Question

use substitution to find the taylor series at x = 0 of the function \\( \ln \left(1 + 3 x^{4}\
ight) \\).
what is the general expression for the nth term in the taylor series at x = 0 for \\( \ln \left(1 + 3 x^{4}\
ight) \\)?
\\( \sum _ { n = 1 } ^ { \infty } \square \\)

Explanation:

Step1: Recall the Taylor series of \(\ln(1 + t)\)

The Taylor series of \(\ln(1 + t)\) is \(\sum_{n = 1}^{\infty}\frac{(- 1)^{n + 1}}{n}t^{n}\), for \(|t|\lt1\).

Step2: Substitute \(t = 3x^{4}\)

Substitute \(t = 3x^{4}\) into the series \(\sum_{n = 1}^{\infty}\frac{(- 1)^{n + 1}}{n}t^{n}\).
We get \(\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}(3x^{4})^{n}\).
Using the power - of - a - product rule \((ab)^{m}=a^{m}b^{m}\), where \(a = 3\), \(b=x^{4}\), and \(m = n\), we have \(\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}\cdot3^{n}}{n}x^{4n}\).

Answer:

\(\sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}\cdot3^{n}}{n}x^{4n}\)