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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?

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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.

What is the series expansion for x * e^(-2x)?

Learn how to find the series expansion of x * e^(-2x) using the Maclaurin series. This detailed solution walks through each step, from recalling the exponential function series to expanding and multiplying. Suitable for advanced high school or undergraduate students.

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