Alternating sum of binomial coefficients $\sum_{k=0}^{49}(-1)^k \binom{99}{2k} = -2^{49}$

Question

How can you prove that $$\sum_{k=0}^{49} \binom{99}{2k}(-1)^k = -2^{49}?$$

A more general formula seems to be $\sum_{k=0}^{n} \binom{2n + 1}{2k}(-1)^k$

For $n = 0:$ it equals $2^0$ For $n = 1: -2^1$ For $n = 2: -2^2$ For $n = 3: 2^3$ For $n = 4: 2^4$ For $n = 5: -2^5$ For $n = 6: -2^6$ For $n = 7: 2^7$ etc.

Answer 1

How to write $(-1)^k = x^{2k}$, to match the $2k$ in the binomial coefficient? $x=i$ or $-i$.

Here only the even $2k$ terms appear; how to introduce the odd terms? By averaging the conjugates: $\frac12 \left[(1+i)^{99}+(1-i)^{99}\right] = \Re\left[(1+i)^{99}\right] = \ldots$

Combining these two ideas,

$$\begin{align*} \sum_{k=0}^{49}\binom{99}{2k}(-1)^k &= \sum_{k=0}^{49}\binom{99}{2k}i^{2k}\\ &= \Re\left[\sum_{h=0}^{99}\binom{99}h i^h\right] &&(h=2k)\\ &= \Re\left[(1+i)^{99}\right]\\ &= \left(\sqrt2\right)^{96}\Re\left[(1+i)^3\right]\\ &= \left(\sqrt2\right)^{96}\left(1-3\right)\\ &= -2^{49} \end{align*}$$

Answer 2

Use binomial theorem as $$F(x)=\frac{(1+x)^{2n+1}+(1-x)^{2n+1}}{2}=\sum_{k=0}^{n} {2n+1 \choose 2k} x^{2k}$$ Then take$x=i$, to get $$\sum_{k=0}^{n} {2n+1 \choose 2k} (-1)^{2k}=F(i)=\frac{(1+i)^{2n+1}+(1-i)^{2n+1}}{2}=2^{n+1/2}\cos[(2n+1)\pi/4]n$$ where we have used $(1\pm i)=\sqrt{2}[\cos(\pi/4)\pm i\sin(\pi/4)]$ and demoivre's theorem.