Find the sum of the geometric series for those x for which the series converges.
-$$\sum _ { n = 0 } ^ { \infty } \left( \frac { x - 9 } { 6 } \right) ^ { n }$$
A) $\frac { 6 } { 15 + x }$
B) $\frac { 9 } { 15 - x }$
C) $\frac { 9 } { 15 + x }$
D) $\frac { 6 } { 15 - x }$

Question

Quizplus · Accepted Answer

The series is geometric with first term 1 and common ratio $\frac{x-9}{6}$. It will converge if $|common\ ratio|<1$. Thus, we have 
\begin{align*}
\left|\frac{x-9}{6}ight|&<1 \
\Rightarrow -1<\frac{x-9}{6}&<1 \
\Rightarrow -6<x-9&<6 \
\Rightarrow 3<x&<15
\end{align*}
Therefore, the series converges for $3<x<15$. The formula for the sum of an infinite geometric series with first term a and common ratio r (assuming $|r|<1$) is $\frac{a}{1-r}$. Thus, in this case, the sum of the series is $\frac{1}{1-\frac{x-9}{6}}=\frac{6}{15-x}$. Therefore, the answer is (D).

Find the Sum of the Geometric Series for Those X ∑n=0∞(x−96)n\sum _ { n = 0 } ^ { \infty } \left( \frac { x - 9 } { 6 } \right) ^ { n }n=0∑∞​(6x−9​)n

Find the Sum of the Geometric Series for Those X $\sum _ { n = 0 } ^ { \infty } \left( \frac { x - 9 } { 6 } \right) ^ { n }$