Find the Maclaurin series for the given function.
-$$\frac { 1 } { 3 + x }$$
A) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { 3 ^ { n + 1 } }$
B) $\sum _ { \mathrm { n } = 0 } ^ { \infty } \frac { ( - 1 ) ^ { \mathrm { n } } \mathrm { x } ^ { \mathrm { n } } } { 3 ^ { \mathrm { n } + 1 } }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n + 1 } } { 3 ^ { n } }$
D) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { n } } { 3 ^ { n } }$

Question

Quizplus · Accepted Answer

The Maclaurin series for $\frac{1}{3+x}$ can be derived by recognizing it as a geometric series with a ratio of $-\frac{x}{3}$, starting from $n=0$. The correct series representation is $\sum_{n=0}^{\infty} \frac{(-1)^n x^n}{3^{n+1}}$, which matches choice B.

Find the Maclaurin Series for the Given Function 13+x\frac { 1 } { 3 + x }3+x1​

Find the Maclaurin Series for the Given Function $\frac { 1 } { 3 + x }$