We are tasked with using the fact that the geometric series representation for:

$\frac{1}{1-x} = \sum_{n=0}^\infty x^n \quad \text{for } |x| < 1$

to derive a power series for:

$\frac{1}{(1-2x)^2}.$

We will solve this step-by-step by term-by-term differentiation.

Step 1: Start with the geometric series

From the geometric series formula:

$\frac{1}{1-2x} = \sum_{n=0}^\infty (2x)^n \quad \text{for } |2x| < 1 \Rightarrow |x| < \frac{1}{2}.$

Simplifying, we get:

$\frac{1}{1-2x} = \sum_{n=0}^\infty 2^n x^n.$

Step 2: Differentiate both sides with respect to $x$

Differentiating $\frac{1}{1-2x}$ with respect to $x$, we get:

$\frac{d}{dx} \left( \frac{1}{1-2x} \right) = \frac{2}{(1-2x)^2}.$

On the other hand, differentiating the series term-by-term:

$\frac{d}{dx} \left( \sum_{n=0}^\infty 2^n x^n \right) = \sum_{n=1}^\infty n \cdot 2^n x^{n-1}.$

Thus:

$\frac{2}{(1-2x)^2} = \sum_{n=1}^\infty n \cdot 2^n x^{n-1}.$

Step 3: Solve for $\frac{1}{(1-2x)^2}$

Divide through by 2:

$\frac{1}{(1-2x)^2} = \frac{1}{2} \sum_{n=1}^\infty n \cdot 2^n x^{n-1}.$

Simplifying the powers of 2, we rewrite the series as:

$\frac{1}{(1-2x)^2} = \sum_{n=1}^\infty n \cdot 2^{n-1} x^{n-1}.$

Final Answer:

The power series for $\frac{1}{(1-2x)^2}$ is:

$\frac{1}{(1-2x)^2} = \sum_{n=1}^\infty n \cdot 2^{n-1} x^{n-1}.$

Do you need further clarification or details on any of the steps?

Related Questions:

1. How would you derive the power series for $\frac{1}{(1-ax)^2}$ for general $a$?
2. What happens when you integrate this power series term-by-term?
3. How can the power series for $\frac{1}{1-2x}$ be used to find a series for $\frac{1}{(1-2x)^3}$?
4. What is the radius of convergence for the series $\frac{1}{(1-2x)^2}$?
5. Can you generalize this method for functions of the form $\frac{1}{(1-ax)^k}$?

Tip: Term-by-term differentiation of power series is valid within the radius of convergence. Always check the convergence before differentiating or integrating.