Đặt A là tên biểu thức
Xét mẫu số, ta có: \(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}\)
\(=1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}\)
\(=\frac{2}{2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}\)\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\left(1-\frac{1}{2013}\right)=2\cdot\frac{2012}{2013}\)
\(\Rightarrow A=\frac{2.2012}{2\cdot\frac{2012}{2013}}=\frac{2012.2013}{2012}=2013\)
\(\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+...+2012}}\)
\(=\frac{2.2012}{1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}}\)
\(=\frac{2.2012}{\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}}\)
\(=\frac{2.2012}{2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(=\frac{2.2012}{2.\left(1-\frac{1}{2013}\right)}=\frac{2.2012}{2.\frac{2012}{2013}}=\frac{2012}{\frac{2012}{2013}}=\frac{2012.2013}{2012}=2013\)
