Use substitution to find the Taylor series at x = 0 of the given function.
-$\ln \left( 1 + x ^ { 2 } \right)$
A) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } x ^ { 2 n } } { ( 2 n ) ! }$
B) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } x ^ { 2 n } } { n }$
C) $\sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n - 1 } x ^ { 2 n }$
D) $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n - 1 } x ^ { 2 n } } { n }$

Question

Quizplus · Accepted Answer

We know that 
$$
\ln(1+x) = \sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n}
$$
Thus, substituting $x^2$ for $x$ gives
$$
\ln(1+x^2) = \sum_{n=1}^\infty\frac{(-1)^{n-1}(x^2)^n}{n}=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^{2n}}{n}
$$
This matches the form of option D.

Use Substitution to Find the Taylor Series at X = ln⁡(1+x2)\ln \left( 1 + x ^ { 2 } \right)ln(1+x2)

Use Substitution to Find the Taylor Series at X = $\ln \left( 1 + x ^ { 2 } \right)$