Chapter 11 - 11.3 - Geometric Sequences and Series - 11.3 Exercises - Page 796: 62

$\displaystyle \sum_{j=0}^{6}500(1.04)^n=1579.66$

$\displaystyle \sum_{j=0}^{6}500(1.04)^n=500(1.04)^0+500(1.04)^1+500(1.04)^2+...500(1.04)^6$ There are 7 terms is this sequence. $a_1=500(1.04)^0=500$ $r=\frac{a_2}{a_1}=\frac{500(1.04)^1}{500(1.04)^0}=1.04$ $S_n=a_1(\frac{1-r^n}{1-r})$ $S_{7}=500(\frac{1-(1.04)^{7}}{1-1.04})=200(\frac{1.04^7-1}{0.04})=1579.66$