For  |x| < 1 , n \in \mathbb{Z}^{+}

Prove that


 \sum_{k=0}^{\infty} ( -\frac{x^n}{4^n})^k \binom{2nk}{nk}

 = \frac{1}{n}[ \sum_{k=1}^{\frac{n-f(n)}{2}}(\frac{\sqrt{2}\sqrt{ 1 - x \cos{ \theta_k} + \sqrt{x^2 - 2x\cos{ \theta_k} + 1}}}{\sqrt{ x^2 -2x\cos{2 \theta_k} + 1}}) + \frac{ f(n)}{\sqrt{x+1}}]

where
 f(n) = 1 when n is odd and
 f(n) = 0 when n is even
(  f(n) = \frac{ 1 - (-1)^n}{2} )

 \theta_k = \frac{ (2k-1)}{n} \pi