Eigenvalues of different symmetric $(2n+1)\times(2n+1)$ matrix 
I ve looked at other similar post but I could not find help with them

The rank of your matrix is $2$, which implies that $\lambda=0$ is an eigenvalue with multiplicity at least $2n+1-rank=2n-1$.
Now, if $\lambda_1, \lambda_2$ are the remaining eigenvalues, since $tr(A)$ is the sum of the eigenvalues you get $\lambda_1+\lambda_2=0$. [I assume the row and column have the same index]. This Yields $\lambda_2=-\lambda_1$.
Finally, the eigenvalues of $A^2$ are $0,0,.., 0, \lambda_1^2, \lambda_2^2$, thus 
$$
tr(A^2)=2 \lambda_1^2 \,.
$$
note that 
$$A^2 =\begin{bmatrix}
0 & 0 & 0 & ..& 1 & ... &0 \\
0 & 0 & 0 & ..& 1 & ... &0 \\
0 & 0 & 0 & ..& 1 & ... &0 \\
... & ... & ... & ...& ... & ... &... \\
1 & 1 & 1 & ..& 0 & ... &1 \\
... & ... & ... & ...& ... & ... &... \\
0 & 0 & 0 & ..& 1 & .. &0 \\
\end{bmatrix}
\begin{bmatrix}
0 & 0 & 0 & ..& 1 & ... &0 \\
0 & 0 & 0 & ..& 1 & ... &0 \\
0 & 0 & 0 & ..& 1 & ... &0 \\
... & ... & ... & ...& ... & ... &... \\
1 & 1 & 1 & ..& 0 & ... &1 \\
... & ... & ... & ...& ... & ... &... \\
0 & 0 & 0 & ..& 1 & .. &0 \\
\end{bmatrix}
=\begin{bmatrix}
1 & 1 & 1 & ..& 0 & ... &1 \\
1 & 1 & 1 & ..& 0 & ... &1 \\
1 & 1 & 1 & ..& 0 & ... &1 \\
... & ... & ... & ...& ... & ... &... \\
0 & 0 & 0 & ..&  2n& ... &0 \\
... & ... & ... & ...& ... & ... &... \\
 1& 1 & 1 & ..& 0 & .. &1 \\
\end{bmatrix}$$
Therefore 
$$2 \lambda_1^2=tr(A^2)=4n$$
