Provide an appropriate response.
-Derive the series for $\frac { 1 } { 1 - x }$ for $x 1$ by first writing $\frac { 1 } { 1 - x } = \frac { 1 } { x } \frac { 1 } { 1 - 1 / x }$.
A) $\sum _ { n = 0 } ^ { \infty } \frac { 1 } { x ^ { n + 1 } }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { 1 } { x ^ { n } }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { x ^ { n } }$
D) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { x ^ { n + 1 } }$

Question

Provide an appropriate response.
-Derive the series for $\frac { 1 } { 1 - x }$ for $x > 1$ by first writing $\frac { 1 } { 1 - x } = \frac { 1 } { x } \frac { 1 } { 1 - 1 / x }$.
A) $\sum _ { n = 0 } ^ { \infty } \frac { 1 } { x ^ { n + 1 } }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { 1 } { x ^ { n } }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { x ^ { n } }$
D) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { x ^ { n + 1 } }$

Quizplus · Accepted Answer

The given expression $\frac { 1 } { 1 - x } = \frac { 1 } { x } \frac { 1 } { 1 - 1 / x }$ can be expanded using the geometric series formula for $|1/x| < 1$, which is valid for $x > 1$. The series expansion is $\sum _ { n = 0 } ^ { \infty } \frac { 1 } { x ^ { n + 1 } }$.

Provide an Appropriate Response 11−x\frac { 1 } { 1 - x }1−x1​

Provide an Appropriate Response $\frac { 1 } { 1 - x }$