$\text { Find a power series for the function } \frac { 9 } { 9 + x ^ { 2 } } \text { centered at } 0 \text {. }$
A) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n + 1 } } { 9 ^ { n } }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { 9 ^ { n } }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n } } { 9 ^ { n } }$
D)$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n + 1 } } { 9 ^ { n } }$
E) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n } } { 9 ^ { n + 1 } }$

Question

$\text { Find a power series for the function } \frac { 9 } { 9 + x ^ { 2 } } \text { centered at } 0 \text {. }$ 
A) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n + 1 } } { 9 ^ { n } }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { 9 ^ { n } }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n } } { 9 ^ { n } }$
D)$\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n + 1 } } { 9 ^ { n } }$
E) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n } } { 9 ^ { n + 1 } }$

Quizplus · Accepted Answer

We can use the geometric series to expand $\frac{1}{1-\frac{-x^2}{9}}$ into a power series. Then, we can simplify to get the coefficient for $x^{2n}$ as $\frac{(-1)^n}{9^n}$. Therefore, the power series for $\frac{9}{9+x^2}$ is $\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{9^n}$. This matches choice C.

Find a power series for the function 99+x2 centered at 0. \text { Find a power series for the function } \frac { 9 } { 9 + x ^ { 2 } } \text { centered at } 0 \text {. } Find a power series for the function 9+x29​ centered at 0.

$\text { Find a power series for the function } \frac { 9 } { 9 + x ^ { 2 } } \text { centered at } 0 \text {. }$