Find the sum of the series.
-$$\sum _ { n = 0 } ^ { \infty } \left( \frac { 1 } { 7 ^ { n } } - \frac { 1 } { 9 ^ { n } } \right)$$
A) $- \frac { 1 } { 3 }$
B) $- \frac { 1 } { 24 }$
C) $\frac { 1 } { 3 }$
D) $\frac { 1 } { 24 }$

Question

Quizplus · Accepted Answer

This series can be split into two separate geometric series: $\sum _ { n = 0 } ^ { \infty } \frac { 1 } { 7 ^ { n } }$ and $-\sum _ { n = 0 } ^ { \infty } \frac { 1 } { 9 ^ { n } }$. The sum of an infinite geometric series is given by $\frac{a}{1 - r}$, where $a$ is the first term and $r$ is the common ratio. For the first series, $a = 1$ and $r = \frac{1}{7}$, giving a sum of $\frac{1}{1 - \frac{1}{7}} = \frac{7}{6}$. For the second series, $a = 1$ and $r = \frac{1}{9}$, giving a sum of $\frac{1}{1 - \frac{1}{9}} = \frac{9}{8}$. Subtracting the second sum from the first gives $\frac{7}{6} - \frac{9}{8} = \frac{56 - 54}{48} = \frac{2}{48} = \frac{1}{24}$.

Find the Sum of the Series ∑n=0∞(17n−19n)\sum _ { n = 0 } ^ { \infty } \left( \frac { 1 } { 7 ^ { n } } - \frac { 1 } { 9 ^ { n } } \right)n=0∑∞​(7n1​−9n1​)

Find the Sum of the Series $\sum _ { n = 0 } ^ { \infty } \left( \frac { 1 } { 7 ^ { n } } - \frac { 1 } { 9 ^ { n } } \right)$