1. **Problem statement:** Find the values of $x \in \mathbb{R}$ for which the series $$\sum_{k=2}^\infty \left(\frac{x+3}{2}\right)^k$$ converges, and find the sum of the series for those values. 2. **Formula and rules:** This is a geometric series with common ratio $$r = \frac{x+3}{2}$$ starting from $k=2$. A geometric series $$\sum_{k=0}^\infty r^k$$ converges if and only if $$|r| < 1$$. The sum of a geometric series starting at $k=0$ is $$\frac{1}{1-r}$$. 3. **Determine convergence:** We require $$\left|\frac{x+3}{2}\right| < 1$$. Multiply both sides by 2: $$|x+3| < 2$$ This inequality means: $$-2 < x+3 < 2$$ Subtract 3 from all parts: $$-5 < x < -1$$ So the series converges for $$x \in (-5, -1)$$. 4. **Find the sum:** The series starts at $k=2$, so write it as: $$\sum_{k=2}^\infty r^k = \sum_{k=0}^\infty r^k - r^0 - r^1 = \frac{1}{1-r} - 1 - r$$ Substitute $$r = \frac{x+3}{2}$$: $$S = \frac{1}{1 - \frac{x+3}{2}} - 1 - \frac{x+3}{2}$$ Simplify the denominator: $$1 - \frac{x+3}{2} = \frac{2 - (x+3)}{2} = \frac{-x -1}{2}$$ So: $$S = \frac{1}{\frac{-x -1}{2}} - 1 - \frac{x+3}{2} = \frac{2}{-x -1} - 1 - \frac{x+3}{2}$$ Rewrite: $$S = \frac{2}{-x -1} - 1 - \frac{x+3}{2}$$ This is the sum for $$x \in (-5, -1)$$. **Final answers:** - The series converges for $$x \in (-5, -1)$$. - The sum of the series is $$S = \frac{2}{-x -1} - 1 - \frac{x+3}{2}$$ for those values of $x$.