The sum $$\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\cdots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}$$ is a root of the quadratic $x^2+x+c$. What is $c$ in terms of $n$?

$\text{(A) }-\frac{n}{2}\qquad\text{(B) }2n\qquad\text{(C) }-2n\qquad\text{(D) }n+\frac{1}{2}\qquad\text{(E) }n-2$
1. We start with the given sum:
   \[
   \sum_{k=1}^{n} \frac{1}{\sqrt{2k-1} + \sqrt{2k+1}}
   \]

2. To simplify each term, we rationalize the denominator:
   \[
   \frac{1}{\sqrt{2k-1} + \sqrt{2k+1}} \cdot \frac{\sqrt{2k+1} - \sqrt{2k-1}}{\sqrt{2k+1} - \sqrt{2k-1}} = \frac{\sqrt{2k+1} - \sqrt{2k-1}}{(\sqrt{2k-1} + \sqrt{2k+1})(\sqrt{2k+1} - \sqrt{2k-1})}
   \]

3. Simplify the denominator using the difference of squares:
   \[
   (\sqrt{2k+1})^2 - (\sqrt{2k-1})^2 = (2k+1) - (2k-1) = 2
   \]

4. Thus, each term becomes:
   \[
   \frac{\sqrt{2k+1} - \sqrt{2k-1}}{2}
   \]

5. The sum of the series is:
   \[
   \sum_{k=1}^{n} \frac{\sqrt{2k+1} - \sqrt{2k-1}}{2}
   \]

6. This is a telescoping series, where most terms cancel out:
   \[
   \frac{1}{2} \left( (\sqrt{3} - \sqrt{1}) + (\sqrt{5} - \sqrt{3}) + (\sqrt{7} - \sqrt{5}) + \cdots + (\sqrt{2n+1} - \sqrt{2n-1}) \right)
   \]

7. After cancellation, we are left with:
   \[
   \frac{1}{2} (\sqrt{2n+1} - 1)
   \]

8. The sum is:
   \[
   \frac{\sqrt{2n+1} - 1}{2}
   \]

9. We are given that this sum is a root of the quadratic equation \(x^2 + x + c = 0\). The roots of this quadratic equation are:
   \[
   x = \frac{\sqrt{2n+1} - 1}{2} \quad \text{and} \quad x = -\frac{\sqrt{2n+1} + 1}{2}
   \]

10. Using Vieta's formulas, the product of the roots of the quadratic equation \(x^2 + x + c = 0\) is \(c\). Therefore:
    \[
    c = \left( \frac{\sqrt{2n+1} - 1}{2} \right) \left( -\frac{\sqrt{2n+1} + 1}{2} \right)
    \]

11. Simplify the product:
    \[
    c = \frac{(\sqrt{2n+1} - 1)(-\sqrt{2n+1} - 1)}{4}
    \]

12. Expand the numerator:
    \[
    (\sqrt{2n+1} - 1)(-\sqrt{2n+1} - 1) = -(\sqrt{2n+1})^2 - \sqrt{2n+1} + \sqrt{2n+1} + 1 = -(2n+1) + 1 = -2n
    \]

13. Thus:
    \[
    c = \frac{-2n}{4} = -\frac{n}{2}
    \]

The final answer is \(\boxed{-\frac{n}{2}}\)