$\text { Find the sum of the convergent series } 8 - 1 + \frac { 1 } { 8 } - \frac { 1 } { 64 } + \cdots \cdots$
A) $\frac { 64 } { 9 }$
B) $\frac { 64 } { 65 }$
C) $\frac { 512 } { 9 }$
D) $\frac { 65 } { 9 }$
E) $\frac { 9 } { 64 }$

Question

$\text { Find the sum of the convergent series } 8 - 1 + \frac { 1 } { 8 } - \frac { 1 } { 64 } + \cdots \cdots$ 
A) $\frac { 64 } { 9 }$
B) $\frac { 64 } { 65 }$
C) $\frac { 512 } { 9 }$
D) $\frac { 65 } { 9 }$
E) $\frac { 9 } { 64 }$

Quizplus · Accepted Answer

This series is a geometric series with the first term $a = 8$ and common ratio $r = -\frac{1}{8}$. The sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$, where $|r| < 1$. Substituting $a = 8$ and $r = -\frac{1}{8}$, we get $S = \frac{8}{1 - (-\frac{1}{8})} = \frac{8}{1 + \frac{1}{8}} = \frac{8}{\frac{9}{8}} = \frac{64}{9}$.

Find the sum of the convergent series 8−1+18−164+⋯⋯\text { Find the sum of the convergent series } 8 - 1 + \frac { 1 } { 8 } - \frac { 1 } { 64 } + \cdots \cdots Find the sum of the convergent series 8−1+81​−641​+⋯⋯

$\text { Find the sum of the convergent series } 8 - 1 + \frac { 1 } { 8 } - \frac { 1 } { 64 } + \cdots \cdots$