[tex]\displaystyle 3+6+9+...+99 -(1+4+7+...+97) = \\ \\ =\sum\limits_{k=1}^{33}(3k)-\sum\limits_{k=1}^{33}(3k-2) = \sum\limits_{k=1}^{33}\Big[3k-(3k-2)\Big] = \\ \\ = \sum\limits_{k=1}^{33}(3k-3k+2)= \sum\limits_{k=1}^{33}2 = 2\cdot 33 = 66[/tex]
