## College Algebra (6th Edition)

The Sum = $\frac{364}{2187}$
Put i = 1 to 6 in $(\frac{1}{3})^{i + 1}$. The Sum = $(\frac{1}{3})^{1 + 1}$ + $(\frac{1}{3})^{2 + 1}$ + $(\frac{1}{3})^{3 + 1}$ + $(\frac{1}{3})^{4 + 1}$ + $(\frac{1}{3})^{5 + 1}$ + $(\frac{1}{3})^{6 + 1}$ = $(\frac{1}{3})^{ 2}$ + $(\frac{1}{3})^{3}$ + $(\frac{1}{3})^{4}$ + $(\frac{1}{3})^{5}$ + $(\frac{1}{3})^{6}$ + $(\frac{1}{3})^{7}$ = $\frac{1}{9}$ + $\frac{1}{27}$ + $\frac{1}{81}$ + $\frac{1}{243}$ + $\frac{1}{729}$ + $\frac{1}{2187}$ = $\frac{(243 + 81 + 27 + 9 + 3 +1)}{2187}$ = $\frac{364}{2187}$