Which of the following is the power series centered at $a = 0$ for $f ( x ) = \frac { 1 } { x ^ { 2 } + 4 }$ ?
1) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 4 ^ { n + 1 } } x ^ { n }$ 2) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 4 ^ { n + 1 } } x ^ { 2 n }$ 3) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 4 ^ { n } } x ^ { 2 n }$
A)None
B)1
C)2
D)3
E)1, 2
F)1, 3
G)2, 3
H)1, 2, 3

Question

Which of the following is the power series centered at $a = 0$ for $f ( x ) = \frac { 1 } { x ^ { 2 } + 4 }$ ?
1) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 4 ^ { n + 1 } } x ^ { n }$ 2) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 4 ^ { n + 1 } } x ^ { 2 n }$ 3) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 4 ^ { n } } x ^ { 2 n }$ 
A)None
B)1
C)2
D)3
E)1, 2
F)1, 3
G)2, 3
H)1, 2, 3

Quizplus · Accepted Answer

The correct power series for $f(x) = \frac{1}{x^2 + 4}$ can be derived by recognizing it as a geometric series with the form $\frac{1}{1 - r}$, where $r = -\frac{x^2}{4}$. The general formula for a geometric series is $\sum_{n=0}^{\infty} r^n$, which in this case translates to $\sum_{n=0}^{\infty} \left(-\frac{x^2}{4}\right)^n$, simplifying to $\sum_{n=0}^{\infty} \frac{(-1)^n}{4^n} x^{2n}$, matching choice 2.

Which of the Following Is the Power Series Centered At a =