Find a geometric power series for the function $\frac { 1 } { 7 - x } \text { centered at } 0 \text {. }$
A)
$\sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { 7 ^ { n } + 1 }$
B)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { x } { 7 } \right) ^ { n } + 1$
C)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { x } { 7 } \right) ^ { n }$
D)
$\sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { 7 ^ { n + 1 } }$
E)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { x } { 7 } \right) ^ { n + 1 }$

Question

Find a geometric power series for the function $\frac { 1 } { 7 - x } \text { centered at } 0 \text {. }$ 
A) 
$\sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { 7 ^ { n } + 1 }$
B)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { x } { 7 } \right) ^ { n } + 1$
C)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { x } { 7 } \right) ^ { n }$
D)
$\sum _ { n = 0 } ^ { \infty } \frac { x ^ { n } } { 7 ^ { n + 1 } }$
E)
$\sum _ { n = 0 } ^ { \infty } \left( \frac { x } { 7 } \right) ^ { n + 1 }$

Quizplus · Accepted Answer

A geometric power series has the form $\sum_{n=0}^{\infty} ar^n$ where a is the constant term and r is the common ratio.
Notice that $\frac{1}{7-x} = \frac{1}{7(1-\frac{x}{7})}$. So the constant term is 1/7 and the common ratio is $\frac{x}{7}$. Therefore, the geometric power series is $\sum_{n=0}^{\infty} \frac{x^n}{7^{n+1}} = \frac{1}{7} + \frac{x}{7^2} + \frac{x^2}{7^3} + \cdots$ which is choice D.

Find a Geometric Power Series for the Function 17−x centered at 0. \frac { 1 } { 7 - x } \text { centered at } 0 \text {. }7−x1​ centered at 0.

Find a Geometric Power Series for the Function $\frac { 1 } { 7 - x } \text { centered at } 0 \text {. }$