We know that $\displaystyle-\frac{2}{1+4x^2}=-2+8x^2-32x^4+128x^6+...$ for $x\in\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$. **Using this fact, find the function that corresponds to the following series.** $\displaystyle -2x+\frac{{8}}{3}x^3-\frac{32}{5}x^5+\frac{128}{7}x^7+...$ [[☃ radio 1]]