Problem 1172

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 $\boxed{-2^{49}}$