[tex]\displaystyle S = \sum\limits_{i=1}^{2013}\Bigg(\Big(1+\dfrac{1}{i}\Big)\sum\limits_{k=1}^{i}k!(k^2+1)\Bigg) \\ \\\\ \sum\limits_{k=1}^{i}k!(k^2+1)= \sum\limits_{k=1}^{i}k!\Big(k(k+1)-(k-1)\Big) =\\ \\ = \sum\limits_{k=1}^{i}\Big((k+1)!k-k!(k-1)\Big) =\\ \\=2!\cdot 1+3!\cdot 2+...+i!\cdot(i-1)+(i+1)!\cdot i -\\ - 1!\cdot 0 - 2!\cdot1 - 3!\cdot 2-...-i!\cdot(i-1) =\\ \\ =(i+1)!\cdot i[/tex]
[tex]\displaystyle S = \sum\limits_{i=1}^{2013}\Bigg(\Big(1+\dfrac{1}{i}\Big)\Big((1+i)!\cdot i\Bigg) \\ \\S = \sum\limits_{i=1}^{2013}\Bigg(\dfrac{i+1}{i}\cdot \Big((1+i)!\cdot i\Big)\Bigg) \\ \\ S =\sum\limits_{i=1}^{2013}(1+i)(1+i)! \\ \\ S =\sum\limits_{i=1}^{2013}(i+1)!(i+1)\\ \\ S = \sum\limits_{i=1}^{2013}(i+1)!(i+2-1)\\ \\ S = \sum\limits_{i=1}^{2013}\Big[(i+2)! - (i+1)!\Big] \\ \\ S = 3!+4!+...+2014!+2015! - 2!-3!-...-2014! \\ \\\Rightarrow \boxed{ S = 2015! - 2}[/tex]
⇒ d) corect
