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The given function \( f(x) \) is defined as a series expansion:

\[
f(x) = \frac{1}{1 + 2x} = \sum_{n=0}^{\infty} (-1)^n 2^n x^n
\]

You are asked to differentiate this series term-by-term to find the series for \( f'(x) \).

The general term of the series is:

\[
(-1)^n 2^n x^n
\]

To differentiate term-by-term with respect to \( x \), we apply the power rule for derivatives:

\[
\frac{d}{dx}(x^n) = n x^{n-1}
\]

So, differentiating each term:

\[
f'(x) = \sum_{n=1}^{\infty} (-1)^n 2^n n x^{n-1}
\]

This is the series expansion for \( f'(x) \).

Let me know if you'd like more details or have questions! Here are some additional questions you might consider:

1. How do we verify that term-by-term differentiation is valid for this series?
2. Can we represent \( f'(x) \) in a closed form?
3. What is the radius of convergence of the original series?
4. How does this series behave near \( x = 0 \)?
5. Can you perform a similar expansion for more complicated functions like \( e^x \) or \( \sin(x) \)?

**Tip:** When working with power series, make sure the series converges within the region where you're differentiating or integrating it!

Differentiate the given series expansion of f term-by-term to obtain the corresponding series expansion for the derivative of f. If f(x) = 1 / (1 + 2x) = Σ (-1)^n 2^n x^n, find f'(x).

Learn how to differentiate a power series term-by-term to find the derivative of f(x) = 1 / (1 + 2x). This example covers key calculus concepts like power series expansion, differentiation rules, and the radius of convergence.

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