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\(A=\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(2A=\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\)
\(2A-A=\left(\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)
\(A=\frac{1}{4}+\frac{1}{4}-\frac{1}{2}+\frac{1}{1024}\)
\(A=\frac{1}{1024}\)
\(B=\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1024}\)
\(=-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
\(=-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
Đặt \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}=A\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\).Thay A vào ta đc: \(B=-\left(1-\frac{1}{2^{10}}\right)\)
\(B=-\left(1-\frac{1}{1024}\right)\)
\(B=-\frac{1023}{1024}\)
Ta có:
\(-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
đặt \(A=1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\)
\(A-\frac{1}{2}A=\frac{1}{2}A\Rightarrow A=\frac{1-\frac{1}{2^{11}}}{\frac{1}{2}}=2-\frac{1}{2^{10}}\)
\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=-1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(A=1-\frac{1}{1024}=\frac{1023}{1024}\)
Vậy, \(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=-1-A=-1-\frac{1023}{1024}=-\frac{2047}{1024}\)
Đặt A = \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\)...\(-\frac{1}{1024}\)
A= \(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)....\(-\frac{1}{2^{10}}\)
2A=\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^9}\)
2A-A=(\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^{10}}\)) \(-\)(\(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)..\(-\frac{1}{2^9}\))
A=\(1+\frac{1}{2^{10}}\)
A= \(\frac{1025}{1024}\)
ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
tách
\(B=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2B-B=\frac{1}{2}-\frac{1}{1024}\)
thay vào B ta có
\(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)
\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)
\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)
\(\Rightarrow A=\frac{513}{1024}\)
ta có : \(\frac{1}{2}=1-\frac{1}{2};\frac{1}{4}=\frac{1}{2}-\frac{1}{4};\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)
\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16};\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)
\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-.....-\frac{1}{1024}\)
\(=1-\frac{1}{2}-\frac{1}{2}-\frac{1}{4}-\frac{1}{4}-\frac{1}{8}-\frac{1}{8}-\frac{1}{16}-\frac{1}{16}-....-\frac{1}{512}-\frac{1}{1024}\)
\(=1-\frac{1}{1024}\)
\(=\frac{1023}{1024}\)
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}\)
\(=1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-.......-\left(\frac{1}{512}-\frac{1}{1024}\right)\)
\(=1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{512}-\frac{1}{1024}\)
\(=-\frac{1}{1024}\)
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> \(A=2A-A=1-\frac{1}{2^{10}}\)
=> \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=1-A=1-\left(1-\frac{1}{2^{10}}\right)=1-1+\frac{1}{2^{10}}\)
\(=\frac{1}{2^{10}}\)
\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-.....-\frac{1}{1024}\)
\(=-1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-.....-\left(\frac{1}{512}-\frac{1}{1024}\right)\)
\(=-1-\left(1-\frac{1}{1024}\right)\)
\(=-1-\frac{1023}{1024}\)
\(=-\frac{2047}{1024}\)
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Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
Ta có:
\(A=\left(-1\right)-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(\left(-1\right)-A=\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(2\left[\left(-1\right)-A\right]=1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\)
\(2\left[\left(-1\right)-A\right]-\left[-1-A\right]=\left(1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)
\(\left[\left(-1\right)-A\right]=1-\frac{1}{1024}=\frac{1023}{1024}\)
\(A=\left(-1\right)-\frac{1023}{1024}\)
\(=\frac{-2047}{1024}\)
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Đặt :
\(A=1+\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{1024}\)
\(\Leftrightarrow A=1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{10}}\)
\(\Leftrightarrow2A=2+1+\frac{1}{2}+......+\frac{1}{2^9}\)
\(\Leftrightarrow2A-A=\left(2+1+\frac{1}{2}+....+\frac{1}{2^9}\right)-\left(1+\frac{1}{2}+....+\frac{1}{2^{10}}\right)\)
\(\Leftrightarrow A=2-\frac{1}{2^{10}}\)
Gọi dãy số trên là A
\(A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{512}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
\(A=2-\frac{1}{1024}\)
\(A=\frac{2048}{1024}-\frac{1}{1024}\)
\(A=\frac{2047}{1024}\)