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?
$$f(x) =x^2\ln(1+2x),\;\vert x\vert\lt\frac{1}{2}$$


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}
