Question

2017/1+2    +   2017/1+2+3      + 2017/ 1+2+3 +4 .......     2017/1+2+3.....+2016

Answer 1

\(=\dfrac{2017}{3}+\dfrac{2017}{6}+...+\dfrac{2017}{2031120}\)

\(=2017\left(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{2031120}\right)\)

\(=2007\cdot2\cdot\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{2016\cdot2015}\right)\)

\(=4014\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)

\(=4014\cdot\dfrac{1007}{2016}=\dfrac{224561}{112}\)