Alternating binomial sum with intervals of two Fix integer $n\geq 1$. Consider the number $$1-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdots$$where the sum continues as long as the lower number in the binomial is $\leq n$. Is there a way to simplify this sum? 
The first few values are $1, 0, -2, -4, -4$.

Note that
$$(1+x)^n = \dbinom{n}0 + x \dbinom{n}1 + x^2 \dbinom{n}2 + x^3 \dbinom{n}3 + \cdots + x^n \dbinom{n}n$$
Hence,
$$(1+i)^n = \dbinom{n}0 + i \dbinom{n}1 + i^2 \dbinom{n}2 + i^3 \dbinom{n}3 + \cdots + i^n \dbinom{n}n$$
where $i^2=-1$.
Hence,
$$\text{Real part of }(1+i)^n = 1 - \dbinom{n}2 + \dbinom{n}4 - \dbinom{n}6 \pm \cdots$$
Also note that
$$(1+i)^n = 2^{n/2} e^{i n \pi/4} = 2^{n/2} \left(\cos(n \pi/4) + i \sin(n \pi/4)\right)$$
Hence,
$$\text{Real part of }(1+i)^n = 2^{n/2} \cos(n \pi/4)$$
This gives us
$$\boxed{\color{blue}{1 - \dbinom{n}2 + \dbinom{n}4 - \dbinom{n}6 \pm \cdots = 2^{n/2} \cos(n \pi/4)}}$$
