Let's solve the problem step by step based on the uploaded image.

(a) Use differentiation to find a power series representation for $f(x) = \frac{1}{(1 + x)^2}$

1. Start with the geometric series: The geometric series for $\frac{1}{1+x}$ is: $\frac{1}{1+x} = \sum_{n=0}^\infty (-1)^n x^n, \quad |x| < 1.$

2. Differentiate both sides: To get $\frac{1}{(1+x)^2}$, differentiate the geometric series term by term with respect to $x$: $\frac{d}{dx} \left( \frac{1}{1+x} \right) = -\frac{1}{(1+x)^2}.$ Differentiating the series: $\frac{d}{dx} \left( \sum_{n=0}^\infty (-1)^n x^n \right) = \sum_{n=1}^\infty n (-1)^n x^{n-1}.$ Therefore: $\frac{1}{(1+x)^2} = \sum_{n=1}^\infty n (-1)^n x^{n-1}.$ To simplify the index, let $m = n-1$, so $n = m+1$. Substituting: $\frac{1}{(1+x)^2} = \sum_{m=0}^\infty (m+1)(-1)^{m+1} x^m.$

3. Final series representation: $\frac{1}{(1+x)^2} = \sum_{n=0}^\infty (n+1)(-1)^{n+1} x^n, \quad |x| < 1.$

Radius of Convergence:

The radius of convergence remains the same as the original geometric series, $|x| < 1$. Therefore: $\text{Radius of Convergence} = 1.$

(b) Use part (a) to find a power series for $f(x) = \frac{1}{(1+x)^3}$

1. Differentiate $\frac{1}{(1+x)^2}$: Differentiating $\frac{1}{(1+x)^2}$ gives: $\frac{d}{dx} \left( \frac{1}{(1+x)^2} \right) = -\frac{2}{(1+x)^3}.$ Using the series representation from part (a): $\frac{d}{dx} \left( \sum_{n=0}^\infty (n+1)(-1)^{n+1} x^n \right) = -\frac{2}{(1+x)^3}.$ Differentiating term by term: $\sum_{n=1}^\infty n(n+1)(-1)^{n+1} x^{n-1}.$

2. Rewrite the series: Factor out $-\frac{1}{2}$ to match $\frac{1}{(1+x)^3}$: $\frac{1}{(1+x)^3} = \sum_{n=1}^\infty \frac{n(n+1)}{2} (-1)^n x^{n-1}.$

(c) Use part (b) to find a power series for $f(x) = \frac{x^2}{(1+x)^3}$

1. Multiply the series in (b) by $x^2$: From part (b), the series for $\frac{1}{(1+x)^3}$ is: $\frac{1}{(1+x)^3} = \sum_{n=1}^\infty \frac{n(n+1)}{2} (-1)^n x^{n-1}.$ Multiply each term by $x^2$: $\frac{x^2}{(1+x)^3} = \sum_{n=1}^\infty \frac{n(n+1)}{2} (-1)^n x^{n+1}.$

2. Reindex the series (if needed): Let $m = n+1$, so $n = m-1$. Adjusting the series: $\frac{x^2}{(1+x)^3} = \sum_{m=2}^\infty \frac{(m-1)m}{2} (-1)^m x^m.$

3. Final power series: $\frac{x^2}{(1+x)^3} = \sum_{n=2}^\infty \frac{(n-1)n}{2} (-1)^n x^n.$

Let me know if you need more details!

Here are 5 questions to expand:

1. Can you rederive part (a) using integration instead of differentiation?
2. What happens to the series representation if the center changes from $x = 0$ to $x = c$?
3. How would the radius of convergence change if $x^2/(1+x)^3$ were modified to $x^3/(1+x)^4$?
4. Can you find the nth term formula for the series representation of $f(x) = \frac{1}{(1+x)^k}$?
5. What are the real-world applications of power series expansions in physics or engineering?

Tip: When working with power series, always verify convergence at the endpoints separately!