Chapter 11 - 11.3 - Geometric Sequences and Series - 11.3 Exercises - Page 796: 55

$\displaystyle \sum_{n=1}^{7}4^{n-1}=5461$

$\displaystyle \sum_{n=1}^{7}4^{n-1}=4^0+4^1+...+4^6$ $a_1=4^0=1$ $r=\frac{a_2}{a_1}=\frac{4^1}{4^0}=4$ $S_7=a_1(\frac{1-r^n}{1-r})=1(\frac{1-4^7}{1-4})=\frac{-16383}{-3}=5461$