A = \(1+\frac{3}{2}+\frac{3}{2^2}+\frac{3}{2^3}+....+\frac{3}{2^{2012}}\)
A - 1 = \(\frac{3}{2}+\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^{2012}}\)
\(\frac{1}{3}\left(A-1\right)=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\frac{2}{3}\left(A-1\right)=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(\frac{2}{3}\left(A-1\right)-\frac{1}{3}\left(A-1\right)=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(\frac{1}{3}\left(A-1\right)=1-\frac{1}{2^{2012}}\)
\(A-1=3\left(1-\frac{1}{2^{2012}}\right)\)
\(A=3-\frac{3}{2^{2012}}+1=4-\frac{3}{2^{2012}}\)
\(A=1+\frac{3}{2}+\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^{2012}}=1+3\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
Đặt B = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
=> 2B = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
Lấy 2B trừ B theo vế ta có
2B - B = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
=> \(B=1-\frac{1}{2^{2012}}\)
Khi đó A = \(1+3\left(1-\frac{1}{2^{2012}}\right)=1+3-\frac{3}{2^{2012}}=4-\frac{3}{2^{2012}}\)
\(A=\frac{1+3}{2}+\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^{2012}}\)
\(A-1=\frac{3}{2}+\frac{3}{2^2}+\frac{3}{2^3}+...+\frac{3}{2^{2012}}\)
\(\frac{1}{3}\left(A-1\right)=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\frac{2}{3}\left(A-1\right)-\frac{1}{3}\left(A-1\right)=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(\frac{1}{3}\left(A-1\right)=1-\frac{1}{2^{2012}}\)
\(A-1=3\left(1-\frac{1}{2^{2012}}\right)\)
\(A=3-\frac{3}{2^{2012}}+1=4-\frac{3}{2^{2012}}\)
