Đặt A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)
2A = \(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
2A - A = \(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}-1-\frac{1}{2}-..-\frac{1}{2^{2011}}-\frac{1}{2^{2012}}\)
A = \(2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)