#### Answer

$${S_7} = 4372$$

#### Work Step by Step

$$\eqalign{
& \sum\limits_{i = 0}^6 {4{{\left( 3 \right)}^i}} \cr
& {S_n}{\text{ we write using summation notation as}}:{\text{ }}\left( {{\text{see page 612}}} \right) \cr
& {S_n} = \sum\limits_{i = 0}^{n - 1} {a{r^i}} \cr
& {\text{comparing the given sumation }}\sum\limits_{i = 0}^6 {4{{\left( 3 \right)}^i}} {\text{ with }}\sum\limits_{i = 0}^{n - 1} {a{r^i}} {\text{ we obtain }} \cr
& a = 4{\text{ and }}r = 3 \cr
& {\text{the summation is from }}i = 0{\text{ to }}n - 1 = 6,{\text{ so }}n = 7 \cr
& {\text{using the formula }}{S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},{\text{ }}r \ne 1{\text{ gives}} \cr
& {S_7} = \frac{{4\left( {{3^7} - 1} \right)}}{{3 - 1}} \cr
& {\text{simplifying}} \cr
& {S_7} = 4372 \cr} $$