Question

Tính \(E=\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2012}}\)

 

Answer 1

Xét mẫu:

\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+2012}\)

= \(1+\frac{1}{3}+\frac{1}{6}+....+\frac{1}{2025078}\)

= \(1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)\)

= \(1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2012}-\frac{1}{2013}\right)\)

= \(1+2.\left(\frac{1}{2013}\right)\)

= \(\frac{4024}{2013}\)

=> E =  \(\frac{2.2012}{\frac{4024}{2013}}\)

=> E = \(4024.\frac{2013}{4024}\)

=> E = 2013