I'm reading a paper (see pages 6 and 7 pf the pdf) which claims that a given $n+1$-dimensional square matrix $C$ is orthogonal. The $kj^{th}$ element of this matrix is given by:
$$C_{kj}=2^{-\frac{n}{2}}\binom{n}{k-1}^{\frac{1}{2}}\binom{n}{j-1}^{\frac{1}{2}}\sum_{\nu=0}^{j-1}(-1)^{n-1-\nu}\binom{n-k+1}{j-\nu-1}\binom{k-1}{\nu}$$
The variable $n$ is a non-negative integer. I didn't find any proper proof of the orthogonality of this matrix in the paper or in anywhere else, so I will try to achieve it by myself (or hopefully with some of your help).
To prove that $C$ is indeed orthogonal, one would need to show that:
$$C^T=C^{-1}\Leftrightarrow (C^TC)_{ij}=\delta_{ij}\Leftrightarrow\sum_{k=1}^{n+1}C_{ki}C_{kj}=\delta_{ij}\Leftrightarrow$$
$$\sum_{k=1}^{n+1}\sum_{\mu=0}^{i-1}\sum_{\nu=0}^{j-1}2^{-n}(-1)^{\mu+\nu}\binom{n}{k-1}\binom{n}{i-1}^{\frac{1}{2}}\binom{n}{j-1}^{\frac{1}{2}}\binom{n-k+1}{i-\mu-1}\binom{n-k+1}{j-\nu-1}\binom{k-1}{\mu}\binom{k-1}{\nu}=\delta_{ij}$$
There are too many binomial coefficients to deal in this equation. I really don't know how to simplify it. Can you help me in this one?