I have
\begin{equation} G(x) = \sum_{i = 0}^{\infty} \sum_{r = 0}^{\infty} (-1)^i \binom{k}{2i} \binom{n-k}{r} x^{2i} x^{r} = \frac{1}{2} \left( (1 + x{\iota})^{k} + (1 - x \iota)^{k} \right)(1 + x)^{n-k} \end{equation}
where $\iota$ is the imaginary unit, i.e., $\iota^2 = -1$. The $l$th term of the above is the alternating sum of even coefficients in the Vandermonde convolution, with $l \le k$. I am interested in obtaining a form of the $l$th term that involves a single binomial coefficient (as in the Vandermonde convolution) as opposed to the convolution of two binomial coefficients. Is there a way to do so?
Edit: Sorry, I needed to explain better. By the $l$th term I mean the coefficient of $x^l$. If we fix such an $l$, then we see that for $i = 0, 1, \ldots $, we have $r = l - 0, l - 2, \ldots $, and therefore fixing $r = l - 2i$, we obtain the following as the coefficient of $x^l$:
$\sum_{i = 0}^{\infty} (-1)^{i} \binom{k}{2i} \binom{n-k}{l - 2i}$
So, for instance, if $l = 3$, we get
$\binom{k}{0} \binom{n-k}{3 - 0} - \binom{k}{2} \binom{n-k}{3 - 2}$, where I have used the convention that $\binom{a}{b} = 0$ if $a < b$.
The Vandermonde convolution is:
$\sum_{i = 0}^{\infty} \binom{k}{i} \binom{n-k}{l - i} = \binom{n}{l}$.
I was hoping to get something like the RHS of the above, i.e., $\binom{n}{l}$, for the $l$th term of the main equation. Note that the inner sum in the main equation has the closed form $(1 + x)^{n-k}$, and the outer sum has the closed form $\frac{1}{2}\left((1 + x\iota)^k + (1 - x\iota)^k\right)$. Please bear in mind the difference between $i$ and $\iota$.