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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 \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
=> \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
=> \(A=2A-A=1-\frac{1}{2^{10}}\)
=> \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-....-\frac{1}{1024}=1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=1-A=1-\left(1-\frac{1}{2^{10}}\right)=1-1+\frac{1}{2^{10}}\)
\(=\frac{1}{2^{10}}\)
\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-.....-\frac{1}{1024}\)
\(=-1-\left(1-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{4}\right)-.....-\left(\frac{1}{512}-\frac{1}{1024}\right)\)
\(=-1-\left(1-\frac{1}{1024}\right)\)
\(=-1-\frac{1023}{1024}\)
\(=-\frac{2047}{1024}\)
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Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
Ta có:
\(A=\left(-1\right)-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(\left(-1\right)-A=\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(2\left[\left(-1\right)-A\right]=1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\)
\(2\left[\left(-1\right)-A\right]-\left[-1-A\right]=\left(1-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{512}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)
\(\left[\left(-1\right)-A\right]=1-\frac{1}{1024}=\frac{1023}{1024}\)
\(A=\left(-1\right)-\frac{1023}{1024}\)
\(=\frac{-2047}{1024}\)
Ai mún nhờ mk giải violympic vòng 1 300 đ ko
10 nghìn 1 lần nhé hoặc là xóa nick facebook (20 nghìn 1 lần)
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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
\(A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{512}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
\(A=2-\frac{1}{1024}\)
\(A=\frac{2048}{1024}-\frac{1}{1024}\)
\(A=\frac{2047}{1024}\)
-1 - 1/2 -1/4- 1/8 - ... - 1/1024
= -(1+1/2+1/4+...+1/1024)
Đặt biểu thức là A => -A = 1+1/2 +...+ 1/1024
=> -2A = 2+1+1/2+....+1/512
=> -A = 2-1/1024
=> A = -2 + 1/1024
\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=2-\frac{1}{1024}=\frac{2047}{1024}=A\)
=> biểu thức ban đầu \(=-\frac{2047}{1024}\)
Đặt \(A=-\left(1+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(-2A=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(-2A+A=2-\frac{1}{2^{10}}\)
\(\Leftrightarrow-A=2-\frac{1}{1024}=\frac{2047}{1024}\)
\(\Rightarrow A=-\frac{2047}{1024}\)
Vậy giá trị của biểu thức là -2047/1024
\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
\(\Rightarrow2A-A=2-\frac{1}{2^{10}}\)
\(A=2-\frac{1}{2^{10}}\)
\(\Rightarrow-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{1024}=-\left(2-\frac{1}{2^{10}}\right)=-2+\frac{1}{2^{10}}\)
Ta có :
\(A=-1-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{1024}\)
\(\Rightarrow\left(-2+1\right)A=\left(-2+1\right)\left(-1-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{1024}\right)\)
\(\Rightarrow-A=\left(-2\right)\left(-1-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{1024}\right)+\left(-1-\frac{1}{2}-\frac{1}{4}-.....-\frac{1}{1024}\right)\)
\(\Rightarrow-A=\left(2+1+\frac{1}{2}+....+\frac{1}{512}\right)+\left(-1-\frac{1}{2}-\frac{1}{4}-.....-\frac{1}{1024}\right)\)
\(\Rightarrow-A=2-\frac{1}{1024}\)
\(\Rightarrow A=-2+\frac{1}{1024}\)
\(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(A=-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt A = -B
\(B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2B=2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2B-B=\left(2+1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(B=2-\frac{1}{1024}=\frac{2047}{1024}\)
=> \(A=-\frac{2047}{1024}\)