\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
\(2A-A=(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}})-(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}})\)
\(A=2-\frac{1}{2^{2012}}\)
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
\(A=2-\frac{1}{2^{2012}}\)
\(A=1+\frac{1}{2}+...+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^{2012}}\right)\)
=> \(A=2-\frac{1}{2^{2012}}\)
Trả lời
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{...1}{2^{2012}}\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2012}}\)
\(\Leftrightarrow2B=\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+...+\frac{2}{2^{2012}}\)
\(\Leftrightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(\Leftrightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{...1}{2^{2012}}\right)\)
\(\Leftrightarrow B=1-\frac{1}{2^{2012}}\)
\(\Leftrightarrow A=1+\left(1-\frac{1}{2^{2012}}\right)\)
\(\Leftrightarrow A=2-\frac{1}{2^{2012}}\)
Vậy \(A=2-\frac{1}{2^{2012}}\)
Ta có : 2A = 2 (1 + 1/2 + 1/2^2 + 1/2^3 +......+1/2^2012)
= 2 + 1 + 1/2 + 1/2^2 + .....+ 1/2^2011
=> A = 2A -A = 2 - 1/2^2012