Answer
$\Sigma_{n=0}^\infty (-1)^n \dfrac{2^{(n+1)} x^{(n+2)}}{(n+1)}$
Work Step by Step
Consider the Maclaurin Series for $e^x$ as follows:
$ \ln (1+x)=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^{n+1}}{(n+1)!}=x-\dfrac{x^2}{2!}+\dfrac{x^3}{3!}-\dfrac{x^4}{4!}+...$
Then,we have $ x \cdot \ln (1+2x)=x \cdot [2x-\dfrac{(2x)^2}{2!}+\dfrac{(2x)^3}{3!}-\dfrac{(2x)^4}{4!}+...]$
$\implies 2x^2-\dfrac{2^2}{2}x^3+\dfrac{2^3}{3}x^{4}-=\Sigma_{n=0}^\infty (-1)^n \dfrac{2^{(n+1)} x^{(n+2)}}{(n+1)}$
