We know that
$\displaystyle-\frac{2}{1-2x}=-2-4x-8{{x}^{2}}-16{{x}^{3}}-...-2{(2x)}^{n-1}+...$
for $x\in\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$.
**Using this fact, find the function that corresponds to the following series.**
$\displaystyle -2x-2x^2-\frac{{8}}{3}x^3-4x^4-...$
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