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This is a bit of a brute force approach, but it's effective. Note that the sum of the reverse diagonal elements of a $2\times 2$ matrix $M$ equals ${\rm tr}\,\sigma M$ with $$\sigma=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$ For the most general form of the matrix $$A=\begin{pmatrix}a&b\\ -b&c\end{pmatrix},\;\;\text{and for}\;\;D=\mathbf c\mathbf b^{\rm T}=\begin{pmatrix}1&-1\\1&-1\end{pmatrix},$$ I calculate $$J(r,n)={\rm tr}\,\sigma A^r DA^{n-1-r}=$$ $$=\frac{2^{-n-1} (a+c-z)^{-r} (a+c+z)^{-r}}{(a-2 b-c) \left(a c+b^2\right)} \left[\left(z (a+c)-(a-c)^2+4 b^2\right) (a+c+z)^n (a+c-z)^{2 r}-\left(z (a+c)+(a-c)^2-4 b^2\right) (a+c-z)^n (a+c+z)^{2 r}\right],$$ with the definition $z=\sqrt{(a-c)^2-4 b^2}$. Then I evaluate for $n\geq 1$ the sum $$\sum_{r=0}^{n-1}J(r,n)=\frac{2^{-n-1} (a+c) \left((a-c)^2-4 b^2-z^2\right) \left((a+c-z)^n-(a+c+z)^n\right)}{z (a-2 b-c) \left(a c+b^2\right)}.$$ Substitution of the definition of $z$ finally gives the desired result $$\sum_{r=0}^{n-1}J(r,n)=0.$$