How to find the MacLaurin series of $x^3 \ln(1+2x^2)$

Question

I was doing some MacLaurin Series problems and came across this one:

$x^3\text{ln}(1+2x^2)$

What I first did was find the MacLaurin Series for $\ln(1+z)$ where z = $2x^2$:

$\sum_{n=0}^\infty \frac{(-1)^n(1+z)^n}{n}$

Then I plugged in my $2x^2$ for z and got this:

$\sum_{n=0}^\infty \frac{(-1)^n(1+2x^2)^n}{n}$

From there I am stuck as I don't know how to do this:

$x^3*\sum_{n=0}^\infty \frac{(-1)^n(1+2x^2)^n}{n}$

Any help is appreciated thanks.

Answer 1

Hint: use correct MacLaurin series for $\log$ function: $$\log(1+x)=\sum_{n\ge1}\frac{(\color{red}{-x})^n}n.$$

Answer 2

$$\log(1+z)=z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+......+(-1)^{n+1}\frac{z^n}{n}+...$$

So $$x^3\log(1+2x^2)=x^3-2x^7+\frac{8}{3}x^9-4x^{11}+...+(-1)^{n+1} 2^n \frac{x^{2n+3}}{n} +...$$