1. **Problem:** Consider the power series $$\sum_{k=0}^{\infty}\left(-\frac{1}{2}\right)^k (x-3)^k.$$ Find the radius of convergence and closed form of the function it represents. 2. **Formula and rules:** This is a geometric series of the form $$\sum_{k=0}^\infty r^k = \frac{1}{1-r}$$ for $$|r|<1$$. 3. **Identify the ratio:** Here, $$r = -\frac{1}{2}(x-3)$$. 4. **Radius of convergence:** We require $$|r| < 1$$, so $$\left| -\frac{1}{2}(x-3) \right| < 1 \implies \frac{|x-3|}{2} < 1 \implies |x-3| < 2.$$ Thus, the radius of convergence is $$2$$. 5. **Closed form of the series:** Using the geometric series formula, $$\sum_{k=0}^\infty \left(-\frac{1}{2}\right)^k (x-3)^k = \frac{1}{1 - \left(-\frac{1}{2}(x-3)\right)} = \frac{1}{1 + \frac{x-3}{2}} = \frac{1}{\frac{2 + x - 3}{2}} = \frac{2}{x - 1}.$$ 6. **Differentiated series:** Differentiate term-by-term: $$\frac{d}{dx} \sum_{k=0}^\infty \left(-\frac{1}{2}\right)^k (x-3)^k = \sum_{k=1}^\infty k \left(-\frac{1}{2}\right)^k (x-3)^{k-1}.$$ 7. **Closed form of derivative:** Differentiate the closed form: $$\frac{d}{dx} \frac{2}{x-1} = -\frac{2}{(x-1)^2}.$$ So the differentiated series sums to $$-\frac{2}{(x-1)^2}$$ for $$|x-3|<2$$. 8. **Integrated series:** Integrate term-by-term: $$\int \sum_{k=0}^\infty \left(-\frac{1}{2}\right)^k (x-3)^k dx = C + \sum_{k=0}^\infty \left(-\frac{1}{2}\right)^k \frac{(x-3)^{k+1}}{k+1}.$$ 9. **Closed form of integral:** Integrate the closed form: $$\int \frac{2}{x-1} dx = 2 \ln|x-1| + C.$$ --- **Final answers:** - Radius of convergence: $$2$$. - Original series sum: $$\frac{2}{x-1}$$ for $$|x-3|<2$$. - Differentiated series sum: $$-\frac{2}{(x-1)^2}$$ for $$|x-3|<2$$. - Integrated series sum: $$2 \ln|x-1| + C$$ for $$|x-3|<2$$.