Question

2/(1/1+2+1/1+2+3+...+1/1+2+3+...+2015) = ?

Answer 1

\(\frac{2}{\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+....+2015}}\)

\(=2:\left[2.\left(\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2015.2016}\right)\right]\)

\(=2:\left[2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\right)\right]\)

\(=2:\left[2.\left(\frac{1}{2}-\frac{1}{2016}\right)\right]=2:\left[\frac{2.1007}{2016}\right]=\frac{2}{\frac{1007}{2008}}=\frac{2016}{1007}\)