Find $\displaystyle \sum_{k=0}^{49}(-1)^k\binom{99}{2k}$, where $\binom{n}{j}=\frac{n!}{j!(n-j)!}$.

$ \textbf{(A)}\ -2^{50} \qquad\textbf{(B)}\ -2^{49} \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 2^{49} \qquad\textbf{(E)}\ 2^{50} $
To solve the problem, we need to find the sum \(\sum_{k=0}^{49}(-1)^k \binom{99}{2k}\). We can use complex numbers and De Moivre's theorem to simplify the computation.

1. **Define the function:**
   Let \( f(x) = (x + i)^{99} \). Expanding this using the binomial theorem, we get:
   \[
   f(x) = \sum_{k=0}^{99} \binom{99}{k} x^{99-k} i^k
   \]
   We are interested in the real part of \( f(1) \), which is \( (1 + i)^{99} \).

2. **Express \( 1 + i \) in polar form:**
   \[
   1 + i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)
   \]
   Using De Moivre's theorem, we can write:
   \[
   (1 + i)^{99} = \left( \sqrt{2} \right)^{99} \left( \cos \frac{99\pi}{4} + i \sin \frac{99\pi}{4} \right)
   \]

3. **Simplify the expression:**
   \[
   \left( \sqrt{2} \right)^{99} = 2^{49.5}
   \]
   Next, we need to find \(\cos \frac{99\pi}{4}\) and \(\sin \frac{99\pi}{4}\). Note that:
   \[
   \frac{99\pi}{4} = 24\pi + \frac{3\pi}{4}
   \]
   Since \(\cos\) and \(\sin\) are periodic with period \(2\pi\), we have:
   \[
   \cos \frac{99\pi}{4} = \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}
   \]
   \[
   \sin \frac{99\pi}{4} = \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2}
   \]

4. **Compute the real part:**
   \[
   (1 + i)^{99} = 2^{49.5} \left( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right)
   \]
   The real part of this expression is:
   \[
   2^{49.5} \cdot -\frac{\sqrt{2}}{2} = -2^{49.5} \cdot \frac{\sqrt{2}}{2} = -2^{49}
   \]

Thus, the sum \(\sum_{k=0}^{49}(-1)^k \binom{99}{2k}\) is \(-2^{49}\).

The final answer is \(\boxed{-2^{49}}\)