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!