Why does $\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$? In the paper "A Parametric Texture Model based on Joint Statistics of Complex Wavelet Coefficients", the authors use this equation for the angular part of the filter in polar coordinates:
$$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right)$$
My friend and I have tested many values of $k > 1$, and in each case this summation is equal to 
$$\frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$
The paper asserts this as well.
We are interested in having an analytic explanation of this equality, if it really holds. How can we derive this algebraically? 
TL;DR

Is this true, and if so, why?
  $$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$


Suppose we seek to verify that
$$\sum_{k=0}^n \cos^{2n}\left(x+\frac{k\pi}{n+1}\right)
= \frac{n+1}{2^{2n}} {2n\choose n}.$$
The LHS is
$$\sum_{k=0}^n \cos^{2n}\left(x+\frac{k\times 2\pi}{2n+2}\right).$$
Observe also that
$$\sum_{k=0}^n 
\cos^{2n}\left(x+\frac{(k+n+1)\times 2\pi}{2n+2}\right)
\\ = \sum_{k=0}^n 
\cos^{2n}\left(x+\pi+\frac{k\times 2\pi}{2n+2}\right)
= \sum_{k=0}^n 
\cos^{2n}\left(x+\frac{k\times 2\pi}{2n+2}\right)$$
because the cosine is raised to an even power.
Therefore the LHS is in fact
$$\frac{1}{2} 
\sum_{k=0}^{2n+1} 
\cos^{2n}\left(x+\frac{k\times 2\pi}{2n+2}\right).$$
Hence we need to prove that
$$\frac{1}{2} 
\sum_{k=0}^{2n+1}
\left(\exp\left(ix+k\times\frac{2\pi i}{2n+2}\right)
+ \exp\left(-ix-k\times\frac{2\pi i}{2n+2}\right)\right)^{2n}
\\ = (n+1)\times {2n\choose n}.$$
Introducing 
$$f(z) = \left(\exp(ix)z+\exp(-ix)/z\right)^{2n}
\frac{(2n+2)z^{2n+1}}{z^{2n+2}-1}$$
We have that the sum is
$$\frac{1}{2} \sum_{k=0}^{2n+1} 
\mathrm{Res}_{z=\exp(2\pi ik/(2n+2))} f(z).$$
The  other potential poles  are at  $z=0$ and  at $z=\infty$  and the
residues must sum to zero. For the candidate pole at zero we write
$$f(z) = \left(\exp(ix)z^2+\exp(-ix)\right)^{2n}
\frac{(2n+2)z}{z^{2n+2}-1}$$
and we see that it vanishes. Therefore the target sum is given by
$$-\frac{1}{2} \mathrm{Res}_{z=\infty} f(z)
\\ = \frac{1}{2} \mathrm{Res}_{z=0} \frac{1}{z^2}
\left(\exp(ix)/z+\exp(-ix)z\right)^{2n}
\frac{1}{z^{2n+1}}
\frac{2n+2}{1/z^{2n+2}-1}
\\ = \frac{1}{2} \mathrm{Res}_{z=0} \frac{1}{z}
\left(\exp(ix)/z+\exp(-ix)z\right)^{2n}
\frac{2n+2}{1-z^{2n+2}}
\\ = \frac{1}{2} \mathrm{Res}_{z=0} \frac{1}{z^{2n+1}}
\left(\exp(ix)+\exp(-ix)z^2\right)^{2n}
\frac{2n+2}{1-z^{2n+2}}.$$
This is
$$(n+1) [z^{2n}] 
\left(\exp(ix)+\exp(-ix)z^2\right)^{2n}
\frac{1}{1-z^{2n+2}}.$$
Now we have
$$\frac{1}{1-z^{2n+2}} =
1 + z^{2n+2} + z^{4n+4} + \cdots$$
and only the first term contributes, leaving
$$(n+1) [z^{2n}] 
\left(\exp(ix)+\exp(-ix)z^2\right)^{2n}
\\ = (n+1) \times {2n\choose n} \exp(ixn)\exp(-ixn)
= (n+1) \times {2n\choose n}.$$
This is the claim.
Remark. Inspired by the work at this 
MSE link.
