\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(2A-A=2-\frac{1}{2^{2012}}\Rightarrow A=2-\frac{1}{2^{2012}}\)
\(A=\frac{2^{2013}}{2^{2012}}-\frac{1}{2^{2012}}=\frac{2^{2012}+1}{2^{2012}}\)
À bạn Yến Nhi, tại sao mà 22013 - 1 lai bằng 22012 + 1 thế ?
\(\frac{con}{\frac{chịu}{\frac{thưa}{\frac{cụ}{hi}}\frac{câc}{ạ}}}\)
\(\frac{\frac{ai}{làm}}{\frac{được}{\frac{ko}{\frac{?}{?}}}}\)
2A thì phải là 2.A chứ sao lại cộng nhỉ