.
Gọi biểu thức là A ta có A.2
A.2=1+1/2+1/4+1/8+1/16+1/32+1/64
A.2-A=(1+1/2+1/4+1/8+1/16+1/32+1/64)-(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128)
A=1-1/128
A=127/128
