Tính S : \(\frac{1}{2}\)+ \(\frac{1}{4}\)+ \(\frac{1}{8}\)+ ... + \(\frac{1}{1024}\).
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\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}\)
\(=1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-.......-\left(\frac{1}{512}-\frac{1}{1024}\right)\)
\(=1-1+\frac{1}{2}-\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+....+\frac{1}{512}-\frac{1}{1024}\)
\(=-\frac{1}{1024}\)
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> \(A=2A-A=1-\frac{1}{2^{10}}\)
=> \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=1-A=1-\left(1-\frac{1}{2^{10}}\right)=1-1+\frac{1}{2^{10}}\)
\(=\frac{1}{2^{10}}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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\(2A=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{512}\Rightarrow2A-A=1-\frac{1}{1024}=\frac{1023}{1024}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=\left[1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right]-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right]\)
\(A=1-\frac{1}{2014}=\frac{2013}{2014}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-.....-\frac{1}{1024}\)
\(=-1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-.....-\left(\frac{1}{512}-\frac{1}{1024}\right)\)
\(=-1-\left(1-\frac{1}{1024}\right)\)
\(=-1-\frac{1023}{1024}\)
\(=-\frac{2047}{1024}\)
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\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=\frac{1}{1024}\)dùng phương pháp loại trừ
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
Đặ A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)(1)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)(2)
Lấy (2) trừ (1) theo vế ta có :
2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
=> A = \(1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{20}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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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![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{512}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
\(A=2-\frac{1}{1024}\)
\(A=\frac{2048}{1024}-\frac{1}{1024}\)
\(A=\frac{2047}{1024}\)
S=\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{8}\)+.............+\(\frac{1}{1024}\)
S=1-1/2+1/2-1/4+1/4-1/8+.........+1/512-1/1024
S=1-1/1024
S=1023/1024
Vậy s=1023/1024
Ta có : \(\frac{1}{2}=1-\frac{1}{2};\frac{1}{4}=\frac{1}{2}-\frac{1}{4};\frac{1}{8}=\frac{1}{4}-\frac{1}{8};...;\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)
Vậy \(S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{512}-\frac{1}{1024}\)
= \(1-\frac{1}{1024}=\frac{1023}{1024}\)