How to solve this determinant: $a_{ij}=|i-j|+1$? I have to solve determinant of the following form:
$$a_{ij}=|i-j|+1$$
It looks like this: 
$$
\begin{pmatrix}
1 & 2 & 3 & 4 & \cdots & n \\
2 & 1 & 2 & 3 & \cdots & n-1 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & n-3 & \cdots & 1
\end{pmatrix}
$$
It looks something like Toeplitz matrix, but I haven't found any method of solving it. I would appreciate also a kind of hint that would help.
EDIT: OEIS gives a formula for absolute value: $(n+1)\cdot2^{n-2}$
http://oeis.org/A001792
Thanks in advance! 
 A: As suggested, add to the $n$th line the 1st one. Divide the resulting $n$th line by $n+1$ (this produces a factor $n+1$ in the determinant) for it to consist from $1$'s. Now for $j=1$ to $n-1$ do the following: subtract from the $j$th line the sum of $(j+1)$th and $n$th. In the end of this procedure, you obtain a lower triangular matrix with diagonal $(\underbrace{-2,\ldots,-2}_{(n-2)\;\mathrm{times}},-1,1)$. Hence, indeed, 
$$\mathrm{det}\,a=(-1)^{n-1}2^{n-2}(n+1).$$
A: What do you get when you substract (i+1)th row - ith row?
$$
\begin{pmatrix}
1 & 2 & 3 & 4 & \cdots & n \\
1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & -1 & -1 & \cdots & -1 \\
1 & 1 & 1 & -1 & \cdots & -1 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & 1 & \cdots & -1
\end{pmatrix}
$$
What do you get when you add nth column to all other columns?
$$\begin{pmatrix}
n+1 & n+2 & n+3 & n+4 & \cdots & n \\
0 & -2 & -2 & -2 & \cdots & -1 \\
0 & 0 & -2 & -2 & \cdots & -1 \\
0 & 0 & 0 & -2 & \cdots & -1 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \cdots & -1
\end{pmatrix}$$
What is determinant of the last matrix?
