$\text { Find a power series for the function } \frac { 1 } { 12 - x } \text { centered at } 1 \text {. }$
A) $\sum _ { n = 0 } ^ { \infty } \frac { ( x - 1 ) ^ { n } } { 11 ^ { n } }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( x - 1 ) ^ { n } } { 11 ^ { n + 1 } }$
C)$\sum _ { n = 0 } ^ { \infty } \frac { ( x - 1 ) ^ { n } } { 11 ^ { n + 1 } }$
D)$\sum _ { n = 0 } ^ { \infty } \frac { ( x - 1 ) ^ { n } } { 11 ^ { n } + 1 }$
E)$\sum _ { n = 0 } ^ { \infty } \frac { ( x - 1 ) ^ { n } } { 11 ^ { n } } + 1$

Question

$\text { Find a power series for the function } \frac { 1 } { 12 - x } \text { centered at } 1 \text {. }$ 
A) $\sum _ { n = 0 } ^ { \infty } \frac { ( x - 1 ) ^ { n } } { 11 ^ { n } }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( x - 1 ) ^ { n } } { 11 ^ { n + 1 } }$
C)$\sum _ { n = 0 } ^ { \infty } \frac { ( x - 1 ) ^ { n } } { 11 ^ { n + 1 } }$
D)$\sum _ { n = 0 } ^ { \infty } \frac { ( x - 1 ) ^ { n } } { 11 ^ { n } + 1 }$
E)$\sum _ { n = 0 } ^ { \infty } \frac { ( x - 1 ) ^ { n } } { 11 ^ { n } } + 1$

Quizplus · Accepted Answer

We can use the formula $\frac{1}{1-x}= \sum_{n = 0}^{\infty} x^{n}$, with $-1<x<1$, to obtain 
$$\frac{1}{12-x}=\frac{1}{11-(x-1)}=\sum_{n=0}^{\infty}\frac{(x-1)^n}{11^{n+1}}.$$Therefore, the power series for $\frac{1}{12-x}$ centered at $1$ is $\boxed{	extbf{(C)}}$.$$\sum _ { n = 0 } ^ { \infty } \frac { ( x - 1 ) ^ { n } } { 11 ^ { n + 1 } }.$$

Find a power series for the function 112−x centered at 1. \text { Find a power series for the function } \frac { 1 } { 12 - x } \text { centered at } 1 \text {. } Find a power series for the function 12−x1​ centered at 1.

$\text { Find a power series for the function } \frac { 1 } { 12 - x } \text { centered at } 1 \text {. }$