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14 tháng 7 2017
\(F=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(F=\left(\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(F=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)-2.\left(\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{100}}\right)\)
\(F=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{50}}\right)\)
\(F=\frac{1}{2^{51}}+\frac{1}{2^{52}}+...+\frac{1}{2^{100}}\)
14 tháng 7 2017
\(E=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2E=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2E-E=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(E=1-\frac{1}{2^{100}}\)
20 tháng 1 2019
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^{3.}}+.............+\frac{1}{2^{100}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+.................+\frac{1}{2^{99}}\)
\(2B-B=1-\frac{1}{2^{100}}\)
\(B=1-\frac{1}{2^{100}}\)
\( C=\frac{1}{2}-\frac{1}{2^2}+.................+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(2 C=1-\frac{1}{2}+......................+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)
\(2 C+C=1-\frac{1}{2^{100}}\)
\(C=\left(1-\frac{1}{2^{100}}\right):3\)
22 tháng 4 2017
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
=>\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
=>\(A=2A-A=2+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
\(A=2+\frac{1}{2^{98}}\)
Vậy: \(A=2+\frac{1}{2^{98}}\)
22 tháng 4 2017
Gọi \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2B=2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2B-B=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow B=2-\frac{1}{2^{100}}\)
\(\Rightarrow A=2\)
Vậy A = 2
11 tháng 3 2019
Tách 100 thành 100 số 1
Ta có: TS=\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=100-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}=\left(1-1\right)+\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+...+\left(1-\frac{1}{100}\right)\)
=\(0+\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}=\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}\)=MS
=> Phân số trên=1
\(A=\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{100}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)
\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}\)
\(A=\frac{2^{100-1}}{2^{100}}\)