Find a geometric power series for the function centered at 0, (i) by the technique shown in Examples 1 and 2 and (ii) by long division. $$f ( x ) = \frac { 10 } { 4 - x }$$
A)
$\sum _ { n = 0 } ^ { \infty } \frac { 5 } { 2 } \left( \frac { x } { 4 } \right) ^ { n } , | x |

Question

Find a geometric power series for the function centered at 0, (i) by the technique shown in Examples 1 and 2 and (ii) by long division. $$f ( x ) = \frac { 10 } { 4 - x }$$
A) 
$\sum _ { n = 0 } ^ { \infty } \frac { 5 } { 2 } \left( \frac { x } { 4 } \right) ^ { n } , | x | < 4$
B)
$\sum _ { n = 0 } ^ { \infty } \frac { 1 } { 4 } \left( \frac { x } { 4 } \right) ^ { n } , | x | < 4$
C)
$\sum _ { n = 0 } ^ { \infty } 10 \left( - \frac { x } { 4 } \right) ^ { n } , | x | < 4$
D)
$\sum _ { n = 0 } ^ { \infty } \frac { 5 } { 2 } ( - 4 x ) ^ { n } , | x | < 4$
E)
$\sum _ { n = 0 } ^ { \infty } \frac { 5 } { 2 } ( - x ) ^ { n } , | x | < 1$

Quizplus · Accepted Answer

The function $\frac{10}{4-x}$ can be rewritten as $\frac{10}{4(1-\frac{x}{4})}$, which resembles the form of a geometric series $\frac{a}{1-r}$ where $a = \frac{10}{4} = \frac{5}{2}$ and $r = \frac{x}{4}$. Therefore, its geometric power series is $\sum_{n=0}^{\infty} \frac{5}{2} \left(\frac{x}{4}\right)^n$, valid for $|x| < 4$ to ensure convergence.

Find a Geometric Power Series for the Function Centered at 0