\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
Nên \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
Suy ra \(2A-A=2-\frac{1}{2^{2012}}\)hay \(A=2-\frac{1}{2^{2012}}\)
Vậy \(A=2-\frac{1}{2^{2012}}\)
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)
=>\(A-\frac{1}{2}A=\left(1+\frac{1}{2}+..+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\right)\)
=>\(\frac{1}{2}A=1-\frac{1}{2^{2013}}\)
=>\(A=2-\frac{1}{2^{2012}}\)
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