<s>[INST] 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.
 [/INST] 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.
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