Given n = 4k. Prove that $\sum_{i=0}^{2k}\binom{n}{2i}(-1)^i=2^{2k}(-1)^k$. 
Given $n = 4k$, where $k$ is a non-negative integer. Prove that $\sum_{i=0}^{2k}\binom{n}{2i}(-1)^i=2^{2k}(-1)^k$.

I have no idea how to prove this equation.
 A: Consider the imaginary number $i$ (with $i^2=-1$). Since $(1+i)^2=2i$ and $(1+i)^4=-4$, the binomial Newton formula gives
$$
\begin{array}{lcl}
(-4)^k &=& (1+i)^{4k} \\
&=& (1+i)^n \\
&=& \sum_{j=0}^n \binom{n}{j} i^j \\
&=&\sum_{t=0}^{2k} \binom{n}{2t} i^{2t}+
\sum_{t=0}^{2k-1} \binom{n}{2t+1} i^{2t+1} \\
&=& \sum_{t=0}^{2k} \binom{n}{2t} (-1)^{t}+
\bigg(\sum_{t=0}^{2k-1} \binom{n}{2t+1} (-1)^{t}\bigg)i
\end{array}
$$
Your identity follows by taking the real part on each side.
A: Let $i=\sqrt{-1}$, and note that $(1+i)^4=(1-i)^4=-4$.  Also, for $j \ge 0$, we have
$$\frac{1+(-1)^j}{2} = \begin{cases}1 &\text{if $2|j$}\\0 &\text{otherwise}\end{cases}.$$
Then
\begin{align}
\sum_{j\ge 0}\binom{4k}{2j}(-1)^j
&= \sum_{j\ge 0}\binom{4k}{2j}i^{2j} \\
&= \sum_{j\ge 0}\frac{1+(-1)^j}{2}\binom{4k}{j}i^{j} \\
&= \frac{1}{2}\sum_{j\ge 0}\binom{4k}{j}i^{j} + \frac{1}{2}\sum_{j\ge 0}\binom{4k}{j}(-i)^{j} \\
&= \frac{1}{2}(1+i)^{4k} + \frac{1}{2}(1-i)^{4k} \\
&= \frac{1}{2}(-4)^k + \frac{1}{2}(-4)^k \\
&= (-4)^k \\
&= 2^{2k}(-1)^k
\end{align}
