Find the sum of the infinite geometric sequence.? $$\sum _ { n = 1 } ^ { \infty } 8 - \left( \frac { 1 } { 6 } \right) ^ { n - 1 }$$
A)?s = $\frac { 5 } { 34 }$
B)?s = $\frac { 34 } { 5 }$
C)?s = $\frac { 42 } { 5 }$
D)?s = $\frac { 46 } { 5 }$
E)?s = $\frac { 1 } { 5 }$

Question

Quizplus · Accepted Answer

The given series can be rewritten as $\sum _ { n = 1 } ^ { \infty } 8 - \sum _ { n = 1 } ^ { \infty } \left( \frac { 1 } { 6 } \right) ^ { n - 1 }$. The first sum is not a geometric series and does not converge. However, the second sum is a geometric series with the first term $a = 1$ (since $\left(\frac{1}{6}\right)^{0} = 1$) and common ratio $r = \frac{1}{6}$. The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$ when $|r| < 1$. Applying this formula, we get $S = \frac{1}{1 - \frac{1}{6}} = \frac{1}{\frac{5}{6}} = \frac{6}{5}$. However, since the first part of the series does not converge, the initial interpretation of the series was incorrect. The correct interpretation should consider the entire expression as part of the geometric sequence, which was not properly analyzed in the initial explanation. The correct approach involves recognizing the entire expression as part of a geometric sequence or correcting the misinterpretation of the series' components.

Find the Sum of the Infinite Geometric Sequence ∑n=1∞8−(16)n−1\sum _ { n = 1 } ^ { \infty } 8 - \left( \frac { 1 } { 6 } \right) ^ { n - 1 }n=1∑∞​8−(61​)n−1

Find the Sum of the Infinite Geometric Sequence $\sum _ { n = 1 } ^ { \infty } 8 - \left( \frac { 1 } { 6 } \right) ^ { n - 1 }$