Use a Maclaurin series in this table to obtain the Maclaurin series for the given function: $f(x) = x^2 \ln(1 + x^3)$

$\displaystyle\sum_{n=1}^{\infty} = \boxed{(\ ? \ )}$

Neetesh Kumar | December 21, 2024

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This is the solution to Math 1c
Assignment: 11.10 Question Number 16
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Step-by-step solution:

Step 1: Recall the Maclaurin Series for $\ln(1+x)$:

The Maclaurin series for $\ln(1+x)$ is given by:

$\ln(1+x) = \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} x^n$

This series converges for $|x| < 1$.

Step 2: Substitute $x^3$ for $x$ in the Series:

To find the Maclaurin series for $f(x) = x^2 \ln(1+x^3)$, we substitute $x^3$ for $x$ in the series for $\ln(1+x)$:

$\ln(1+x^3) = \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} (x^3)^n = \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} x^{3n}$

Step 3: Multiply by $x^2$:

Multiplying each term by $x^2$ gives the Maclaurin series for $f(x)$:

$x^2 \ln(1+x^3) = x^2 \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} x^{3n} = \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} x^{3n+2}$

Conclusion:

The Maclaurin series for the function $f(x) = x^2 \ln(1 + x^3)$ is:

$f(x) = \displaystyle\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} x^{3n+2}$

This series represents the function for all $x$ such that $x^{3n+2}$ converges, which is true for $|x| < 1$.

Final Answer:

$\displaystyle\sum_{n=1}^\infty \boxed{\frac{(-1)^{n-1}}{n} x^{3n+2}}$

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