Find Maclaurin series for $x^2\ln(1+2x)$ We recently started learning about Maclaurin series. I feel confused a little, can anyone help me with this?
$$f(x) =x^2\ln(1+2x),\;\vert x\vert\lt\frac{1}{2}$$

 A: First start with the Mclaurin series for $\ln(1+x)$ which is: $$\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$$
Now replace the "$x$" with "$2x$" which gives: $$\ln(1+2x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(2x)^n}{n}$$
Finally, multiply the series by $x^2$ to obtain: $$x^2 \ln(1+2x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{2^n(x)^{n+2}}{n}$$
A: Hint. Find the series for the logarithm. That is, you may obtain the series for $$-\log(1-2x)$$ first. This is the same as integrating the geometrical series whose sum is $$\frac{1}{1-2x}.$$ This will converge for $|2x|<1,$ or in other words for $|x|<1/2.$ To find the value of the constant of integration, substitute a well known value for the series. For example, you may take $x=0.$ Finally, you have the series for $-\log(1-2x).$ To get your own series, replace $x$ by $-x$ everywhere in your equation. Then multiply every term by $-x^2,$ and then you'd be done!
A: One way to assess your problem is to it the following way by using the following equation: \begin{equation}f(x)=f(0)+f'(0)\frac{x}{1!}+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+\dots\end{equation}
So when assessing the problems like one has to realize that $x^2$ is a polynomial form, and $\ln(1+2x)$. So the natural log is what one is going to take the Maclaurin these that are Maclaurin series one can do the following by creating a table of derivatives, and values: \begin{array}{|c|c|}\hline f(x)=\ln(1+2x)& f(0)=\ln(1)=0\\ \hline f'(x)=\frac{2}{1+2x} &f'(0)=2 \\ \hline f''(x)=\frac{-4}{(1+2x)^2} & f''(0)= -4\\ \hline f'''(x) =\frac{16}{(1+2x)^3} & f'''(0)=16 \\ \hline f''''(x)=\frac{-96}{(1+2x)^4} & f''''(0)=-96 \end{array}
\begin{equation}\ln(1+2x)=0+2x-\frac{4}{2!}x^2+\frac{16}{3!}x^3-\frac{96}{4!}x^4+\dots \end{equation}
This equation can then be reduced to:
\begin{equation}\ln(1+2x)=2x-\frac{(2x)^2}{2}+\frac{(2x)^3}{3}-\frac{(2x)^4}{4}+\dots \end{equation}
So one can then write this:
\begin{equation}\sum_\limits{n=0}^\infty (-1)^n\frac{(2x)^n}{n}\end{equation}
All you have to now is multiply it by $x^2$, and get the following:
\begin{equation}\sum_\limits{n=0}^\infty (-1)^n\frac{2^nx^{n+2}}{n}\end{equation}
