(Suppose $n =4k$ for some positive integer $k$ )
$\sum_\limits{i=0}^{2k}\binom{n}{2i}{(-1)}^i = 2^{2k}(-1)^k$
$LHS = \binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdot\cdot\cdot+\binom{n}{n}$
To prove above identity I had derived below identity:
$\binom{n}{0}+\binom{n}{2}+\cdot\cdot\cdot\binom{n}{n} = 2^{n-1}$
However, still missing how to configure each $(4k+2)$th term to be $(-1)$ coefficient.
any advice would be appreciate.
\begin{eqnarray*} (1+i)^{4k}&=& \sum_{j=0}^{4k} \binom{4k}{i} i^{j} \\ (1-i)^{4k}&=& \sum_{j=0}^{4k} \binom{4k}{i} (-i)^{j} \\ \end{eqnarray*} Add these together & use $ (1+i)^2=2i$.
We can write, $$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\cdots+\binom{n}{n}x^n$$$$$$
Now put $x=i$
$$(1+i)^n=\binom{n}{0}+\binom{n}{1}i+\binom{n}{2}i^2+\binom{n}{3}i^3+\cdots+\binom{n}{n}i^n$$$$$$
Dividing and multiplying by $\sqrt{2}$ in LHS
$$(\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}))^n=\binom{n}{0}+\binom{n}{1}i+\binom{n}{2}i^2+\binom{n}{3}i^3+\cdots+\binom{n}{n}i^n$$$$$$
Using Euler's identity $e^{i\theta}=\cos\theta+i\sin\theta$
$$\sqrt{2}^n(e^{i\frac{π}{4}})^n=\binom{n}{0}+\binom{n}{1}i+\binom{n}{2}i^2+\binom{n}{3}i^3+\cdots+\binom{n}{n}i^n$$$$$$ $$\sqrt{2}^ne^{in\frac{π}{4}}=\binom{n}{0}+\binom{n}{1}i+\binom{n}{2}i^2+\binom{n}{3}i^3+\cdots+\binom{n}{n}i^n$$$$$$
Using Euler's identity again
$$\sqrt{2}^n\cos(\frac{nπ}{4})+i\sqrt{2}^n\sin(\frac{nπ}{4})=(\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\cdots)+i(\binom{n}{1}-\binom{n}{3}+\binom{n}{5}-\cdots)$$$$$$ Equating the real parts, $$\sqrt{2}^n\cos\frac{nπ}{4}=\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\cdots$$$$$$ Now substitute $n=4k$ and get the answer using, $$\binom{4k}{2k-\gamma}=\binom{4k}{2k+\gamma}$$
