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

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