[tex] 1+4+7+...+31= \sum\limits_{k=1}^{11}(3k-2) = \sum\limits_{k=1}^{11}3k-\sum\limits_{k=1}^{11}2 = \\ \\ = 3\cdot \sum\limits_{k=1}^{11}k-2\cdot 11 = 3\cdot \dfrac{11\cdot (11+1)}{2}-2\cdot 11 = \\ \\ = 3\cdot\dfrac{11\cdot 12}{2}-2\cdot 11 = 3\cdot 11\cdot 6-2\cdot 11 = \\ \\ = 11\cdot (3\cdot 6-2) = 11\cdot 16 = 176[/tex]
