matrices - 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!

Answer

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,

$$\mathrm{det}\,a=(-1)^{n-1}2^{n-2}(n+1).$$