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\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow A=1+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2B=2\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\right)\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(B=1-\frac{1}{2^{2012}}\)
\(\Rightarrow A=1+\left(1-\frac{1}{2^{2012}}\right)\)
\(\Rightarrow A=2-\frac{1}{2^{2012}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
\(2A-A=(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}})-(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}})\)
\(A=2-\frac{1}{2^{2012}}\)
Vậy A = \(2-\frac{1}{2^{2012}}\)
~Chúc bạn học tốt~
Xét\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
Lấy 2A - A Ta được
\(A=2-\frac{1}{2^{2012}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2011}}\)
\(A=2-\frac{1}{2^{2012}}\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
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 \(A=2-\frac{1}{2^{2012}}\)
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\)
=>\(A-\frac{1}{2}A=\left(1+\frac{1}{2}+..+\frac{1}{2^{2012}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2013}}\right)\)
=>\(\frac{1}{2}A=1-\frac{1}{2^{2013}}\)
=>\(A=2-\frac{1}{2^{2012}}\)
Cô mình chữa bài này rồi nên bạn cứ yên tâm
\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+...+\frac{2}{2^{2011}}\)
\(2A-A=\left(2+1+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
Ta có: \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2012}}\)
=> \(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
=> \(2A=2+1+...+\frac{2}{2^{2011}}\)
=> \(2A-A=\left(2+1+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
=> \(A=2-\frac{1}{2012}\)
