Solve the problem.
-Using the Maclaurin series for $\tan ^ { - 1 } x$, obtain a series for $\frac { \tan ^ { - 1 } x ^ { 2 } } { x ^ { 2 } }$.
A) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 4 n } } { 2 n + 2 }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n } } { 2 n + 1 }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n } } { 2 n + 2 }$
D) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 4 n } } { 2 n + 1 }$

Question

Quizplus · Accepted Answer

The Maclaurin series for $\tan^{-1}x$ is $\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}$. Substituting $x^2$ for $x$ and dividing by $x^2$ gives $\sum_{n=0}^{\infty}\frac{(-1)^n x^{4n}}{2n+1}$, which matches option D.

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