Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 19

$\Sigma_{n=0}^\infty 2^n x^{n+2}$

Since, we know $\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n=1+x+x^2+x^3+x^4+...$ Now, the given equation can be written as: $\dfrac{1}{2-x}=\dfrac{1}{2} [\dfrac{1}{1-\dfrac{x}{2}}]$ or, $\dfrac{1}{1-2x}=1+(2x)+(2x)^2+(2x)^3+(2x)^4+...$ or, $\dfrac{x^2}{1-2x}=x^2[1+(2x)+(2x)^2+(2x)^3+(2x)^4+...]$ or, $=\Sigma_{n=0}^\infty 2^n x^{n+2}$