In my work, I have come across the following equation
$$ \sum_{k = 0}^m \sum_{l = 0}^{n-m} {m \choose k} {n - m \choose l} (-1)^k r_{k + l} $$
where $r_n = 1$ if $n$ is congruent to 0 or 1 modulo 4, and $r_n = -1$ if $n$ is congruent to 2 or 3 modulo 4. I've seen some identities involving the products ${m \choose k} {n - m \choose l}$, but I can't seem to find an algebraic trick to reduce this form to a standard identity. Any tips?
This is
$$\sum_{k=0}^m {m\choose k} (-1)^k \sum_{l=0}^{n-m} {n-m\choose l} r_{k+l}.$$
We have that
$$r_{k+l} = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k+l+1}} (1+w - w^2-w^3 + \cdots) \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^{k+l+1}} \frac{1+w - w^2-w^3}{1-w^4} \; dw.$$
We get for the sum
$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w} \frac{1+w - w^2-w^3}{1-w^4} \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{w^k} \sum_{l=0}^{n-m} {n-m\choose l} \frac{1}{w^l} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w} \frac{1+w - w^2-w^3}{1-w^4} \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{w^k} \left(1+\frac{1}{w}\right)^{n-m} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{n-m}}{w^{n-m+1}} \frac{1+w - w^2-w^3}{1-w^4} \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{w^k} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{n-m}}{w^{n-m+1}} \frac{1+w - w^2-w^3}{1-w^4} \left(1-\frac{1}{w}\right)^m \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{n-m} (w-1)^m}{w^{n+1}} \frac{1+w - w^2-w^3}{1-w^4} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{n-m} (w-1)^m}{w^{n+1}} \frac{1+w}{1+w^2} \; dw \\ = \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{(1+w)^{n-m+1} (w-1)^m}{w^{n+1}} \frac{1}{1+w^2} \; dw.$$
Now checking on the residue at infinity by a circular contour of radius $R$ we obtain a term of order $2\pi R \times R^{n+1}/R^{n+1}/R^2$ which is $2\pi/R$ and vanishes so that residue is zero. This leaves for the sum the residues at $w=\pm i$ and we get (remember to flip the sign at the end)
$$\exp(-(n+1)\pi i/2) \sqrt{2}^{n+1} \frac{1}{2i} \exp(+(n-m+1)\pi i/4)\exp(+3m\pi i/4) \\ - \exp(+(n+1)\pi i/2) \sqrt{2}^{n+1} \frac{1}{2i} \exp(-(n-m+1)\pi i/4)\exp(-3m\pi i/4) \\ = \sqrt{2}^{n+1} \sin(-(2n+2)\pi/4 +(n-m+1)\pi/4 + 3m\pi/4).$$
This yields $$- \sqrt{2}^{n+1} \sin((2m-n-1)\pi/4)$$ or
$$\bbox[5px,border:2px solid #00A000]{ \sqrt{2}^{n+1} \sin((n+1-2m)\pi/4).}$$