Solution
= Solution
On the first and third coordinates, the identity $SJS=J$ from part (b) gives $e^{xB}Ce^{xB}=C$; the second coordinate is also unchanged. Thus
$$
\boxed{\sum_{n=1}^N\left(e^{xB}Ce^{xB}\right)^n
=\operatorname{diag}\left(N,\frac{(-1)^N-1}{2},\frac{(-1)^N-1}{2}\right)}.
$$