Question

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2011}\left(1+2+3+...+2011\right)\)

Answer 1

\(\frac{1+2+...+n}{n}=\frac{n\left(n+1\right)}{2n}=\frac{n+1}{2}\)

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

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

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

\(=\frac{1}{2}.\frac{2012.2013}{2}-\frac{1}{2}=503.2013-\frac{1}{2}=...\)