The problem asks for the series expansion of $x e^{-2x}$.

Step-by-Step Solution:

1. Recall the series expansion for $e^x$:
   The exponential function $e^x$ has the Maclaurin series expansion: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$

2. Substitute $-2x$ in place of $x$ in the expansion of $e^x$: $e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!}$ Simplifying this: $e^{-2x} = 1 - 2x + \frac{(2x)^2}{2!} - \frac{(2x)^3}{3!} + \dots$ $= 1 - 2x + 2x^2 - \frac{4x^3}{3} + \dots$

3. Multiply the expansion by $x$:
   Now, multiply the entire series by $x$: $x e^{-2x} = x \left( 1 - 2x + 2x^2 - \frac{4x^3}{3} + \dots \right)$ Expanding this: $x e^{-2x} = x - 2x^2 + 2x^3 - \frac{4x^4}{3} + \dots$

Final Result:

The series expansion of $x e^{-2x}$ is: $x e^{-2x} = x - 2x^2 + 2x^3 - \frac{4x^4}{3} + \dots$

Would you like further details or have any questions?

Here are 5 related questions to deepen your understanding:

1. What is the series expansion of $e^{3x}$?
2. How do you derive the Maclaurin series for a general exponential function?
3. What is the radius of convergence of the series for $e^{-2x}$?
4. How would you apply the series expansion to approximate $x e^{-2x}$ for small values of $x$?
5. How does the series expansion of $e^{-2x}$ change if you replace $-2x$ with $-kx$ for some constant $k$?

Tip: When manipulating series expansions, ensure that you multiply or divide term by term to maintain accuracy.