$\displaystyle A=\begin{pmatrix}a & 0 & b\\ 0 & a+b & 0\\ b & 0 & a\end{pmatrix}$

Um, well, I know I should find a formula for $\displaystyle A^n$ and prove it using induction. I calculated the first four powers of A, but still didn't observe any rule >.>

$\displaystyle A^2=\begin{pmatrix}a^2+b^2 & 0 & 2ab\\ 0 & (a+b)^2 & 0\\2ab & 0 & a^2+b^2\end{pmatrix}$

$\displaystyle A^3=\begin{pmatrix}a^3+3ab^2 & 0 & b^3+3a^2b\\ 0 & (a+b)^3 & 0\\b^3+3a^2b & 0 & a^3+3ab^2\end{pmatrix}$

$\displaystyle A^3=\begin{pmatrix}a^3+3ab^2 & 0 & b^3+3a^2b\\ 0 & (a+b)^3 & 0\\b^3+3a^2b & 0 & a^3+3ab^2\end{pmatrix}$

$\displaystyle A^4=\begin{pmatrix}a^4+b^4+6a^2b^2 & 0 & 4a^3b+4ab^3\\ 0 & (a+b)^3 & 0\\4a^3b+4ab^3 & 0 & a^4+b^4+6a^2b^2\end{pmatrix}$

Actually, it's not really nothing, but still useless:

$\displaystyle a^2+b^2+2ab=(a+b)^2$

$\displaystyle a^3+3ab^2+b^3+3a^2b =(a+b)^3$

$\displaystyle a^4+b^4+6a^2b^2+4a^3b+4ab^3=(a+b)^4$

Thanks in advance for any ideas! ^^