Part 1: Power Series Representation of $f(2x) = \frac{1}{1 - 2x}$

The given geometric series formula is: $\frac{1}{1 - x} = \sum_{k=0}^\infty x^k \quad \text{for } |x| < 1.$

To find the series representation for $f(2x) = \frac{1}{1 - 2x}$, substitute $2x$ for $x$ in the formula: $f(2x) = \frac{1}{1 - 2x} = \sum_{k=0}^\infty (2x)^k = \sum_{k=0}^\infty 2^k x^k.$

So, the correct answer is:

$\boxed{\text{B. } \sum_{k=0}^\infty (2x)^k.}$

Part 2: Interval of Convergence

The geometric series $\sum_{k=0}^\infty x^k$ converges if $|x| < 1$. For $f(2x)$, replace $x$ with $2x$ in the convergence condition: $|2x| < 1 \implies |x| < \frac{1}{2}.$

Thus, the interval of convergence is: $\boxed{(-\frac{1}{2}, \frac{1}{2})}.$

Final Answers:

1. Power Series Representation: $\sum_{k=0}^\infty (2x)^k$ (Answer: B)
2. Interval of Convergence: $(-\frac{1}{2}, \frac{1}{2})$

Would you like more details on the derivation or convergence tests? Here are some related questions:

1. How does substitution affect the interval of convergence of a geometric series?
2. What is the significance of centering a power series at 0?
3. How would the series change if $f(x) = \frac{1}{1 + 2x}$?
4. How do you test for convergence of power series using the ratio test?
5. What happens to the radius of convergence if you scale $x$ by a constant?

Tip: When substituting $cx$ into a geometric series, always adjust the interval of convergence accordingly!