Chapter 11 - 11.3 - Geometric Sequences and Series - 11.3 Exercises - Page 796: 57

$\displaystyle \sum_{n=1}^{6}(-7)^{n-1}=-14706$

$\displaystyle \sum_{n=1}^{6}(-7)^{n-1}=(-7)^0+(-7)^1+...+(-7)^5$ $a_1=(-7)^0=1$ $r=\frac{a_2}{a_1}=\frac{(-7)^1}{(-7)^0}=-7$ $S_{6}=a_1(\frac{1-r^n}{1-r})=1(\frac{1-(-7)^{6}}{1-(-7)})=\frac{1-117649}{8}=-14706$