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Gọi biểu thức trên là A Ta có :
A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......+\frac{1}{2048}\)
=> A : 2 = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{2048}+\frac{1}{4096}\)
=> \(\frac{1}{2}\)A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+......+\frac{1}{2048}\)- \(\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-......-\frac{1}{2048}-\frac{1}{4096}\)
=> A : 2 = \(\frac{1}{2}-\frac{1}{4096}\)
=> A : 2 = \(\frac{2047}{4096}\)
=> A = \(\frac{2047.2}{4096}\)
=> A = \(\frac{4094}{4096}\)
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2048
2A = 1 + 1/2 + 1/4 + 1/8 + ... + 1/1024
2A - A = (1 + 1/2 + 1/4 + 1/8 + ... + 1/1024) - (1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2048)
A = 1 - 1/2048
A = 2047/2048
A = 1/2+1/4+...+1/2048
2A= 1+ 1/2+ 1/4+...+1/1024
2A-A= ( 1+ 1/2+...+1/1024 ) - (1/2+1/4+...+2048)
A= 1- 1/2048
A= 2047/2048
Ta co: 2S=2+1+1/2+1/4+...+1/2048
2S-S=2+1+1/2+1/4+...+1/2048-1-1/2-1/4-...-1/2048-1/4096
\(\Rightarrow\)S=2-1/4096 =8191/4096
Đặt \(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..-\frac{1}{2048}\)
\(\Rightarrow A=1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-..-\left(\frac{1}{1024}-\frac{1}{2048}\right)\)
\(\Rightarrow A=1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-..-\frac{1}{1024}+\frac{1}{2018}\)
\(\Rightarrow A+\frac{1}{2018}\)
1-1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256-1/512-1/1024-1/2048 =0.00048828125
Đặt : \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......+\frac{1}{2048}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+.......+\frac{1}{1024}\)
\(2A-A=1-\frac{1}{2048}\)
\(A=\frac{2047}{2048}\)
Good
\(\text{Đặt }S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}.\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
\(\Rightarrow2S-S=S=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2048}\right)\)
\(\Rightarrow S=1-\frac{1}{2048}=\frac{2047}{2048}\)
Gọi A = 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2048
=> 2A = 1 + 1/2 + 1/4 + 1/8 + ...... + 1/1024
=> 2A - A = 1 - 1/2048
=> A = 2047/2048
Vậy A = 2047/2048
\(\text{Đặt }A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+......+\frac{1}{2048}\)
\(\frac{1}{2}A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{2048}+\frac{1}{4096}\)
\(A-\frac{1}{2}A=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+......+\frac{1}{2048}\right)-\left(\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-......-\frac{1}{2048}-\frac{1}{4096}\right)\)
\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{4096}=\frac{2047}{4096}\)
\(\Rightarrow A=\frac{2047.2}{4096}=\frac{4094}{4096}=\frac{2047}{2048}\)