[tex]\displaystyle \sum\limits_{i=1}^{2013}\Big(\sum\limits_{j=1}^i j\Big) = \sum\limits_{i=1}^{2013}\Big(1+2+3+...+i\Big) = \\ \\ = \sum\limits_{i=1}^{2013}\Big[\dfrac{i(i+1)}{2}\Big] = \dfrac{1}{2}\sum\limits_{i=1}^{2013}(i^2+i) = \dfrac{1}{2}\sum\limits_{i=1}^{2013} i^2 +\dfrac{1}{2}\sum\limits_{i=1}^{2013} i =\\ \\ \\=\dfrac{1}{2}\cdot \dfrac{2013(2013+1)(2\cdot 2013+1)}{6}+\dfrac{1}{2}\cdot \dfrac{2013(2013+1)}{2} =[/tex]
[tex]=\dfrac{2013(2013+1)(2\cdot 2013+1)}{12}+\dfrac{2013(2013+1)}{4} =\\ \\ = \dfrac{2013(2013+1)}{4}\Big(\dfrac{2\cdot 2013+1}{3}+1\Big) = \\ \\ = \dfrac{2013\cdot 2014\cdot 4030}{12}[/tex]
