Construction of generating function from identity I am trying to solve identity involving binomials and Fibonacci numbers by using generating functions:
$$\sum_{k=0}^n{n \choose k}{n+k\choose k}f_{k+1}=\sum_{k=0}^n{n \choose k}{n+k\choose k}(-1)^{n-k}f_{2k+1}$$
My computational approach by Mathematica lead me to derive this generating function:
$$\frac{\sqrt{3x^2-2x+3+2\sqrt{x^4-8x^3-2x^2-8x+1}}}{\sqrt{5}\sqrt{x^4-8x^3-2x^2-8x+1}}$$
Can someone show how to transform both or any of the identity sides to obtain (coefficiens of) this generating function.
 A: Here is a proof using complex variables. We seek to show that
$$\sum_{k=0}^n {n\choose k} {n+k\choose k} F_{k+1}
=\sum_{k=0}^n {n\choose k} {n+k\choose k} (-1)^{n-k} F_{2k+1}.$$
Start from
$${n+k\choose k}
= \frac{1}{2\pi i} 
\int_{|z|=1} \frac{1}{z^{k+1}} (1+z)^{n+k} \; dz.$$
This yields the following expression for the sum on the LHS
$$\frac{1}{2\pi i} 
\int_{|z|=1}
\sum_{k=0}^n {n\choose k} \frac{1}{z^{k+1}} (1+z)^{n+k}
\frac{\varphi^{k+1} - (-1/\varphi)^{k+1}}{\sqrt{5}} \; dz$$
This simplifies to
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i} 
\int_{|z|=1} \frac{(1+z)^n}{z}
\sum_{k=0}^n {n\choose k} 
\left(\varphi\left(\varphi\frac{1+z}{z}\right)^k
+\frac{1}{\varphi}\left(-\frac{1}{\varphi}\frac{1+z}{z}\right)^k
\right)\; dz$$
This finally yields
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i} 
\int_{|z|=1} \frac{(1+z)^n}{z}
\left(
\varphi\left(1+\varphi\frac{1+z}{z}\right)^n
+\frac{1}{\varphi}\left(1-\frac{1}{\varphi}\frac{1+z}{z}\right)^n
\right) \; dz$$
or
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i} 
\int_{|z|=1} \frac{(1+z)^n}{z^{n+1}}
\left(
\varphi\left(z+\varphi(1+z)\right)^n
+\frac{1}{\varphi}\left(z-\frac{1}{\varphi}(1+z)\right)^n
\right) \; dz$$
Continuing we have the following expression for the sum on the RHS
$$\frac{1}{2\pi i} 
\int_{|z|=1}
\sum_{k=0}^n {n\choose k} (-1)^{n-k} \frac{1}{z^{k+1}} (1+z)^{n+k}
\frac{\varphi^{2k+1} - (-1/\varphi)^{2k+1}}{\sqrt{5}} \; dz$$
This simplifies to
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i} 
\int_{|z|=1} \frac{(1+z)^n}{z}
\\ \times \sum_{k=0}^n {n\choose k} (-1)^{n-k}
\left(\varphi\left(\varphi^2\frac{1+z}{z}\right)^{k}
+\frac{1}{\varphi}\left(\frac{1}{\varphi^2}\frac{1+z}{z}\right)^{k}
\right)\; dz$$
This finally yields
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i} 
\int_{|z|=1} \frac{(1+z)^n}{z}
\left(
\varphi\left(-1+\varphi^2\frac{1+z}{z}\right)^n
+\frac{1}{\varphi}\left(-1+\frac{1}{\varphi^2}\frac{1+z}{z}\right)^n
\right) \; dz$$
or
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i} 
\int_{|z|=1} \frac{(1+z)^n}{z^{n+1}}
\left(
\varphi\left(-z+\varphi^2(1+z)\right)^n
+\frac{1}{\varphi}\left(-z+\frac{1}{\varphi^2}(1+z)\right)^n
\right) \; dz$$
Apply the substitution $z=1/w$ to this integral to obtain
(the sign to correct the reverse orientation of the circle is 
canceled by the minus on the derivative)
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i} 
\int_{|w|=1} \left(1+\frac{1}{w}\right)^n w^{n+1}
\\ \times \left(
\varphi\left(-\frac{1}{w}+\varphi^2(1+\frac{1}{w})\right)^n
+\frac{1}{\varphi}
\left(-\frac{1}{w}+\frac{1}{\varphi^2}(1+\frac{1}{w})\right)^n
\right) \frac{1}{w^2} \; dw$$
which is
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i} 
\int_{|w|=1} \left(1+\frac{1}{w}\right)^n \frac{1}{w}
\\ \times \left(
\varphi\left(-1+\varphi^2(w+1)\right)^n
+\frac{1}{\varphi}
\left(-1+\frac{1}{\varphi^2}(w+1)\right)^n
\right) \; dw$$
which finally yields
$$\frac{1}{\sqrt{5}}\frac{1}{2\pi i} 
\int_{|w|=1} \frac{(1+w)^n}{w^{n+1}}
\\ \times \left(
\varphi\left(-1+\varphi^2(w+1)\right)^n
+\frac{1}{\varphi}
\left(-1+\frac{1}{\varphi^2}(w+1)\right)^n
\right) \; dw$$
This shows that the LHS is the same as the RHS
because
$$-1 + \varphi^2(w+1) = -1 + (1+\varphi)(w+1)
= w + \varphi(w+1)$$
and
$$-1+\frac{1}{\varphi^2}(w+1)
= -1 + (1-\frac{1}{\varphi}) (w+1)
\\ = -1 + (w+1) - \frac{1}{\varphi} (w+1)
= w - \frac{1}{\varphi} (w+1).$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
