How to compute this determinant 
I think adding some rows or column to each other will get a triangular matrix which is easy to compute but i cant see how

\begin{align} 
\Delta_n&= \left|\begin{matrix}
1 & 2 & 3 & \cdots  & n-1 & n\\
-1 & x & 0 & \cdots & 0 & 0 \\
0 & -1 & x & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & x & 0 \\
0 & 0 & 0 & \cdots & -1 & x
\end{matrix}\right|
\tag{1}\label{1}
\end{align}
By expanding the determinant $\Delta_n$
with respect to the $n$-th column,
we get a recurrence relation
\begin{align} 
\Delta_n&=x\cdot\Delta_{n-1}+(-1)^{n-1} n\cdot(-1)^{n-1}
=x\cdot\Delta_{n-1}+n
\end{align}
with initial condition $\Delta_1=1$, which has a solution
in a compact form
\begin{align} 
\Delta_n&=\frac{x(x^n-1)-n(x-1)}{(x-1)^2}
\quad \text{for }\quad x\ne 1
\tag{2}\label{2}
.
\end{align}
In the limiting case for $x=1$ we have
\begin{align} 
\Delta_n&=\lim_{x=1}
\frac{x(x^n-1)-n(x-1)}{(x-1)^2}
=\tfrac12n(n+1)
\tag{3}\label{3}
.
\end{align}
Expression \eqref{2} can be also expressed in an expanded form
as follows
\begin{align} 
\Delta_n
&=\frac{(x\sum_{k=0}^{n-1} x^k)-n}{x-1}
\tag{4}\label{4}
\\
&=
\frac1{x-1}\sum_{k=1}^n (x^k-1)
\tag{5}\label{5}
\\
&=
\sum_{k=1}^n \sum_{j=0}^{k-1} x^j
=
\sum_{k=1}^n k x^{n-k}
\tag{6}\label{6}
.
\end{align}
