Answer:
To find the series representation of the function \(f(x) = \frac{(1-x)^2}{x^3}\) using the derivatives of \(\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n\) for \(|x| < 1\), we can make use of the properties of power series and the rules for differentiation. We'll start by finding the derivative of \(\frac{1}{1-x}\) and then use it to find the derivative of \(f(x)\).
First, we know that:
\[
\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n \quad \text{for } |x| < 1.
\]
Taking the derivative of both sides with respect to \(x\):
\[
\frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{d}{dx}\left(\sum_{n=0}^{\infty}x^n\right).
\]
Let's find the derivative of the left-hand side first:
\[
\frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{d}{dx}\left((1-x)^{-1}\right) = (1-x)^{-2} \cdot (-1) = -\frac{1}{(1-x)^2}.
\]
Now, find the derivative of the right-hand side:
\[
\frac{d}{dx}\left(\sum_{n=0}^{\infty}x^n\right) = \sum_{n=0}^{\infty}\frac{d}{dx}(x^n).
\]
The derivative of \(x^n\) with respect to \(x\) is \(nx^{n-1}\), so:
\[
\frac{d}{dx}(x^n) = nx^{n-1}.
\]
Therefore, the right-hand side becomes:
\[
\sum_{n=0}^{\infty}nx^{n-1}.
\]
Now, we equate the two derivatives:
\[
-\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty}nx^{n-1}.
\]
Next, we can rewrite the series on the right-hand side with an index shift, so that it starts at \(n = 0\):
\[
-\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty}nx^{n-1} = \sum_{n=0}^{\infty}(n+1)x^n.
\]
Now, let's multiply both sides of this equation by \((1-x)^2\):
\[
-\frac{1}{(1-x)^2}(1-x)^2 = \sum_{n=0}^{\infty}(n+1)x^n(1-x)^2.
\]
Simplify:
\[
-1 = \sum_{n=0}^{\infty}(n+1)x^n(1-x)^2.
\]
Finally, we can use this result to find the series representation of \(f(x) = \frac{(1-x)^2}{x^3}\). We'll multiply both sides by \(\frac{(1-x)^2}{x^3}\):
\[
f(x) = \frac{(1-x)^2}{x^3} = -\frac{1}{x^3}\sum_{n=0}^{\infty}(n+1)x^n(1-x)^2.
\]
So, the series representation of \(f(x)\) is:
\[
f(x) = -\sum_{n=0}^{\infty}(n+1)x^{n-3}(1-x)^2 \quad \text{for } |x| < 1.
\]
This is the desired series representation of \(f(x)\).
Explanation: