Alternating sum of binomial coefficients $1 - \binom{n}{2} + \binom{n}{4} - \binom{n}{6} + ...$ I want to find the sum: $1 - \binom{n}{2} + \binom{n}{4} - \binom{n}{6} + ...$
but i'm not entirely sure where to start. I know that alternating sums of binomial coefficients is zero, but don't know if that will help at the moment.
Any ideas?
 A: This is $(1+i)^n+(1-i)^n$ divided by $2$. Simplify it by converting the complex numbers to polar form and then split into cases depending on $n$ mod $4$.

Here's an explanation of how I got this (beyond "I've seen this type of thing before and know how to deal with it now"). One may write the sum as $\sum_{k\ge0}\binom{n}{k} f(k)$ where $f(k)$ is $0$ if $k$ is of the form $4r+1$ or $4r+3$, is $-1$ if $k$ is of the form $4r+2$ and is $+1$ if $k$ is of the form $4r$. This function is periodic.
According to discrete Fourier analysis, every function $f:\Bbb Z/m\Bbb Z\to\Bbb C$ is expressible as
$$f(k)=\sum_\zeta\color{Blue}{a_\zeta} \color{Green}{\zeta^k} $$
where $\zeta$ in the sum ranges over all $m$th roots of unity, for some constant coefficients $a_\zeta$ (these are the "Fourier amplitudes," one for every "harmonic" $\zeta^k$). The $\zeta$ in $a_\zeta$ is an index.
Then, after "twisting" any series $\sum_{k\ge0}c_k$ termwise with this periodic function $f$, we get
$$\sum_{k\ge0} c_k f(k)=\sum_{k\ge0}c_k \sum_\zeta a_\zeta \zeta^k=\sum_\zeta a_\zeta\left(\sum_{k\ge0}c_k\zeta^k\right).$$
In particular, here we have $c_k=\binom{n}{k}$, and our $f(k)$ may be written as
$$\begin{array}{ll} f(k) & =\color{Blue}{0}\cdot\color{Green}{1^k}+\color{Blue}{\frac{1}{2}}\cdot\color{Green}{i^k}+\color{Blue}{0}\cdot\color{Green}{(-1)^k}+\color{Blue}{\frac{1}{2}}\cdot\color{Green}{(-i)^k}  \\ & \displaystyle = \frac{i^k+(-i)^k}{2}.\end{array}$$
The $4$th roots of unity are $1,i,-1,-i$. One can solve for the constant coefficients (in blue) by letting the coefficients be indeterminates,  writing down the equation for $f(k)$ and setting $k=0,1,2,3$. This yields a linear system of four equations in four unknowns, which one can solve with linear algebra.
Therefore, the sum is
$$\begin{array}{ll} \displaystyle \sum_{k\ge0}\binom{n}{k}f(k) & \displaystyle =\sum_{k\ge0}\binom{n}{k}\frac{i^k+(-i)^k}{2} \\ & \displaystyle =\frac{1}{2}\sum_{k\ge0}\binom{n}{k}i^k+\frac{1}{2}\sum_{k\ge0}\binom{n}{k}(-i)^k \\ & = \frac{1}{2}(1+i)^n+\frac{1}{2}(1-i)^n \end{array}$$
by the binomial theorem.
