Inverse of tridiagonal Toeplitz matrix Consider  the following tridiagonal Toeplitz matrix. Let $n$ be even.
$${A_{n \times n}} = \left[ {\begin{array}{*{20}{c}}
{0}&{1}&{}&{}&{}\\
{1}&{0}&{1}&{}&{}\\
{}&{1}&{\ddots}&{\ddots}&{}\\
{}&{}&{\ddots}&{\ddots}&{1}\\
{}&{}&{}&{1}&{0}
\end{array}} \right]$$
What is the inverse $A^{-1}$? 
Clearly, $A^{-1}$ is symmetric.
I look for a proof of the following conjecture that $A^{-1}$ is given as follows:
If $A_{i, j}^{-1}$ such that $j$ is odd and $i =1+j + 2 m$ with $m\in \cal N_0$, then $A_{i, j}^{-1} = (-1)^m$. From which follows by symmetry:
If $A_{i, j}^{-1}$ such that $j$ is even and $i =-1+j - 2 m$ with $m\in \cal N_0$, then $A_{i, j}^{-1} = (-1)^m$. 
All other $A_{i, j}^{-1} = 0$.
Here is an example, computed with Matlab, for $n=10$ which shows the structure: 
$${A_{10 \times 10}^{-1}} = \left[ {\begin{array}{*{20}{r}}
0 &     1 &     0 &    -1 &      0 &     1 &     0 &    -1 &      0 &     1 \\
     1 &     0 &     0 &     0 &     0 &     0 &     0 &     0 &     0 &     0 \\
     0 &     0 &     0 &     1 &     0 &    -1 &     0 &     1 &     0 &    -1  \\
    -1 &      0 &     1 &     0 &     0 &     0 &     0 &     0 &     0 &     0 \\
     0 &     0 &     0 &     0 &     0 &     1 &     0 &    -1 &      0 &     1 \\
     1 &     0 &    -1 &     0 &     1 &     0 &     0 &     0 &     0 &     0 \\
     0 &     0 &     0 &     0 &     0 &     0 &     0 &     1 &     0 &    -1  \\
    -1 &     0 &     1 &     0 &    -1 &     0 &     1 &     0 &     0 &     0 \\
     0 &     0 &     0 &     0 &     0 &     0 &     0 &     0 &     0 &     1 \\
     1 &     0 &    -1 &     0 &     1 &     0 &    -1 &     0 &     1 &     0 
\end{array}} \right]$$

Firstly Matrix is Toeplitz. This means it represents multiplication by power series expansion. This means matrix inversions corresponds to multiplicative inversion
Therefore, consider $$x+x^{-1}=\frac{x^2+1}x$$ Now it's multiplicative inverse:
$$(x+x^{-1})^{-1}=\frac x{x^2+1}$$
Now you can expand with geometric series / Taylor expansion for $$\frac 1{x^2+1}=\frac 1{1-(-1\cdot x^2)}$$
And substitute with
$$\frac{1}{1-t}=1+t+t^2+\cdots, t=-x^2$$
and then finish. You will notice the flipping sign pattern and that the odd exponents disappear when you substitute and do the expansion.
