1. We are given a series with terms from $k=2n$ to $k=n^2$, and there are 9 terms in this series. 2. Recall that the number of terms in a series from $k=a$ to $k=b$ is given by $b - a + 1$. 3. Applying this to our series, the number of terms is: $$n^2 - 2n + 1$$ 4. Since the series has 9 terms, set this expression equal to 9: $$n^2 - 2n + 1 = 9$$ 5. Simplify the equation: $$n^2 - 2n + 1 - 9 = 0 \implies n^2 - 2n - 8 = 0$$ 6. Solve the quadratic equation $n^2 - 2n - 8 = 0$. Using the quadratic formula: $$n = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}$$ 7. The solutions are: $$n = \frac{2 + 6}{2} = 4 \quad \text{or} \quad n = \frac{2 - 6}{2} = -2$$ 8. Since $n$ represents an index (likely positive integer), we discard the negative solution. **Final answer:** $$\boxed{4}$$