$\text { Find a geometric power series for the function } \frac { 1 } { 6 + x } \text { centered at } 0 \text {. }$
A)
$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { 6 ^ { n + 1 } }$
B)
$\sum _ { n = 0 } ^ { \infty } \frac { - x ^ { n } } { 6 ^ { n + 1 } }$
C)
$\sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { 6 ^ { n + 1 } }$
D)
$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { 6 ^ { n } } + 1$
E)
$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { 6 ^ { n } + 1 }$

Question

$\text { Find a geometric power series for the function } \frac { 1 } { 6 + x } \text { centered at } 0 \text {. }$ 
A) 
$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { 6 ^ { n + 1 } }$
B)
$\sum _ { n = 0 } ^ { \infty } \frac { - x ^ { n } } { 6 ^ { n + 1 } }$
C) 
$\sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { 6 ^ { n + 1 } }$
D)
$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { 6 ^ { n } } + 1$
E) 
$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { 6 ^ { n } + 1 }$

Quizplus · Accepted Answer

The function $\frac{1}{6+x}$ can be rewritten as $\frac{1}{6(1+\frac{x}{6})}$, which resembles the form of a geometric series $\frac{1}{1-r}$ with $r = -\frac{x}{6}$. The series expansion is $\sum_{n=0}^{\infty}(-1)^n\left(\frac{x}{6}\right)^n$, simplifying to $\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{6^{n+1}}$.

Find a geometric power series for the function 16+x centered at 0. \text { Find a geometric power series for the function } \frac { 1 } { 6 + x } \text { centered at } 0 \text {. } Find a geometric power series for the function 6+x1​ centered at 0.

$\text { Find a geometric power series for the function } \frac { 1 } { 6 + x } \text { centered at } 0 \text {. }$