Proving an equality involving binomial coefficients and summations Question:
$$\sum_{k=0}^{n}\left ( -1 \right )^{k}\binom{2n}{k}\binom{2n-k}{2n-2k}=\sum_{2n}^{k=0}\binom{2n}{k}^{2}\left ( \frac{1+\sqrt{5}}{2} \right )^{2n-k}\left ( \frac{1-\sqrt{5}}{2} \right )^{k}$$
Attempt:
It looks like I need to start with a quadratic function like $x^{2}-x-1$ since the solutions of this function are $\frac{1\pm \sqrt{5}}{2}$. Is that the correct approach?
How do I get the LHS using this function?
 A: Suppose we  seek to verify that
$$\sum_{k=0}^n (-1)^k {2n\choose k} {2n-k\choose 2n-2k}
= \sum_{k=0}^{2n} {2n\choose k}^2 
\left(\frac{1+\sqrt{5}}{2}\right)^{2n-k}
\left(\frac{1-\sqrt{5}}{2}\right)^{k}.$$
For the LHS introduce
$${2n-k\choose 2n-2k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n-2k+1}} (1+z)^{2n-k} \; dz.$$
Observe that when  $2n\ge k\gt n$ the integral vanishes  so we may use
it to control the  range, extending the sum in $k$ on  the LHS to $2n$
to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} (1+z)^{2n} 
\sum_{k=0}^{2n} (-1)^k {2n\choose k}
\frac{z^{2k}}{(1+z)^k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}} (1+z)^{2n} 
\left(1-\frac{z^2}{1+z}\right)^{2n}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2n+1}}
(1+z-z^2)^{2n} \; dz.$$
This means the LHS is
$$[z^{2n}] (1+z-z^2)^{2n}.$$
On the other hand we may re-write the RHS as
$$\sum_{k=0}^{2n} {2n\choose k} {2n\choose 2n-k} 
\left(\frac{1+\sqrt{5}}{2}\right)^{2n-k}
\left(\frac{1-\sqrt{5}}{2}\right)^{k}.$$
By inspection this is
$$[w^{2n}]
\left.(1+z)^{2n}\right|_{z=w(1+\sqrt{5})/2}
\left.(1+z)^{2n}\right|_{z=w(1-\sqrt{5})/2}
\\ = [w^{2n}]
\left(1+\frac{1+\sqrt{5}}{2}w\right)^{2n}
\left(1+\frac{1-\sqrt{5}}{2}w\right)^{2n}
\\ = [w^{2n}]
(1 + w + (1-5)/4 \times w^2)^{2n}
\\ = [w^{2n}] (1+w-w^2)^{2n}.$$
We have equality for the LHS and the RHS, QED.
