[tex]\displaystyle c). 1+3+5+...+2011+2013 \\ 2013=1+(n-1) \cdot 2 \\ 2013=1+2n-2 \\ 2n=2013-1+2 \\ 2n=2014 \\ n=2014:2 \\ n=1007 \\ S_{1007}= \frac{2+1006 \cdot 2}{2} \cdot 1007 \\ \\ S_{1007}= \frac{2+2012}{2} \cdot 1007 \\ \\ S_{1007}= \frac{2014}{2} \cdot 1007 \\ \\ S_{1007}=1007 \cdot 1007 \\ S_{1007}=1014049[/tex]
[tex]\displaystyle d). 1+6+11+...+2006+2011 \\ 2011=1+(n-1) \cdot 5 \\ 2011=1+5n-5 \\ 5n=2011-1+5 \\ 5n=2015 \\ n=2015:5 \\ n=403 \\ S_{403}= \frac{2+402 \cdot 5}{2} \cdot 403 \\ \\ S_{403}= \frac{2+2010}{2} \cdot 403 \\ \\ S_{403}= \frac{2012}{2} \cdot 403 \\ \\ S_{403}= 1006 \cdot 403 \\ S_{403}=405418[/tex]
[tex]\displaystyle e).1+4+7+...+2011+2014 \\ 2014=1+(n-1) \cdot 3 \\ 2014=1+3n-3 \\ 3n=2014-1+3 \\ 3n=2016 \\ n=2016:3 \\ n=672 \\ S_{672}= \frac{2+671 \cdot 3}{2} \cdot 672 \\ \\ S_{672}= \frac{2+2013}{2} \cdot 672 \\ \\ S_{672}= \frac{2015}{2} \cdot 672 \\ \\ S_{672}=2015 \cdot 336 \\ S_{672}=677040[/tex]
[tex]\displaystyle f).4+9+14+...+2009+2014 \\ 2014=4+(n-1) \cdot 5 \\ 2014=4+5n-5 \\ 5n=2014-4+5 \\ 5n=2015 \\ n=2015:5 \\ n=403 \\ S_{403}= \frac{8 +402 \cdot 5}{2} \cdot 403 \\ \\ S_{403}= \frac{8+2010}{2} \cdot 403 \\ \\ S_{403}= \frac{2018}{2} \cdot 403 \\ \\ S_{403}=1009 \cdot 403 \\ S_{403}=406627[/tex]
