Question

2015+2×(1+2+3.....+2014)​

Answer 1

 

$\displaystyle\bf\\2015+2\times(1+2+3+...+2014)=\\\\=2015+2\left[\frac{2014(2014+1)}{2} \right]=\\\\=2015+2014(2014+1)=\\\\=2015+2014\times2015=\\\\=2015(1+2014)=2015\times2015=\boxed{\bf2015^2}$