$\text { Find a power series for the function } \frac { 1 } { 1 - 9 x } \text { centered at } 0 \text {. }$
A) $\sum _ { n = 0 } ^ { \infty } \left( \frac { x } { 9 } \right) ^ { n }$
B) $\sum _ { n = 0 } ^ { \infty } ( - 9 x ) ^ { n }$
C) $\sum _ { n = 0 } ^ { \infty } \left( \frac { 9 } { x } \right) ^ { n }$
D)$\sum _ { n = 0 } ^ { \infty } ( 9 x ) ^ { n }$
E) $\sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } ( 9 x ) ^ { n }$

Question

$\text { Find a power series for the function } \frac { 1 } { 1 - 9 x } \text { centered at } 0 \text {. }$ 
A) $\sum _ { n = 0 } ^ { \infty } \left( \frac { x } { 9 } \right) ^ { n }$
B) $\sum _ { n = 0 } ^ { \infty } ( - 9 x ) ^ { n }$
C) $\sum _ { n = 0 } ^ { \infty } \left( \frac { 9 } { x } \right) ^ { n }$
D)$\sum _ { n = 0 } ^ { \infty } ( 9 x ) ^ { n }$
E) $\sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } ( 9 x ) ^ { n }$

Quizplus · Accepted Answer

We want to find a power series centered at 0 for $\frac{1}{1-9x}$. Recall that the formula for a geometric series is $\frac{1}{1-r}=\sum_{n=0}^\infty r^n$ when $|r|<1$. Thus, we can rewrite $\frac{1}{1-9x}$ as $\frac{1}{1-(9x)}=\sum_{n=0}^\infty (9x)^n$. Therefore, the power series we seek is $\boxed{\sum _ { n = 0 } ^ { \infty } ( 9 x ) ^ { n }}$.

Find a power series for the function 11−9x centered at 0. \text { Find a power series for the function } \frac { 1 } { 1 - 9 x } \text { centered at } 0 \text {. } Find a power series for the function 1−9x1​ centered at 0.

$\text { Find a power series for the function } \frac { 1 } { 1 - 9 x } \text { centered at } 0 \text {. }$