1. **State the problem:** We want to 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 then find the sum of the series for those values. 2. **Recall the 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 $$|r| < 1$$, and its sum is $$\frac{1}{1-r}$$. 3. **Apply convergence condition:** For our series starting at $k=2$, the convergence condition is still $$|r| < 1$$, so $$\left|\frac{x+3}{2}\right| < 1$$ 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 of the series:** The series starts at $k=2$, so $$\sum_{k=2}^\infty r^k = \sum_{k=0}^\infty r^k - r^0 - r^1 = \frac{1}{1-r} - 1 - r$$ 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: $$S = \frac{2}{-x -1} - 1 - \frac{x+3}{2}$$ Find a common denominator 2 for the last two terms: $$S = \frac{2}{-x -1} - \frac{2}{2} - \frac{x+3}{2} = \frac{2}{-x -1} - \frac{2 + x + 3}{2} = \frac{2}{-x -1} - \frac{x + 5}{2}$$ Thus, $$\boxed{\text{The series converges for } x \in (-5, -1) \text{ and its sum is } S = \frac{2}{-x -1} - \frac{x + 5}{2}}$$