While reading a paper on supersymmetry (by Peter West) i faced the following problem. Its about the symmetry property of charge conjugation matrix in different space time dimension. The charge conjugation matrix is defined as

\begin{equation}

C^{T} = -\epsilon C

\end{equation}

Problem is to find out the $\epsilon$ as a function of space-time dimension.

If the spacetime dimension $\mathit{D}$ is even, the finite group generated by Clifford algebra $\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=2\eta_{\mu\nu}$ (Where , and $\eta_{\mu\nu}=\text{Diag}(-,_,+,\cdots,+)$) consists of elements

\begin{equation}

I, \gamma_{m}, \gamma_{m_{1}m_{2}},\gamma_{{m_{1}m_{2}}m_{3}},\dots\gamma_{{m_{1}m_{2}}m_{3}\dots m_{D}}

\end{equation}

There are altogether $2^D$ such matrices.

Following little algebra it can be shown that

\begin{equation}

(C\gamma_{{m_{1}m_{2}}m_{3}\dots m_{D}})^{T} = \epsilon (-1)^{\frac{(p-1)(p-2)}{2}} C\gamma_{{m_{1}m_{2}}m_{3}\dots m_{D}}

\end{equation}

Which means $C\gamma_{{m_{1}m_{2}}m_{3}\dots m_{D}}$ either symmetric or anti symmetric.

It says number of anti symmetric matrices in this set is equal to

\begin{equation}

\sum_{p=0}^{D}\frac{1}{2}(1-\epsilon(-1)^{\frac{(p-1)(p-2)}{2}})\binom{D}{p}

\end{equation}

I know how to prove this result by calculating the (anti)symmetry property of $C\gamma_{m_{1}m_{2}\cdots m_{D}}$ for even dimensions by explicit calculations. But unable to prove the result mentioned.

Can anyone help to derive the last expression? It will be a great help. Thanks in advance.