Question

\(\frac{2\cdot2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}}\)

Answer 1

\(A=\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}}\)

\(A=\frac{4024}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{2012.2013:2}}\)

\(A=\frac{4024}{1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}}\)

\(A=\frac{4024}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)}\)

\(A=\frac{4024}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)

\(A=\frac{4024}{1+2\left(\frac{1}{2}-\frac{1}{2013}\right)}\)

\(A=\frac{4024}{1+1-\frac{2}{2013}}=\frac{4024}{2-\frac{2}{2013}}=4024:\frac{4024}{2013}=\frac{4024.2013}{4024}=2013\)