Calculate determinant with induction I need to prove the following, with induction to every $1 \leq n$:
$$D(a_1,...,a_n) = \left| \begin{array}{ccc}
a_1+x& a_2 & a_3 & \cdots & a_n \\ 
a_1& a_2+x & a_3 & \cdots & a_n \\
a_1& a_2 & a_3+x & \cdots & a_n \\
\vdots & \vdots & \vdots &   & \vdots \\
a_1& a_2 & a_3 & \cdots & a_n + x
\end{array} \right| = x^n + (a_1 + \cdots + a_n)x^{n-1}$$
I played with it a bit and couldn't find a way to prove it.
This is what I did: I assumed that it is correct for $n$, and tried to solve it for $n+1$
$$D(a_1, \ldots , a_n, a_{n+1}) =  \left| \begin{array}{ccc}
a_1+x& a_2 & a_3 & \cdots & a_{n+1} \\ 
a_1& a_2+x & a_3 & \cdots & a_{n+1} \\
a_1& a_2 & a_3+x & \cdots & a_{n+1} \\
\vdots & \vdots & \vdots &   & \vdots \\
a_1& a_2 & a_3 & \cdots & a_{n+1} + x
\end{array} \right| $$
and I did the following operation on the determinant  ($R_{n+1} \to R_{n+1} - R_1$) and got:
$$ \left| \begin{array}{ccc}
a_1+x& a_2 & a_3 & \cdots & a_{n+1} \\ 
a_1& a_2+x & a_3 & \cdots & a_{n+1} \\
a_1& a_2 & a_3+x & \cdots & a_{n+1} \\
\vdots & \vdots & \vdots &  & \vdots \\
-x& 0 & \cdots & 0 &  x
\end{array} \right| $$
And I wasn't sure on how to proceed from here, or even if I'm on the right path.
 A: Developing with respect to the last row, after performing those elementary row operations (that don't change the determinant), you get
$$
D(a_1,\dots,a_n,a_{n+1})=\\
xD(a_1,\dots,a_n)+(-1)^{(n+1)+1}(-x)\det\begin{bmatrix}
a_2 & a_3 & \dots & a_n & a_{n+1} \\
a_2+x & a_3 & \dots & a_n & a_{n+1} \\
a_2 & a_3+x & \dots & a_n & a_{n+1} \\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_2 & a_3 & \dots & a_n+x & a_{n+1}
\end{bmatrix}
$$
Doing $n-1$ row swaps, the determinant we need is
\begin{multline}
\det\begin{bmatrix}
a_2+x & a_3 & \dots & a_n & a_{n+1} \\
a_2 & a_3+x & \dots & a_n & a_{n+1} \\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_2 & a_3 & \dots & a_n+x & a_{n+1} \\
a_2 & a_3 & \dots & a_n & a_{n+1}
\end{bmatrix}=\\
\det\begin{bmatrix}
a_2+x & a_3 & \dots & a_n & a_{n+1} \\
a_2 & a_3+x & \dots & a_n & a_{n+1} \\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_2 & a_3 & \dots & a_n+x & a_{n+1} \\
a_2 & a_3 & \dots & a_n & a_{n+1}+x
\end{bmatrix}-\\
\det\begin{bmatrix}
a_2+x & a_3 & \dots & a_n & 0 \\
a_2 & a_3+x & \dots & a_n & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_2 & a_3 & \dots & a_n+x & 0 \\
a_2 & a_3 & \dots & a_n & -x
\end{bmatrix}=\\[6px]
D(a_2,\dots,a_{n+1})-xD(a_2,\dots,a_n)
\end{multline}
Therefore
$$
D(a_1,\dots,a_n,a_{n+1})=
xD(a_1,\dots,a_n)+
x(D(a_2,\dots,a_{n+1})-xD(a_2,\dots,a_n))
$$
By the induction hypothesis,
\begin{multline}
xD(a_1,\dots,a_n)+
x(D(a_2,\dots,a_{n+1})-xD(a_2,\dots,a_n))=\\
x(x^n+(a_1+\dots+a_n)x^{n-1})+\\
\qquad x(x^n+(a_2+\dots+a_{n+1})x^{n-1}-x^n-(a_2+\dots+a_n)x^{n-1})=\\
x^{n+1}+(a_1+\dots+a_n+a_{n+1})x^n
\end{multline}
Note that you have to check the induction basis for $n=1$ and $n=2$, which is easy.
A: First, developing with respect to the last column, your $D_{n+1}$ is of the form $Aa_{n+1}+B$. If you put $a_{n+1}=0$, you get $B=xD_n$. Now divide $D_{n+1}$ by $a_{n+1}$ (divide only the entries of the last column)  and let $a_{n+1} \to +\infty$. The last column in this new determinant $T$ has now only $1$ as entries. Now in $T$ subtract to the first column $a_1$ times the last column, and do the same for the other columns. Then you find easily that this determinant $T$ is $x^n$, hence $A=x^n$ and $D_{n+1}=x^na_{n+1}+xD_n$, and we are done.  
